RD Sharma Class 12 Exercise 19.1 Definite Integrals Solutions Maths - Download PDF Free Online
Many CBSE board students widely use the RD Sharma books to clarify their doubts. Following the chapter Indefinite Integrals, the Definite Integrals is also a challenging portion. Students need to be taught this concept and work it out without confusing the previous chapter. And as the first exercise needs help, the RD Sharma Class 12th Exercise 19.1 solution book is essential.
RD Sharma Class 12 Solutions Chapter19 Definite Integrals - Other Exercise
Definite Integrals Excercise:19.1
Definite Integrals Exercise 19.1 Question 1
Answer: 2Hint:Use indefinite formula to solve the integral and then put the value of limit to get the required answer.
Given: $\int_{4}^{9}\frac{1}{\sqrt{x}}dx$
Solution:
$\begin{aligned} &\int_{4}^{9} \frac{1}{\sqrt{x}} d x=\int_{4}^{9} x^{-\frac{1}{2}} d x=\left[\frac{x^{\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_{4}^{9} \quad\left[\int x^{n+1} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}}\right]_{4}^{9} \\ &=2\left[x^{\frac{1}{2}}\right]_{4}^{9} \\ &=2\left[9^{\frac{1}{2}}-4^{\frac{1}{2}}\right] \\ &=2\left[3^{2 \times \frac{1}{2}}-2^{2 \times \frac{1}{2}}\right] \\ &=2[3-2]=2 \end{aligned}$
Definite Integrals Exercise 19.1 Question 2
Answer: log 2Hint:Use indefinite formula to solve the integral and then put the value of limit to get the required answer.
Given: $\int_{-2}^{3}\frac{1}{x+7}dx$
Solution:$\int_{-2}^{3}\frac{1}{x+7}dx$
Putting $x+7=t\Rightarrow dx=dt$
And when $x=-2$ then $t=-2+7=5$
$x=3$ then $t=3+7=10$
Then,$\int_{-2}^{3}\frac{1}{x+7}dx=\int_{5}^{10}\frac{1}{t}dt=\left [ log|t| \right ]^{10}_{5}$ $\left [ \int \frac{1}{x}dx-log|x| \right ]$
$=\left [ log|10|-log|5| \right ]$ $\left [ log\: a-log\: b=log\frac{a}{b} \right ]$
$=log10-log5$
$=log\frac{10}{5}$
$=log 2$
Definite Integrals Exercise 19.1 Question 3
Answer: $\frac{\pi }{6}$Hint:: Use indefinite formula to solve the integral and then put the value of limit to get the required answer
Given: $\int_{0}^{\frac{1}{2}}\frac{1}{\sqrt{1-x^{2}}}dx$
Solution:
$\int_{0}^{\frac{1}{2}}\frac{1}{\sqrt{1-x^{2}}}dx=\left [ \sin ^{-1}x \right ]^{\frac{1}{2}}_{0}$ $\left [ \int \frac{1}{\sqrt{1-x^{2}}}dx=\sin ^{-1}x \right ]$
$=\sin ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}(0) \quad\left[\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \Rightarrow \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}, \sin 0=0 \Rightarrow \sin ^{-1}(0)=0\right]$
$=\left [ \frac{\pi }{6}-0 \right ]=\frac{\pi }{6}$
Definite Integrals Exercise 19.1 Question 4
Answer: $\frac{\pi }{4}$Hint: Use indefinite formula to solve the integral and then put the value of limit to get the required answer
Given: $\int_{0}^{1}\frac{1}{1+x^{2}}dx$
Solution: $\int_{0}^{1}\frac{1}{1+x^{2}}dx=\int \tan ^{-1}xdx$ $\left [ \int \frac{1}{1+x^{2}}dx=\tan ^{-1}x \right ]$
$=\tan ^{-1}(1)-\tan ^{-1}(0) \quad\left[\tan \left(\frac{\pi}{4}\right)=1 \Rightarrow \tan ^{-1}(1)=\frac{\pi}{4}, \tan 0=0 \Rightarrow \tan ^{-1}(0)=0\right]$
$=\left [ \frac{\pi }{4}-0 \right ]=\frac{\pi }{4}$
Definite Integrals Exercise 19.1 Question 5
Answer: $log\left ( \sqrt{2} \right )$Hint:Use indefinite formula to solve the integral and then put the value of limit to get the required answer.
Given: $\int_{2}^{3}\frac{x}{x^{2}+1}dx$
Solution:$\int_{2}^{3}\frac{x}{x^{2}+1}dx$
Put $x^{2}+1=t\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{dt}{2}$
When $x=2$ then$t=3^{2}+1=4+1=5$
When $x=3$ then $t=3^{2}+1=9+1=10$
Then$\int_{2}^{3}\frac{x}{x^{2}+1}dx=\frac{1}{2}\int_{5}^{10}\frac{1}{t}dt=\frac{1}{2}\left [ log|t| \right ]^{10}_{5}$ $\left [ \int \frac{1}{x}dx=log|x| \right ]$
$=\frac{1}{2}\left [ log10-log5 \right ]$ $\left [ log\; \; a -log\; b=log\frac{a}{b}\right ]$
$=\frac{1}{2}log\frac{10}{5}$
$=\frac{1}{2}log2$ $\left [ log\: a^{m}=m\: log\: a \right ]$
$=log\left ( 2^{\frac{1}{2}} \right )$
$=log\sqrt{2}$
Definite Integrals Exercise 19.1 Question 6
Answer: $\frac{\pi }{2ab}$Hint:Use indefinite formula to solve the integral and then put the value of limit to get the required answer
Given:$\int_{0}^{\infty }\frac{1}{a^{2}+b^{2}x^{2}}dx$
Solution:
$\int_{0}^{\infty }\frac{1}{a^{2}+b^{2}x^{2}}dx$
Definite Integrals Exercise 19.1 Question 7
Answer:$\frac{\pi }{2}$Hint:Use indefinite formula to solve the integral and then put the value of limit to get the required answer
Given: $\int_{-1}^{1}\frac{1}{1+x^{2}}dx$
Solution:$\int_{-1}^{1}\frac{1}{1+x^{2}}dx$ $\left [ \int \frac{1}{1+x^{2}}dx=\tan ^{-1}x \right ]$
$=\tan ^{-1}(1)-\tan ^{-1}(-1)$ $\left[\begin{array}{l} \tan \left(\frac{\pi}{4}\right)=1 \Rightarrow \tan ^{-1}(1)=\frac{\pi}{4} \\ \tan ^{-1}(-1)=-\frac{\pi}{4} \end{array}\right]$
$=\left [ \frac{\pi }{4}+\frac{\pi }{4} \right ]=\frac{2\pi }{4}=\frac{\pi }{2}$
Definite Integrals Exercise 19.1 Question 8
Answer:1Hint: Use indefinite formula to solve the integral and then put the value of limit to get the required answer
Given: $\int_{0}^{\infty }e^{-x}dx$
Solution:
$\int_{0}^{\infty }e^{-x}dx=\left [ \frac{e^{-x}}{-1} \right ]^{\infty }_{0}$ $\left ( \int e^{ax}=dx\frac{e^{ax}}{a} \right )$
$\begin{aligned} &=-1\left[e^{-x}\right]_{0}^{\infty} \\ &=-\left[e^{-\infty}-e^{-0}\right]=-\left[e^{-\infty}-e^{0}\right] \\ &=-1[0-1]=-1(-1) \\ &=1 \end{aligned} \quad\left[e^{-\infty}=0, e^{0}=1\right]$
Definite Integrals Exercise 19.1 Question 10
Answer: 2Hint: Use indefinite formula to solve the integral and then put the value of limit to get the required answer
Given: $\int_{0}^{\frac{\pi }{2}}\left ( \sin x+\cos x \right )dx$
Solution:
$\int_{0}^{\frac{\pi }{2}}\left ( \sin x+\cos x \right )dx=\int_{0}^{\frac{\pi }{2}}\sin xdx+\int_{0}^{\frac{\pi }{2}}\cos \: xdx$
$=[-\cos x]_{0}^{\frac{\pi}{2}}+[\sin x]_{0}^{\frac{\pi}{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \int \sin x d x=-\cos x \\ \int \cos x d x=\sin x \end{array}\right]$
$=-\left[\cos \frac{\pi}{2}-\cos 0\right]+\left[\sin \frac{\pi}{2}-\sin 0\right]\left[\begin{array}{l} \cos \frac{\pi}{2}=0, \cos 0=1 \\ \sin \frac{\pi}{2}=1, \sin 0=0 \end{array}\right]$
$=-\left ( 0-1 \right )+\left ( 1-0 \right )$
$=-\left ( -1 \right )+1$
$=1+1=2$
Definite Integrals Exercise 19.1 Question 11
Answer: $\frac{1}{2}log2$Hint: Use indefinite formula to solve the integral and then put the value of limit to get the required answer
Given: $\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cot xdx$
Solution:$\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}\cot xdx=\left [ log|\sin x| \right ]^{\frac{\pi }{2}}_{\frac{\pi }{4}}$ $\left ( \int \cot xdx=log|\sin x| \right )$
$=\left [ log|\sin \frac{\pi }{2} |-|\sin \frac{\pi }{4}|\right ]$
$=log1-log\frac{1}{\sqrt{2}}$ $\left [ \sin \frac{\pi }{2} =1,\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right ]$
$\begin{aligned} &=\log \frac{1}{\frac{1}{\sqrt{2}}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\log a-\log b=\log \frac{a}{b}\right] \\ &=\log \sqrt{2}=\log 2^{\frac{1}{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\quad\left[\log a^{m}=m \log a\right] \end{aligned}$
$=\frac{1}{2}log2$
Definite Integrals Exercise 19.1 Question 12
Answer:$log\left ( \sqrt{2}+1 \right )$Hint:: Use indefinite formula to solve the integral and then put the value of limit to get the required answer
Given: $\int_{0}^{\frac{\pi }{4}}\sec xdx$
Solution: $\int_{0}^{\frac{\pi}{4}} \sec x d x=[\log |\sec x+\tan x|]_{0}^{\frac{\pi}{4}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \sec x d x=\log |\sec x+\tan x|\right]$
$\left.\begin{array}{l} =\left[\log \left|\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right|-\log |\sec 0-\tan 0|\right] \\ =[\log |\sqrt{2}+1|-\log |1-0|] \end{array}\right]\left[\begin{array}{l} \sec \frac{\pi}{4}=\sqrt{2} \\ \tan \frac{\pi}{4}=1 \ \\ \sec 0=1 \\ \tan 0=0 \end{array}\right]$
$=log\left ( \sqrt{2}+1 \right )-log1$
$=log\left ( \sqrt{2}+1 \right )-0$ $\left [ log1=0 \right ]$
$=log\left ( \sqrt{2}+1 \right )$
Definite Integrals Exercise 19.1 Question 13
Answer: $log\left ( \frac{\sqrt{2}-1}{2-\sqrt{3}} \right )$Hint: Use indefinite formula then put the limit to get the required answer
Given:$\int_{\frac{\pi }{6}}^{\frac{\pi }{4}}\cos ecxdx$
Solution:$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x=[\log |\cos e c x-\cot x|]_{\frac{\pi}{6}}^{\frac{\pi}{4}}\: \: \: \: \: \: \: \: \: \quad\left[\int \cos \operatorname{ecx} d x=\log |\operatorname{cosec} x-\cot x|\right]$
$=\left[\log \left|\cos e c \frac{\pi}{4}-\cot \frac{\pi}{4}\right|-\log \left|\operatorname{cosec} \frac{\pi}{6}-\cot \frac{\pi}{6}\right|\right]\left[\begin{array}{l} \cos e c \frac{\pi}{4}=\sqrt{2} \\ \cot \frac{\pi}{4}=2 \\ \cos \operatorname{ec} \frac{\pi}{6}=2 \\ \cot \frac{\pi}{6}=\sqrt{3} \end{array}\right]$
$\begin{aligned} &=[\log (\sqrt{2}-1)-\log (2-\sqrt{3})] \\ &=\log (\sqrt{2}-1)-\log (2-\sqrt{3}) \end{aligned} \quad\left[\log a-\log b=\log \frac{a}{b}\right]$
$=log\left ( \frac{\sqrt{2}-1}{2-\sqrt{3}} \right )$
Definite Integrals Exercise 19.1 Question 14
Answer:$2log2-1$Hint: Use indefinite formula then put the limit to get the required answer
Given: $\int_{0}^{1}\left ( \frac{1-x}{1+x} \right )dx$
Solution:
$\int_{0}^{1}\left ( \frac{1-x}{1+x} \right )dx$
Put
$\begin{aligned} &x=\cos 2 \theta \Rightarrow 2 \theta=\cos ^{-1} x \Rightarrow \theta=\frac{\cos ^{-1} x}{2} \\ &d x=-\sin 2 \theta(2) d \theta \Rightarrow d x=-2 \sin 2 \theta d \theta \end{aligned}$
When$x=0$ then$\theta =\frac{\pi }{2 }\times \frac{1}{2}=\frac{\pi }{4}$
When $x=1$ then $\theta =0$
Then
$\begin{aligned} &\int_{0}^{1}\left(\frac{1-x}{1+x}\right) d x=\int_{\frac{\pi}{4}}^{0} \frac{1-\cos 2 \theta}{1+\cos 2 \theta}(-2 \sin 2 \theta) d \theta \\ &=-\int_{\frac{\pi}{4}}^{0} \frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta} 2 \sin 2 \theta d \theta \end{aligned}$ $\left[\begin{array}{l} 1+\cos 2 \theta=2 \cos ^{2} \theta \\ 1-\cos 2 \theta=2 \sin ^{2} \theta \end{array}\right]$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} \theta}{\cos ^{2} \theta} 2 \sin 2 \theta d \theta \quad\left[\int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x\right] \\ &=\int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} \theta}{\cos ^{2} \theta} 2.2 \sin \theta \cos \theta d \theta \quad[\sin 2 \theta=2 \sin \theta \cos \theta] \\ &=4 \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{3} \theta}{\cos \theta} d \theta \end{aligned}$
Again put $\cos \theta =t\Rightarrow \sin \theta d\theta =dt\Rightarrow d\theta =\frac{dt}{-\sin \theta }$
When $\theta =0$ then $t=2$
And When $\theta =\frac{\pi }{4}$ then $t=\frac{1}{\sqrt{2}}$
Then
$\begin{aligned} &\int_{0}^{1} \frac{1-x}{1+x} d x=4 \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{3} \theta}{t} \frac{d t}{-\sin \theta}=-4 \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{2} \theta}{t} d t \\ &=-4 \int_{2}^{\frac{1}{\sqrt{2}} 1-\cos ^{2} t}{t} d t \end{aligned}$
$\left [ \because \sin ^{2}\theta =1-\cos ^{2}\theta \right ]$
$\begin{aligned} &=-4 \int_{2}^{\frac{1}{\sqrt{2}} 1-t^{2}}{t} d t \\ &=-4 \int_{1}^{\frac{1}{\sqrt{2}}}\left(\frac{1}{t}-\frac{t^{2}}{t}\right) d t \\ &=-4 \int_{1}^{\frac{1}{\sqrt{2}}}\left(\frac{1}{t}-t\right) d t \\ &=-4 \int_{1}^{\frac{1}{\sqrt{2}} \frac{1}{t} d t+4 \int_{1}^{\frac{1}{\sqrt{2}}} t d t} \\ &=-4[\log |t|]_{1}^{\frac{1}{\sqrt{2}}}+4\left[\frac{t^{1+1}}{1+1}\right]_{1}^{\frac{1}{\sqrt{2}}} \end{aligned}$
$\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \int \frac{1}{x} d x=\log |x| \end{array}\right]$
$\begin{aligned} &=-4[\log |t|]_{1}^{\frac{1}{\sqrt{2}}}+4\left[\frac{t^{2}}{2}\right]_{1}^{\frac{1}{\sqrt{2}}} \\ &=-4\left[\log \frac{1}{\sqrt{2}}-\log 1\right]+2\left[\left(\frac{1}{\sqrt{2}}\right)^{2}-1^{2}\right] \\ &=-4 \log \frac{1}{\sqrt{2}}+2\left(\frac{1}{2}-1\right) \end{aligned}$ $\left [ log\; \; a^{m}=mlog\: a \right ]$
$\begin{aligned} &=-4 \log (\sqrt{2})^{-1}+2\left(-\frac{1}{2}\right) \\ &=-4(-\log \sqrt{2})-1 \\ &=4 \log \sqrt{2}-1=2 \log 2^{\frac{1}{2} \times 2}-1 \\ &=2 \log 2-1 \end{aligned}$
Definite Integrals Exercise 19.1 Question 15
Answer: 2Hint: Use indefinite formula then put the limit to get the required answer
Given:$\int_{\pi }^{0}\frac{1}{1+\sin x}dx$
Solution:$\int_{\pi }^{0}\frac{1}{1+\sin x}dx$
Rationalizing,$\int_{0}^{\pi}\left(\frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}\right) d x=\int_{0}^{\pi}\left(\frac{1-\sin x}{(1+\sin x)(1-\sin x)}\right)$
$\begin{aligned} &=\int_{0}^{\pi} \frac{1-\sin x}{1-\sin ^{2} x} d x \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right] \\ &=\int_{0}^{\pi} \frac{1-\sin x}{\cos ^{2} x} d x \quad\left[1-\sin ^{2} x=\cos ^{2} x\right] \end{aligned}$
$\begin{aligned} &=\int_{0}^{\pi}\left(\frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}\right) d x \\ &=\int_{0}^{\pi}\left(\sec ^{2} x-\frac{\sin x}{\cos x} \frac{1}{\cos x}\right) d x \end{aligned}$
$\begin{aligned} &=\int_{0}^{\pi}\left(\sec ^{2} x-\tan x \sec x\right) d x \quad\left[\frac{1}{\cos x}=\sec x, \frac{\sin x}{\cos x}=\tan x\right] \\ &=\int_{0}^{\pi} \sec ^{2} x-\int_{0}^{\pi} \tan x \sec x d x \quad\left[\begin{array}{l} \int \sec ^{2} x d x=\tan x \\ \int \sec x \tan x d x=\sec x \end{array}\right] \end{aligned}$
$\left.\begin{array}{l} =[\tan x]_{0}^{\pi}-[\sec x]_{0}^{\pi} \\ =[\tan \pi-\tan 0]-[\sec \pi-\sec 0] \end{array}\right]\left[\begin{array}{l} \tan \pi=\tan 0=0 \\ \sec \pi=-1, \sec 0=1 \end{array}\right]$
$=\left [ 0-0 \right ]-\left [ -1-1 \right ]$
$=0-\left ( -2 \right )$
$=2$
Definite Integrals Exercise 19.1 Question 16
Answer: 2Hint: Use indefinite formula then put the limit to get the required answer
Given: $\int_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1+\sin x}dx$
Solution:
$\int_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1+\sin x}dx$
$\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}\right) d x \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1-\sin x}{1-\sin ^{2} x}\right) d x \end{aligned}$$\left [ \left ( a+b \right )\left ( a-b \right )=a^{2}-b^{2}\right ]$
$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1-\sin x}{\cos ^{2} x}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1-\sin ^{2} x=\cos ^{2} x\right]$
$\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{1}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}\right) d x \text {. } \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sec ^{2} x-\frac{\sin x}{\cos x} \frac{1}{\cos x}\right) d x \end{aligned} \quad\left[\begin{array}{l} \frac{1}{\cos x}=\sec x \\ \frac{\sin x}{\cos x}=\tan x \end{array}\right]$
$\begin{aligned} &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\sec ^{2} x-\tan x \sec x\right) d x \\ &=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec x \tan x d x \end{aligned}$ $\left[\begin{array}{l} \int \sec x \tan x d x=\sec x \\ \int \sec ^{2} x d x=\tan x \end{array}\right]$
$=[\tan x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}-[\sec x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}$
$\begin{aligned} &=\left[\tan \frac{\pi}{4}-\tan \left(-\frac{\pi}{4}\right)\right]-\left[\sec \frac{\pi}{4}-\sec \left(\frac{-\pi}{4}\right)\right] \\ &=\left[\tan \frac{\pi}{4}+\tan \left(\frac{\pi}{4}\right)\right]-\left[\sec \frac{\pi}{4}-\sec \left(\frac{\pi}{4}\right)\right] \end{aligned} \quad\left[\begin{array}{l} \tan (-\theta)=-\tan \theta] \\ \sec (-\theta)=\sec \theta \end{array}\right]$
$=\left[\tan \frac{\pi}{4}+\tan \left(\frac{\pi}{4}\right)\right] \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan \frac{\pi}{4}=1\right]$
$=2\tan \frac{\pi }{4}$
$=2.1$
$=2$
Definite Integrals Exercise 19.1 Question 17
Answer:$\frac{\pi }{4}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{2}}\cos ^{2}xdx$
Solution:$\int_{0}^{\frac{\pi }{2}}\cos ^{2}xdx$
$=\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\cos 2 x}{2}\right) d x \quad\left[\begin{array}{l} 2 \cos ^{2} \theta=1+\cos 2 \theta \\ \cos ^{2} \theta=\frac{1+\cos 2 \theta}{2} \end{array}\right]$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}+\frac{\cos 2 x}{2}\right) d x=\int_{0}^{\frac{\pi}{2}} \frac{1}{2} d x+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x \end{aligned} \quad\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]$
$\begin{aligned} &=\frac{1}{2}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2}[x]_{0}^{\frac{\pi}{2}}+\frac{1}{4}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2}\left[\frac{\pi}{2}-0\right]+\frac{1}{4}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right] \end{aligned}$ $\left [ \sin \pi =\sin 0=0 \right ]$
$=\frac{\pi }{4}+\frac{1}{4}\left [ \sin \pi -\sin 0 \right ]$
$=\frac{\pi }{4}$
Definite Integrals Exercise 19.1 Question 18
Answer: $\frac{2}{3}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{2}}\cos ^{3}xdx$
Solution:$\int_{0}^{\frac{\pi}{2}} \cos ^{3} x d x=\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos 3 x+3 \cos x}{4}\right) d x \quad\left[\begin{array}{l} \cos 3 x=4 \cos ^{3} x-3 \cos x \\ \Rightarrow \cos 3 x+3 \cos x=4 \cos ^{3} x \\ \Rightarrow \cos ^{3} x=\frac{\cos 3 x+3 \cos x}{4} \end{array}\right]$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{3}}\left(\frac{\cos 3 x}{4}+\frac{3}{4} \cos x\right) d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \cos 3 x+\frac{3}{4} \int_{0}^{\frac{\pi}{2}} \cos x d x \\ &=\frac{1}{4}\left[\frac{\sin 3 x}{3}\right]_{0}^{\frac{\pi}{2}}+\frac{3}{4}[\sin x]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{4 \times 3}\left[\sin 3 \times \frac{\pi}{2}-\sin 3 \times 0\right]+\frac{3}{4}\left[\sin \frac{\pi}{2}-\sin 0\right] \quad\left[\begin{array}{l} \left.\sin \frac{3 \pi}{2}=\sin \left(\pi+\frac{\pi}{2}\right)\right] \\ =-\sin \frac{\pi}{2}=-1 \\ \sin \frac{\pi}{2}=1 \end{array}\right. \end{aligned}$
$=\frac{1}{12}\left [ -1-0 \right ]+\frac{3}{4}\left [ 1-0 \right ]$
$=\frac{-1}{12}+\frac{3}{4}=\frac{-1+9}{12}$
$=\frac{8}{12}=\frac{2}{3}$
Definite Integrals Exercise 19.1 Question 19
Answer: $\frac{5}{12}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{6}}\cos x\cos 2xdx$
Solution:$\int_{0}^{\frac{\pi }{6}}\cos x\cos 2xdx=\frac{2}{2}\int_{0}^{\frac{\pi }{6}}\cos x\cos 2xdx$
$\begin{aligned} &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}} 2 \cos x \cos 2 x d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}}[\cos (x+2 x)+\cos (x-2 x)] d x[2 \cos C \cos D=\cos (C+D)+\cos (C-D)] \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}}(\cos 3 x+\cos (-x)) d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}}(\cos 3 x+\cos x) d x \end{aligned}$$\left [ \cos \left ( -\theta \right )=\cos \theta \right ]$
$\begin{aligned} &=\frac{1}{2} \int_{0}^{\frac{\pi}{6}} \cos 3 x d x+\frac{1}{2} \int_{0}^{\frac{\pi}{6}} \cos x d x \\ &=\frac{1}{2}\left[\frac{\sin 3 x}{3}\right]_{0}^{\frac{\pi}{6}}+\frac{1}{2}[\sin x]_{0}^{\frac{\pi}{6}} \\ &=\frac{1}{6}[\sin 3 x]_{0}^{\frac{\pi}{6}}+\frac{1}{2}[\sin x]_{0}^{\frac{\pi}{6}} \\ &=\frac{1}{6}\left[\sin 3 \times \frac{\pi}{6}-\sin 3 \times 0\right]+\frac{1}{2}\left[\sin \frac{\pi}{6}-\sin 0\right] \end{aligned}$
$\begin{aligned} &=\frac{1}{6}\left[\sin \frac{\pi}{2}-\sin 0\right]+\frac{1}{2}\left[\sin \frac{\pi}{6}\right] \\ &=\frac{1}{6} \cdot 1+\frac{1}{2} \cdot \frac{1}{2} \\ &=\frac{1}{6}+\frac{1}{4} \\ &=\frac{4+6}{24}=\frac{10}{24}=\frac{5}{12} \end{aligned}$
Definite Integrals Exercise 19.1 Question 20
Answer: $\frac{2}{3}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{2}}\sin x\sin 2xdx$
Solution:$\int_{0}^{\frac{\pi }{2}}\sin x\sin 2xdx=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\2sin x\sin 2xdx$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\{\cos (x-2 x)-\cos (x+2 x)\} d x$ $\quad[2 \sin A \sin B=\cos (A-B)-\cos (A+B)]$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\{\cos (-x)-\cos (3 x)\} d x \quad[\cos (-\theta)=\cos \theta]$
$\begin{aligned} &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\{\cos x-\cos 3 x\} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos x d x-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 3 x d x \end{aligned} \quad\left[\begin{array}{l} \int \cos x d x=\sin x \\ \left.\int \cos a x d x=\frac{\sin a x}{a}\right] \end{array}\right.$
$\begin{aligned} &=\frac{1}{2}[\sin x]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin 3 x}{3}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2}\left[\sin \frac{\pi}{2}-\sin 0\right]-\frac{1}{3 \times 2}\left[\sin \frac{3 \pi}{2}-\sin 0\right] \quad[\sin 0=0] \end{aligned}$
$\begin{aligned} &=\frac{1}{2} \sin \frac{\pi}{2}-\frac{1}{6} \sin \frac{3 \pi}{2} \\ &=\frac{1}{2} \sin \frac{\pi}{2}-\frac{1}{6} \sin \left(\pi+\frac{\pi}{2}\right) \\ &=\frac{1}{2} \sin \frac{\pi}{2}-\frac{1}{6} \sin \left(-\frac{\pi}{2}\right) \quad\left[\sin \frac{\pi}{2}=1\right] \end{aligned}$
$=\frac{1}{2}\times 1-\frac{1}{6}\left ( -1 \right )$
$=\frac{1}{2}+\frac{1}{6}=\frac{6+2}{12}=\frac{8}{12}$
$=\frac{2}{3}$
Definite Integrals Exercise 19.1 Question 21
Answer: $\frac{-2}{\sqrt{3}}$
Given: $\int_{\frac{\pi }{3}}^{\frac{\pi }{4}}\left ( \tan x+\cot x \right )^{2}dx$
Solution:$\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}(\tan x+\cot x)^{2} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]^{2} d x \quad\left[\tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
$=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right]^{2} d x \quad\left[\sin ^{2} x+\cos ^{2} x=1\right]$
$\begin{aligned} &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{1}{\sin x \cos x}\right]^{2} d x \\ &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{2}{2 \sin x \cos x}\right]^{2} d x \end{aligned}$ $\left [ 2\sin x\cos x=\sin 2x \right ]$
$\begin{aligned} &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{2}{2 \sin x \cos x}\right]^{2} d x \\ &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left[\frac{2}{\sin 2 x}\right]^{2} d x \\ &=\int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \frac{4}{\sin ^{2} 2 x} d x \end{aligned}$ $\left [ \frac{1}{\sin x} =\cos ecx\right ]$
$\begin{aligned} &=4 \int_{\frac{\pi}{3}}^{\frac{\pi}{4}} \operatorname{cosec}^{2}2 x d x \\ &=4\left[\frac{-\cot 2 x}{2}\right]_{\frac{\pi}{3}}^{\frac{\pi}{4}} \end{aligned}$ $\left[\int \operatorname{cosec}^{2} \theta d \theta=-\cot \theta\right]$
$\begin{aligned} &=\frac{-4}{2}[\cot 2 x]_{\frac{\pi}{3}}^{\frac{\pi}{4}} \\ &=-2\left[\cot 2 \times \frac{\pi}{4}-\cot 2 \times \frac{\pi}{3}\right] \\ &=-2\left[\cot \frac{\pi}{2}-\cot \frac{2 \pi}{3}\right]=-2\left[\cot \frac{\pi}{2}-\cot \left(\pi-\frac{\pi}{3}\right)\right] \end{aligned}$ $\left[\begin{array}{l} \cot \frac{\pi}{3}=\frac{1}{\sqrt{3}} \\ \cot \frac{\pi}{2}=0 \end{array}\right]$
$\begin{aligned} &=-2\left[\cot \frac{\pi}{2}-\left(-\cot \frac{\pi}{3}\right)\right]=-2\left[0-\left(\frac{-1}{\sqrt{3}}\right)\right] \\ &=-2\left(\frac{1}{\sqrt{3}}\right) \\ &=\frac{-2}{\sqrt{3}} \end{aligned}$
Definite Integrals Exercise 19.1 Question 22
Answer: $\frac{3\pi }{16}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{2}}\cos ^{4}xdx$
Solution:
$\int_{0}^{\frac{\pi }{2}}\cos ^{4}xdx=\int_{0}^{\frac{\pi }{2}}\left ( \cos ^{2}x \right )^{2}dx$
$=\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\cos 2 x}{2}\right)^{2} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} 2 \cos ^{2} x=1+\cos 2 x \\ \Rightarrow \cos ^{2} x=\frac{1+\cos 2 x}{2} \end{array}\right]$
$\begin{aligned} &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 x)^{2} d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}\left[1+\cos ^{2} 2 x+2 \cos 2 x\right] d x \end{aligned}$ $\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}\left[1+\frac{1+\cos 4 x}{2}+2 \cos 2 x\right] d x \quad\left[\begin{array}{l} 2 \cos ^{2} x=1+\cos 2 x \\ \Rightarrow \cos ^{2} x=\frac{1+\cos 2 x}{2} \end{array}\right]$
$\begin{aligned} &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}\left[1+\frac{1}{2}+\frac{\cos 4 x}{2}+2 \cos 2 x\right] d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{2}}\left[\frac{3}{2}+\frac{\cos 4 x}{2}+2 \cos 2 x\right] d x \\ &=\frac{1}{4} \times \frac{3}{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x+\frac{1}{4 \times 2} \int_{0}^{\frac{\pi}{2}} \cos 4 x d x+\frac{2}{4} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x \end{aligned}$ $\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]$
$\begin{aligned} &=\frac{3}{8}[x]_{0}^{\frac{\pi}{2}}+\frac{1}{8}\left[\frac{\sin 4 x}{4}\right]_{0}^{\frac{\pi}{2}}+\frac{2}{4}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{3}{8}\left[\frac{\pi}{2}-0\right]+\frac{1}{32}\left[\sin 4 \times \frac{\pi}{2}-\sin 4 \times 0\right]+\frac{1}{2 \times 2}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right] \\ &=\frac{3}{8}\left[\frac{\pi}{2}\right]+\frac{1}{32}[\sin 2 \pi-\sin 0]+\frac{1}{4}[\sin \pi-\sin 0] \\ &=\frac{3 \pi}{16}+0+0 \quad[\sin 2 \pi=\sin 0=\sin \pi=0] \\ &=\frac{3 \pi}{16} \end{aligned}$
Definite Integrals Exercise 19.1 Question 23
Answer: $\frac{\pi }{4}\left ( a^{2}+b^{2} \right )$Hint: Use indefinite formula then put the limit to solve this integral
Given:$\int_{0}^{\frac{\pi }{2}}\left ( a^{2}\cos ^{2}x+b^{2}\sin ^{2}x \right )dx$
Solution:
$\int_{0}^{\frac{\pi }{2}}\left ( a^{2}\cos ^{2}x+b^{2}\sin ^{2}x \right )dx$
$=\int_{0}^{\frac{\pi}{2}}\left(a^{2} \cos ^{2} x+b^{2}\left(1-\cos ^{2} x\right)\right) d x \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \cos ^{2} x+\sin ^{2} x=1 \\ \Rightarrow \sin ^{2} x=1-\cos ^{2} x \end{array}\right]$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(a^{2} \cos ^{2} x+b^{2}-b^{2} \cos ^{2} x\right) d x \\ &=\int_{0}^{\frac{\pi}{3}}\left(a^{2}-b^{2}\right) \cos ^{2} x d x+b^{2} \int d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}} 2 \cos ^{2} x d x+b^{2} \int_{0}^{\frac{\pi}{2}} 1 d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 x) d x+b^{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x \end{aligned}$ $\left[\begin{array}{l} \cos ^{2} x+\sin ^{2} x=1 \\ \Rightarrow \sin ^{2} x=1-\cos ^{2} x \end{array}\right]$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(a^{2} \cos ^{2} x+b^{2}-b^{2} \cos ^{2} x\right) d x \\ &=\int_{0}^{\frac{\pi}{3}}\left(a^{2}-b^{2}\right) \cos ^{2} x d x+b^{2} \int d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}} 2 \cos ^{2} x d x+b^{2} \int_{0}^{\frac{\pi}{2}} 1 d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}}(1+\cos 2 x) d x+b^{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x \end{aligned}$ $\left [ 2\cos ^{2}\theta =1+\cos 2\theta \right ]$
$\begin{aligned} &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}} 1 d x+\frac{\left(a^{2}-b^{2}\right)^{\frac{\pi}{2}}}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x+b^{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x \\ &=\frac{\left(a^{2}-b^{2}\right)}{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x+\frac{\left(a^{2}-b^{2}\right)^{\frac{\pi}{2}}}{2} \int_{0}^{\frac{\pi}{3}} \cos 2 x d x+b^{2} \int_{0}^{\frac{\pi}{2}} x^{0} d x \end{aligned} \quad\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]$
$\begin{aligned} &=\frac{\left(a^{2}-b^{2}\right)}{2}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\frac{\pi}{2}}+\frac{\left(a^{2}-b^{2}\right)}{2}\left[\frac{\sin 2 x}{2}\right]+b^{2}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{\left(a^{2}-b^{2}\right)}{2}\left[\frac{\pi}{2}-0\right]+\frac{\left(a^{2}-b^{2}\right)}{2 \times 2}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right]+b^{2}\left[\frac{\pi}{2}-0\right] \end{aligned}$ $\left [ \sin \pi =\sin 0=0 \right ]$
$\begin{aligned} &=\frac{\left(a^{2}-b^{2}\right)}{2} \times \frac{\pi}{2}+\frac{\left(a^{2}-b^{2}\right)}{2 \times 2}[\sin \pi-\sin 0]+b^{2}\left[\frac{\pi}{2}-0\right] \\ &=\frac{\left(a^{2}-b^{2}\right) \pi}{4}+0+\frac{b^{2} \pi}{2} \\ &=\frac{a^{2} \pi-b^{2} \pi+2 b^{2} \pi}{4}=\frac{a^{2} \pi+b^{2} \pi}{4} \\ &=\left(a^{2}+b^{2}\right) \frac{\pi}{4} \end{aligned}$
Definite Integrals Exercise 19.1 Question 24
Answer: 2Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{2}}\sqrt{1+\sin xdx}$
Solution:
$\int_{0}^{\frac{\pi }{2}}\sqrt{1+\sin xdx}$
$=\int_{0}^{\frac{\pi}{2}}\left(\sqrt{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) d x \quad\left[\begin{array}{l} \sin ^{2} \theta+\cos ^{2} \theta=1 \\ \sin 2 \theta=2 \sin \theta \cos \theta \end{array}\right]$
$=\int_{0}^{\frac{\pi}{2}} \sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}} d x \; \; \; \; \; \;\; \; \; \; \; \; \; \; \quad\left[a^{2}+b^{2}+2 a b=(a+b)^{2}\right]$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\ &=\int_{0}^{\frac{\pi}{3}} \sin \frac{x}{2} d x+\int_{0}^{\frac{\pi}{3}} \cos \frac{x}{2} d x \end{aligned}$ $\left[\begin{array}{l} \int \sin a x d x=\frac{-\cos a x}{a} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]$
$\begin{aligned} &=\left[\frac{-\cos \frac{x}{2}}{\frac{1}{2}}\right]_{0}^{\frac{\pi}{2}}+\left[\frac{\sin \frac{x}{2}}{\frac{1}{2}}\right]_{0}^{\frac{\pi}{2}} \\ &=-2\left[\cos \frac{x}{2}\right]_{0}^{\frac{\pi}{2}}+2\left[\sin \frac{x}{2}\right]_{0}^{\frac{\pi}{2}} \end{aligned}$
$\begin{aligned} &=-2\left[\cos \frac{\pi}{2 \times 2}-\cos 0 \times \frac{1}{2}\right]+2\left[\sin \frac{\pi}{2 \times 2}-\sin 0 \times \frac{1}{2}\right\rfloor \\ &=-2\left[\cos \frac{\pi}{4}-\cos 0\right]+2\left[\sin \frac{\pi}{4}-\sin 0\right] \\ &=-2\left[\frac{1}{\sqrt{2}}-1\right]+\left[\frac{1}{\sqrt{2}}-0\right] \end{aligned}$ $\left[\begin{array}{l} \sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\ \sin 0=0 \end{array}\right]$
$=\frac{-2}{\sqrt{2}}+2+\frac{2}{\sqrt{2}}=2$
Definite Integrals Exercise 19.1 Question 25
Answer: 2Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{2}}\sqrt{1+\cos xdx}$
Solution:
$\int_{0}^{\frac{\pi }{2}}\sqrt{1+\cos xdx}$
$=\int_{0}^{\frac{\pi }{2}}\sqrt{2\cos ^{2}\frac{x}{2}dx}$ $\left [ 1+\cos 2\theta =2\cos ^{2}\theta \right ]$
$=\int_{0}^{\frac{\pi }{2}}\sqrt{2}\cos \frac{x}{2}dx$ $\left [ \int \cos \: axdx=\frac{\sin \: ax}{a} \right ]$
$\begin{aligned} &=\sqrt{2} \int_{0}^{\frac{\pi}{2}} \cos \frac{x}{2} d x \\ &=\sqrt{2}\left[\frac{\sin \frac{x}{2}}{\frac{1}{2}}\right]_{0}^{\frac{\pi}{2}} \\ &=\sqrt{2} .2\left[\sin \frac{x}{2}\right]_{0}^{\frac{\pi}{2}} \\ &=2 \sqrt{2}\left[\sin \frac{\pi}{2 \times 2}-\sin \frac{0}{2}\right] \end{aligned}$ $\left[\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 0=0\right]$
$\begin{aligned} &=2 \sqrt{2}\left[\sin \frac{\pi}{4}-0\right] \\ &=2 \sqrt{2}\left[\frac{1}{\sqrt{2}}-0\right] \\ &=2 \sqrt{2} \frac{1}{\sqrt{2}}=2 \end{aligned}$
Definite Integrals Exercise 19.1 Question 26
Answer:$\pi -2$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{2}}x^{2}\sin xdx$
Solution:$\int_{0}^{\frac{\pi }{2}}x^{2}\sin xdx$
Integrating by parts =>Let $x^{2}$ be the first part and sin x be the 2nd part
$\begin{aligned} &=x^{2} \int_{0}^{\frac{\pi}{2}} \sin x d x-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d x^{2}}{dx} \int \sin x d x\right\} d x \\ &=\left[x^{2}(-\cos x)\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x(-\cos x) d x \end{aligned}$ $\left [ \int \sin xdx=-\cos x \right ]$
$\begin{aligned} &=-\left[x^{2}(\cos x)\right]_{0}^{\frac{\pi}{2}}+2 \int_{0}^{\frac{\pi}{2}} x \cos x d x \\ &=-\left[\left(\frac{\pi}{2}\right)^{2} \cos \frac{\pi}{2}-0^{2} \cos 0\right]+2 \int_{0}^{\frac{\pi}{2}} x \cos x d x \\ &=-\left[\frac{\pi^{2}}{4} \times 0-0\right]+2\left[x \int \cos x d x\right]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}}\left[\frac{d(x)}{d x} \int \cos x d x\right] d x \end{aligned}$ $\left [ \int \cos xdx=\sin x \right ]$$=2[x(\sin x)]_{0}^{\frac{\pi}{2}}-2 \int_{0}^{\frac{\pi}{2}} 1 \sin x d x$
(Again using integration by parts method)
$\begin{array}{ll} =2\left[\frac{\pi}{2} \sin \frac{\pi}{2}-0 \times \sin 0\right]-2[-\cos x]_{0}^{\frac{\pi}{2}} & {\left[\int \sin x d x=-\cos x\right]} \\ =2\left[\frac{\pi}{2} \times 1-0\right]+2[\cos x]_{0}^{\frac{\pi}{2}} & {\left[\sin \frac{\pi}{2}=1, \sin 0=0\right]} \end{array}$
$\begin{aligned} &=\pi+2\left[\cos \frac{\pi}{2}-\cos 0\right] \\ &=\pi+2[0-1] \\ &=\pi-2 \end{aligned}$ $\left[\begin{array}{c} \cos \frac{\pi}{2}=0 \\ \cos 0=1 \end{array}\right]$
Definite Integrals Exercise 19.1 Question 27
Answer: $\frac{\pi }{2}-1$Hint:Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{2}}x\cos xdx$
Solution:
$\int_{0}^{\frac{\pi }{2}}x\cos xdx$
Integrating by parts => Let x be the 1st part and cos x be the 2nd part
$\begin{aligned} &=\left[x \int \cos x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d(x)}{d x} \int \cos x d x\right\} d x \\ &=[x \sin x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}(1 \cdot \sin x) d x \end{aligned} \quad\left[\int \cos x d x=\sin x\right]$
$=\left[\frac{\pi}{2} \sin \frac{\pi}{2}-0 \times \sin 0\right]-\int_{0}^{\frac{\pi}{2}} \sin x d x \quad\left[\sin \frac{\pi}{2}=1, \sin 0=0\right]$
$\begin{aligned} &=\left[\frac{\pi}{2} \cdot 1-0\right]-[-\cos x]_{0}^{\frac{\pi}{2}} \; \; \; \; \; \; \; \; \; \quad\left[\int \sin x d x=-\cos x\right] \\ &=\frac{\pi}{2}+[\cos x]_{0}^{\frac{\pi}{2}} \\ &=\frac{\pi}{2}+\left[\cos \frac{\pi}{2}-\cos 0\right] \end{aligned} \quad\left[\cos \frac{\pi}{2}=0, \cos 0=1\right]$
$=\frac{\pi }{2}+\left [ 0-1 \right ]$
$=\frac{\pi }{2}-1$
Definite Integrals Exercise 19.1 Question 29
Answer: $\sqrt{2}+\frac{\pi }{2\sqrt{2}}-\frac{\pi ^{2}}{16\sqrt{2}}-2$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi }{4}}x^{2}\sin xdx$
Solution:
$\int_{0}^{\frac{\pi }{4}}x^{2}\sin xdx$
Integrating by parts then, $x^{2}$ and sin x be the 2 parts
$\begin{aligned} =&\left[x^{2} \int \sin x d x\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\left(\frac{d\left(x^{2}\right)}{d x} \int \sin x d x\right) d x \\ =&\left[x^{2}(-\cos x)\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}[2 x(-\cos x) d x] \end{aligned} \quad\left[\int \sin x d x=-\cos x\right]$
$=-\left[x^{2} \cos x\right]_{0}^{\frac{\pi}{4}}+2 \int_{0}^{\frac{\pi}{4}} x \cos x d x$
(Again using integration by parts method x and cos x be the 2 terms)
$=-\left[\frac{\pi^{2}}{4^{2}} \cos \frac{\pi}{4}-0 \times \cos 0\right]+2\left[x \int \cos x d x\right]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}}\left(\frac{d(x)}{d x} \int \cos x d x\right) d x$
$=-\left[\frac{\pi^{2}}{16} \frac{1}{\sqrt{2}}-0\right]+2[x \sin x]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}} 1 \cdot \sin x d \quad\left[\begin{array}{l} \int \cos x d x=\sin x \\ \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1 \end{array}\right]$
$\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4} \sin \frac{\pi}{4}-0 \times \sin 0\right]-2 \int_{0}^{\frac{\pi}{4}} \sin x d x \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4} \frac{1}{\sqrt{2}}-0\right]-2[-\cos x]_{0}^{\frac{\pi}{4}} \end{aligned}\left[\begin{array}{l} \int \sin x d x=\cos x \\ \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 0=0 \end{array}\right]$
$\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4 \sqrt{2}}\right]+2[\cos x]_{0}^{\frac{\pi}{4}} \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+2\left[\cos \frac{\pi}{4}-\cos 0\right] \quad\left[\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1\right] \end{aligned}$
$\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+2\left[\frac{1}{\sqrt{2}}-1\right] \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}-2 \\ &=\sqrt{2}+\frac{\pi}{2 \sqrt{2}}-\frac{\pi^{2}}{16 \sqrt{2}}-2 \end{aligned}$
Definite Integrals Exercise 19.1 Question 30
Answer: $\frac{-\pi }{4}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x$
Solution:
$\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x$
Integrating by parts then
$\begin{aligned} &=\left[x^{2} \int \cos 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d\left(x^{2}\right)}{d x} \int \cos 2 x d x\right\} d x \\ &=\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x \frac{\sin 2 x}{2} \end{aligned} \quad\left[\int \cos a x d x=\frac{\sin a x}{a}\right]$
$=\frac{1}{2}\left[x^{2} \sin 2 x\right]-\frac{2}{2} \int_{0}^{\frac{\pi}{2}} x \sin 2 x d x$
Again using integrating on by parts method
$=\frac{1}{2}\left[\left(\frac{\pi}{2}\right)^{2} \sin 2 \times \frac{\pi}{2}-0^{2} \sin 0\right]-\left[x \int \sin 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(\frac{d(x)}{d x} \int \sin 2 x d x\right) d x$
$\begin{aligned} &=\frac{1}{2}\left[\frac{\pi^{2}}{4} \sin \pi-\sin 0 \times 0\right]-\left[x\left(\frac{-\cos 2 x}{2}\right)\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\left(\frac{-\cos 2 x}{2}\right) d x \\ &=\frac{1}{2} \times 0+\frac{1}{2}[x \cos 2 x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \frac{\cos 2 x}{2} d x & {\left[\begin{array}{l} \int \sin a x d x=\frac{-\cos a x}{a} \\ \sin \pi=0, \sin 0=0 \end{array}\right]} \end{aligned}$
$=\frac{1}{2}\left[\frac{\pi}{2} \cos 2 \times \frac{\pi}{2}-0 \times \cos 2 \times 0\right]-\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \quad\left[\int \cos a x d x=\frac{\sin a x}{a}\right]$
$\begin{aligned} &=\frac{1}{2}\left[\frac{\pi}{2} \cos \pi-0\right]-\frac{1}{2 \times 2}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2}\left[\frac{\pi}{2}\right](-1)-\frac{1}{4}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right] \end{aligned}$ $\left [ \cos \pi =-1 \right ]$
$\begin{aligned} &=\frac{1}{2}\left(\frac{-\pi}{2}\right)-\frac{1}{4}[\sin \pi-\sin 0] \\ &=-\frac{\pi}{4}-\frac{1}{4}[0-0] \end{aligned}$ $\left [ \sin \pi =\sin 0=0 \right ]$
$=\frac{-\pi }{4}$
Definite Integrals Exercise 19.1 Question 31
Answer:$\frac{\pi ^{3}}{48}-\frac{\pi }{8}$
Hint: Use indefinite integral formula then put the limits to solve this integral
Given:
$\int_{0}^{\frac{\pi }{2}}x^{2}\cos ^{2}xdx$
Solution:
$\begin{aligned} &\int_{0}^{\frac{\pi}{2}} x^{2} \cos ^{2} x d x=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 2 x^{2} \cos ^{2} x d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2}\left(2 \cos ^{2} x\right) d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2}(1+\cos 2 x) d x \end{aligned}$
$\begin{aligned} &\left[\because 2 \cos ^{2} \theta=1+\cos 2 \theta\right]\\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(x^{2}+x^{2} \cos 2 x\right) d x\\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} x^{2} d x+\frac{1}{2} \int_{0}^{\frac{\pi}{3}}\left(x^{2} \cos 2 x\right) d x \ldots \ldots . \end{aligned}$ (1)
Now
$\begin{aligned} &\int_{0}^{\frac{\pi}{2}} x^{2} d x=\left[\frac{x^{2+1}}{2+1}\right]_{0}^{\frac{\pi}{2}} \\ &=\left[\frac{x^{3}}{3}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{3}\left[x^{3}\right]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{3}\left[\left(\frac{\pi}{2}\right)^{3}-(0)^{3}\right] \\ &=\frac{1}{3}\left[\frac{\pi^{3}}{8}-0\right] \end{aligned}$
$=\frac{\pi ^{3}}{24}$ .................(ii)
And
$\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(x^{2} \cos 2 x\right) d x$
Integrating by parts, then
$\int_{0}^{\frac{\pi}{2}}\left(x^{2} \cos 2 x\right) d x=\left[x^{2} \int \cos 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d}{d x}\left(x^{2}\right) \int \cos 2 x d x\right\} d x$
$=\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x^{2-1} \frac{\sin 2 x}{2} d x$
$\left[\because \int \cos a x d x=\frac{\sin a x}{a}, \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$ $\begin{aligned} &=\frac{1}{2}\left[x^{2} \sin 2 x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} x \sin 2 x d x \\ &=\frac{1}{2}\left[\left(\frac{\pi}{2}\right)^{2} \sin 2 \times \frac{\pi}{2}-(0)^{2} \sin 2 \times 0\right]-\left\{\left[x \int \sin 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \frac{d}{d x}(x) \int \sin 2 x d x d x\right\} \end{aligned}$
[By integrating in parts method]
$=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^{2} \sin \pi-0\right]-\left[x\left(-\frac{\cos 2 x}{2}\right)\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}} 1\left(-\frac{\cos 2 x}{2}\right) x$
$\left[\because \int \sin a x d x=-\frac{\cos a x}{a}, \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right]$
$=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^{2} \times 0\right]+\left[\frac{x \cos 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2 x d x$ $\left [ \because \sin \pi =0 \right ]$
$=0+\frac{1}{2}[x \cos 2 x]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}$
$\left [ \because \int \cos axdx=\frac{\sin ax}{a} \right ]$
$\begin{aligned} &=\frac{1}{2}\left[\frac{\pi}{2} \cdot \cos 2 \times \frac{\pi}{2}-0 \cdot \cos 2 \times 0\right]-\frac{1}{4}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2} \cdot \frac{\pi}{2} \cdot \cos \pi-\frac{1}{4}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right] \end{aligned}$
$[\because \cos \pi=-1 \& \sin \pi=0=\sin 0]$
$\begin{aligned} &=\frac{\pi}{4}(-1)-\frac{1}{4}[\sin \pi-\sin 0] \\ &=-\frac{\pi}{4}-\frac{1}{4}[0-0] \\ &=-\frac{\pi}{4} \ldots \ldots . . . .(3) \end{aligned}$
Putting the value of integrals from eq(2) and (3) in (1) then
$\begin{aligned} &\int_{0}^{\frac{\pi}{2}}\left(x^{2} \cos 2 x\right) d x=\frac{1}{2} \times \frac{\pi}{24}^{3}+\frac{1}{2}\left(-\frac{\pi}{4}\right) \\ &=\frac{\pi^{3}}{48}-\frac{\pi}{8} \end{aligned}$
Definite Integrals Exercise 19.1 Question 32
Answer:$2log2-1$
Hint: Use indefinite integral formula and then put the limits to solve this integral
Given:
$\int_{1}^{2}log\; xdx$
Solution:
$\int_{1}^{2}log\; xdx$
Integrating by parts, then
$\begin{aligned} &\int_{1}^{2} \log x d x=\left[\log x \int 1 d x\right]_{1}^{2}-\int_{1}^{2}\left[\frac{d}{d x}(\log x) \int 1 d x\right] d x \\ &=[\log x x]_{1}^{2}-\int_{1}^{2} \frac{1}{x} x d x \quad\left[\because \int 1 d x=x ; \frac{d}{d x}(\log x)=\frac{1}{x}\right] \\ &=[2 \log 2-11 \log 1]-\int_{1}^{2} 1 d x \quad\left[\because \int 1 d x=x, \log 1=0\right] \end{aligned}$
$\begin{aligned} &=[2 \log 2-1(0)]-[x]_{1}^{2} \\ &=2 \log 2-[2-1] \\ &=2 \log 2-1 \end{aligned}$
Definite Integrals Exercise 19.1 Question 33
Answer:$\frac{3}{4}log3-log2$
Hint:Use indefinite integral formula then put the limits to solve this integral
Given:
$\int_{1}^{3} \frac{\log x}{(x+1)^{2}} d x=\int_{1}^{3} \log x \frac{1}{(x+1)^{2}} d x$
Integrating by parts then
$\begin{aligned} &\int_{1}^{3} \log x \cdot \frac{1}{(x+1)^{2}} d x=\left[\log x \int \frac{1}{(x+1)^{2}} d x\right]_{1}^{3}-\int_{1}^{3}\left\{\frac{d}{d x}(\log x) \cdot \int \frac{1}{(x+1)^{2}} d x\right\} d x \\ &=\left[\log x \int(x+1)^{-2} d x\right]_{1}^{3}-\int_{1}^{3}\left\{\frac{1}{x} \cdot \int(x+1)^{-2} d x\right\} d x \\ &=\left[\log x\left(\frac{(x+1)^{-2+1}}{-2+1}\right)\right]_{1}^{3}-\int_{1}^{3} \frac{1}{x}\left(\frac{(x+1)^{-2+1}}{-2+1}\right) d x \end{aligned}$
$\begin{aligned} &=\left[\log x\left(\frac{(x+1)^{-1}}{-1}\right)\right]_{1}^{3}-\int_{1}^{3} \frac{1}{x}\left(\frac{(x+1)^{-1}}{-1}\right) d x \\ &=-\left[\frac{\log x}{x+1}\right]_{1}^{3}+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\left[\frac{\log 3}{3+1}-\frac{\log 1}{1+1}\right]_{1}^{3}+\int_{1}^{3} \frac{1}{x(x+1)} d x \end{aligned}$
$\begin{aligned} &=-\left[\frac{\log 3}{4}-\frac{\log 1}{2}\right]+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\left[\frac{\log 3}{4}-0\right]+\int_{1}^{3} \frac{1}{x(x+1)} d x \\ &=-\frac{1}{4} \log 3+\int_{1}^{3} \frac{1}{x(x+1)} d x \ldots . .(1) \\ &\therefore \int_{1}^{3} \frac{1}{x(x+1)} d x \end{aligned}$
To solve this integral first we need to find its partial fraction then we will integrate and put the limits. So$\begin{aligned} &\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1} \\ &\Rightarrow 1=\frac{A}{x}(x+1)(x)+\frac{B}{x+1} x(x+1) \\ &\Rightarrow 1=A(x+1)+B x \\ &\Rightarrow 1=A x+A+B x \\ &\Rightarrow 1=(A+B) x+A \end{aligned}$
Equating coefficient of x from both sides
$0=A+B\Rightarrow A=-B\Rightarrow B=-A$
Again Equating coefficient of constant term from both side, then
$1=A$
$\Rightarrow B=-A=-1$
$\therefore A=1,B=-1$
Thus
$\frac{1}{x\left (x+1 \right )}=\frac{1}{x}-\frac{1}{x+1}$
Now
$\int_{1}^{3}\left(\frac{1}{x}-\frac{1}{x+1}\right) d x=\int_{1}^{3} \frac{1}{x} d x-\int_{1}^{3} \frac{1}{x+1} d x$
put $x+1=u$
$\Rightarrow dx=du$in the 2nd integral, then when x=1
$\Rightarrow u=2$and when x=3,u=4 then
$\begin{aligned} &\int_{1}^{3}\left(\frac{1}{x}-\frac{1}{x+1}\right) d x=\int_{1}^{3} \frac{1}{x} d x-\int_{2}^{4} \frac{1}{u} d u \\ &=[\log |x|]_{1}^{3}-[\log |u|]_{2}^{4} \\ &{\left[\because \int \frac{1}{x} d x=\log |x|\right]} \end{aligned}$
$\begin{aligned} &=[\log 3-\log 1]-[\log 4-\log 2] \\ &=[\log 3-0]-\left[\log \frac{4}{2}\right] \\ &=\log 3-\log 2 \ldots \ldots . .(2) \end{aligned}$
putting the value of this integral eq(2) in eq(1) then
$\int_{1}^{3} \frac{\log x}{(x+1)^{2}} d x=-\frac{1}{4} \log 3+\log 3-\log 2$
$\begin{aligned} &=\left(-\frac{1}{4}+1\right) \log 3-\log 2 \\ &=\left(\frac{-1+4}{4}\right) \log 3-\log 2 \\ &=\frac{3}{4} \log 3-\log 2 \end{aligned}$
Definite Integrals Exercise 19.1 Question 34
Answer:$e^{e}$
Hint: Use indefinite formula and put limits to solve this integral
Given:
$\int_{1}^{e} \frac{e^{x}}{x}(1+x \log x) d x$
Solution:
$\begin{aligned} &\int_{1}^{e} \frac{e^{x}}{x}(1+x \log x) d x=\int_{1}^{e}\left(\frac{e^{x}}{x}+\frac{e^{x}}{x} \log x\right) d x \\ &=\int_{1}^{e} \frac{e^{x}}{x} d x+\int_{1}^{e} e^{x} \log x d x \end{aligned}$
applying integration by parts method in 1st integral then
$\begin{aligned} &=\left[e^{x} \int \frac{1}{x} d x\right]_{1}^{e}-\int_{1}^{e}\left[\frac{d}{d x}\left(e^{x}\right) \int \frac{1}{x} d x\right] d x+\int_{1}^{e} e^{x} \log x d x \\ &=\left[e^{x} \log x\right]_{1}^{e}-\int_{1}^{e} e^{x} \log x d x+\int_{1}^{e} e^{x} \log x d x \\ &=\left[e^{e} \log e-e^{1} \log 1\right] \\ &{[\because \log e=1,1 \log 1=0]} \\ &=\left[e^{e} .1-0\right] \end{aligned}$
$=e^{e}$
Definite Integrals Exercise 19.1 Question 35
Answer:$\frac{1}{2}$
Hint: Use indefinite formula and put limits to solve this integral
Given:
$\int_{1}^{e}\frac{log\: x}{x}dx$
Putting $log x=t$
$\Rightarrow \frac{1}{x}dx=dt$
$\Rightarrow dx=xdt$
and when $x=1$ then$t=log1=0$
$\left [ \because log\; 1=0 \right ]$
When $x=e$ then $t=loge=1$
$\left [ \because log\: e=1 \right ]$
Then
$\begin{aligned} &\int_{1}^{e} \frac{\log x}{x} d x=\int_{0}^{1} \frac{t}{x} x d t \\ &=\int_{0}^{1} t d t \\ &=\left[\frac{t^{1+1}}{1+1}\right]_{0}^{1} \\ &=\left[\frac{t^{2}}{2}\right]_{0}^{1} \end{aligned}$
$\begin{aligned} &=\frac{1}{2}\left[t^{2}\right]_{0}^{1} \\ &=\frac{1}{2}\left[1^{2}-0^{2}\right] \\ &=\frac{1}{2}[1-0] \\ &=\frac{1}{2} \end{aligned}$
Definite Integrals Exercise 19.1 Question 36
Answer:$\frac{e^{2}}{2}-e$
Hint: Use indefinite integral formula and put limits to solve this integral
Given:
$\int_{e}^{e^{2}}\left \{ \frac{1}{log\: x}-\frac{1}{\left ( log\: x \right )^{2}} \right \}dx$
Solution:
$\begin{aligned} &\int_{e}^{e^{2}}\left\{\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right\} d x=\int_{e}^{e^{2}} \frac{1}{\log x} d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &=\int_{e}^{e^{2}} \frac{1}{\log x} \cdot 1 d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &=\left[\frac{1}{\log x} \int 1 d x\right]_{e}^{e^{2}}-\int_{e}^{e^{2}}\left\{\frac{d}{d x}\left(\frac{1}{\log x}\right) \int 1 d x\right\} d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &\quad\left[\because \int 1 d x=x \& \frac{d}{d x}\left(\frac{1}{x}\right)=\frac{d}{d x}\left(x^{-1}\right)=-1 x^{-1-1}=-1 \cdot x^{-2}=-\frac{1}{x^{2}}\right] \end{aligned}$
$\begin{aligned} &=\left[\frac{1}{\log x} x\right]_{\theta}^{e^{2}}-\int_{\theta}^{e}\left(-\frac{1}{(\log x)^{2}}\right) \frac{1}{x} x d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &=\left[\frac{x}{\log x}\right]_{e}^{e^{2}}+\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x-\int_{e}^{e^{2}} \frac{1}{(\log x)^{2}} d x \\ &=\left[\frac{e^{2}}{\log e^{2}}-\frac{e}{\log e}\right] \\ &=\left[\frac{e^{2}}{2 \log e}-\frac{e}{\log e}\right] \; \; \; \; \; \; \; \; \; \; \; \quad[\because \log e=1] \\ &=\left[\frac{e^{2}}{2}-e\right] \end{aligned}$
Definite Integrals Exercise 19.1 Question 37
Answer:$\frac{1}{2}log\; \; 6$
Hint: Use indefinite formula and put limits to solve this integral
Given:
$\int_{1}^{2}\frac{x+3}{x\left ( x+2 \right )}dx$
Solution:
$\begin{aligned} &\int_{1}^{2} \frac{x+3}{x(x+2)} d x=\int_{1}^{2} \frac{x}{x+2} d x+\int_{1}^{2} \frac{3}{x(x+2)} d x \\ &=\int_{1}^{2} \frac{1}{x+2} d x+3 \int_{1}^{2} \frac{1}{x(x+2)} d x \end{aligned}$ ..............(1)
Now
$\int_{1}^{2}\frac{1}{x+2}dx$
Putting $x+2=u$
$\Rightarrow dx=du$
when x=1 then u=1+2=3and when x=2 then u=2+2=4
Then
$\begin{aligned} \int_{1}^{2} \frac{1}{x+2} d x &=\int_{3}^{4} \frac{1}{u} d u=[\log u]_{3}^{4} \quad\left[\because \int \frac{1}{x} d x=\log x\right] \\ &=[\log 4-\log 3] \end{aligned}$ ...............(2)
and
$\int_{1}^{2}\frac{1}{x\left ( x+2 \right )}dx$
To solve this integral, first we need to find its partial fraction then integrate it and put the given limits. So
$\begin{aligned} &\frac{1}{x(x+2)}=\frac{A}{x}+\frac{B}{(x+2)} \\ &\Rightarrow 1=\frac{A}{x}(x)(x+2)+\frac{B}{(x+2)} x(x+2) \\ &\Rightarrow 1=(x+2) A+B x \\ &\Rightarrow 1=A x+2 A+B x \\ &\Rightarrow 1=(A+B) x+2 A \end{aligned}$
Equating coefficient of x from both sides, then$0=A+B\Rightarrow B=-A$
Again, Equating coefficient of constant term from both sides then
$1=2A\Rightarrow A=\frac{1}{2}$
$\Rightarrow B=-\frac{1}{2}$
$\therefore A=\frac{1}{2}$&$B=-\frac{1}{2}$
Then
$\begin{aligned} &\frac{1}{x(x+2)}=\frac{1}{2 x}-\frac{1}{2(x+2)} \\ &\therefore \int_{1}^{2} \frac{1}{x(x+2)} d x=\int_{1}^{2}\left(\frac{1}{2 x}-\frac{1}{2(x+2)}\right) d x \\ &=\frac{1}{2} \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{(x+2)}\right) d x \\ &=\frac{1}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{(x+2)} d x \end{aligned}$
Put $x+2=u$
$\Rightarrow dx=du$ in the 2nd integral then
$\begin{aligned} &\int_{1}^{2} \frac{1}{x(x+2)} d x=\frac{1}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{u} d x \\ &=\frac{1}{2}[\log x]_{1}^{2}-\frac{1}{2}[\log u]_{1}^{2} \\ &{\left[\because \int \frac{1}{x} d x=\log x\right]} \\ &=\frac{1}{2}[\log x]_{1}^{2}-\frac{1}{2}[\log (x+2)]_{1}^{2} \\ &{[\because u=x+2]} \end{aligned}$
$\begin{aligned} &=\frac{1}{2}[\log 2-\log 1]-\frac{1}{2}[\log (2+2)-\log (1+2)] \\ &=\frac{1}{2}[\log 2-0]-\frac{1}{2}[\log (4)-\log (3)] \\ &=\frac{1}{2}[\log 2]-\frac{1}{2} \log 4+\frac{1}{2} \log 3 \end{aligned}$ ..............(3)
Putting the value of integrals from eq(2) and (3) in (1), we get
$\begin{aligned} &\int_{1}^{2} \frac{x+3}{x(x+2)} d x=\log 4-\log 3+3\left[\frac{1}{2} \log 2-\frac{1}{2} \log 4+\frac{1}{2} \log 3\right] \\ &=\log 4-\log 3+\frac{3}{2} \log 2-\frac{3}{2} \log 4+\frac{3}{2} \log 3 \\ &=\left(1-\frac{3}{2}\right) \log 4-\left(1-\frac{3}{2}\right) \log 3+\frac{3}{2} \log 2 \\ &=\left(\frac{2-3}{2}\right) \log 4-\left(\frac{2-3}{2}\right) \log 3+\frac{3}{2} \log 2 \end{aligned}$
$\begin{aligned} &=-\frac{1}{2} \log 4-\left(-\frac{1}{2}\right) \log 3+\frac{3}{2} \log 2 \\ &=\frac{1}{2}\left[-\log 2^{2}+\log 3+3 \log 2\right] \\ &=\frac{1}{2}[-2 \log 2+\log 3+3 \log 2] \\ &=\frac{1}{2}[\log 2+\log 3] \end{aligned}$
$$$ \begin{aligned} &=\frac{1}{2} \log (2 \times 3) \\ &{[\because \log (m n)=\log m+\log n]} \\ &=\frac{1}{2} \log 6 \end{aligned}$
Definite Integrals Exercise 19.1 Question 38
Answer:$\frac{1}{5}log6+\frac{3}{\sqrt{5}}\tan ^{-1}\left ( \sqrt{5} \right )$
Hint: Use indefinite integral formula and put the limits to solve this integral
Given:
$\int_{0}^{1}\frac{2x+3}{5x^{2}+1}dx$
Solution:
$$$ \begin{aligned} &\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x=\int_{0}^{1}\left(\frac{2 x}{5 x^{2}+1}+\frac{3}{5 x^{2}+1}\right) d x \\ &=\int_{0}^{1} \frac{2 x}{5 x^{2}+1} d x+\int_{0}^{1} \frac{3}{5 x^{2}+1} d x \\ &=\frac{1}{5} \int_{0}^{1} \frac{5 \times 2 x}{5 x^{2}+1} d x+3 \int_{0}^{1} \frac{1}{5 x^{2}+1} d x \\ &=\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x+3 \int_{0}^{1} \frac{1}{5 x^{2}+1} d x \end{aligned}$ ..........(1)
Now,
$\frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2}+1} d x$
Put
$5x^{2}+1=u\Rightarrow 10xdx=du$
$$$ \begin{array}{ll} \frac{1}{5} \int_{0}^{1} \frac{1}{u} d u=\frac{1}{5}[\log u]_{0}^{1} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\int \frac{1}{x} d x=\log x\right]} \\ =\frac{1}{5}\left[\log \left(5 x^{2}+1\right)\right]_{0}^{1} &\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\left[u=5 x^{2}+1\right]} \end{array}$
$$$ \begin{aligned} &=\frac{1}{5}\left[\log \left(5 \cdot 1^{2}+1\right)-\log \left(5.0^{2}+1\right)\right] \\ &=\frac{1}{5}[\log (5+1)-\log (0+1)] \\ &=\frac{1}{5}[\log 6-\log 1] \end{aligned}$
$\left [ \because log\: 1=0 \right ]$
$=\frac{1}{5}log\: 6$ .........................(2)
And
$$$ \begin{aligned} &\int_{0}^{1} \frac{1}{5 x^{2}+1} d x=\int_{0}^{1} \frac{1}{5\left(x^{2}+\frac{1}{5}\right)} d x \\ &=\frac{1}{5} \int_{0}^{1} \frac{1}{\left(x^{2}+\left(\frac{1}{5}\right)^{2}\right)} d x \\ &=\frac{1}{5}\left[\frac{1}{\sqrt{5}} \tan ^{-1}\left(\frac{x}{\frac{1}{\sqrt{5}}}\right)\right]_{0}^{1} \\ &=\frac{1}{5} \frac{1}{\frac{1}{\sqrt{5}}}\left[\tan ^{-1}\left(\frac{\sqrt{5} x}{1}\right)\right]_{0}^{1} \end{aligned}$
$$$ \begin{aligned} &=\frac{1}{\sqrt{5}}\left[\tan ^{-1} \sqrt{5} \cdot 1-\tan ^{-1} \sqrt{5} .0\right] \\ &=\frac{1}{\sqrt{5}}\left[\tan ^{-1} \sqrt{5}-\tan ^{-1} 0\right] \\ &=\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5})-0 \\ &=\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5}) \end{aligned}$ .....................(3)
Put the value of the integrals from (2) and (3) in (1) then
$$$ \begin{aligned} &\int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x=\frac{1}{5} \log 6+3 \cdot \frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5}) \\ &=\frac{1}{5} \log 6+\frac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5}) \end{aligned}$
Definite Integrals Exercise 19.1 Question 39
Answer:$\frac{1}{\sqrt{17}}log\left ( \frac{21+5\sqrt{7}}{4} \right )$
Hint: Use indefinite formula and then put limits to solve this integral
Given:
$\int_{0}^{2}\frac{1}{4+x-x^{2}}dx$
Solution:
$$$ \begin{aligned} &\int_{0}^{2} \frac{1}{4+x-x^{2}} d x=\int_{0}^{2} \frac{1}{-\left(x^{2}-x-4\right)} d x \\ &=-\int_{0}^{2} \frac{1}{x^{2}-2 x \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-4} d x \\ &=-\int_{0}^{2} \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1}{4}\right)-4} d x \\ &=-\int_{0}^{2} \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{1+16}{4}\right)} d x \end{aligned}$
$$$ \begin{aligned} &=-\int_{0}^{2} \frac{1}{\left(x-\frac{1}{2}\right)^{2}-\left(\frac{17}{4}\right)} d x \\ &=\int_{0}^{2} \frac{1}{\left(\frac{17}{4}\right)-\left(x-\frac{1}{2}\right)^{2}} d x \end{aligned}$
Putting $\left ( x-\frac{1}{2} \right )=t$
$\Rightarrow dx=dt$
When $x=0$ then
$t=-\frac{1}{2}$
and when $x=2$ then
$t=2-\frac{1}{2}=\frac{4-1}{2}=\frac{3}{2}$
Then
$$$ \begin{aligned} &\int_{0}^{2} \frac{1}{4+x-x^{2}} d x=\int_{-\frac{1}{2}}^{\frac{3}{2}} \frac{1}{\left(\frac{\sqrt{17}}{2}\right)^{2}-t^{2}} d x \\ &=\left[\frac{1}{2 \times \frac{\sqrt{17}}{2}} \log \left|\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right|\right]_{\frac{-1}{2}}^{\frac{3}{2}} \end{aligned}$
$$$ \begin{aligned} &\left.=\frac{1}{\sqrt{17}}\left[\log \left|\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}\right|-\log \mid \frac{\frac{\sqrt{17}}{2}+\left(-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}-\left(-\frac{1}{2}\right)}\right)\right] \\ &=\frac{1}{\sqrt{17}}\left[\log \left|\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}-3}{2}}\right|-\log \left|\frac{\frac{\sqrt{17}}{2}+\left(-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right|\right] \end{aligned}$
$=\frac{1}{\sqrt{17}}\left [ log|\frac{\sqrt{17}+3}{\sqrt{17}-3}|-log|\frac{\frac{\sqrt{17}-1}{2}}{\frac{\sqrt{17}+1}{2}}| \right ]$
$=\frac{1}{\sqrt{17}}\left[\log \left|\frac{\sqrt{17}+3}{\sqrt{17}-3}\right|-\log \left|\frac{\sqrt{17}-1}{\sqrt{17}+1}\right|\right]$
$\left[\because \log m-\log n=\log \left(\frac{m}{n}\right)\right]$
$=\frac{1}{\sqrt{17}} \log \left[\frac{\left(\frac{\sqrt{17}+3}{\sqrt{17}-3}\right)}{\left(\frac{\sqrt{17}-1}{\sqrt{17}+1}\right)}\right]$
$=\frac{1}{\sqrt{17}} \log \left(\frac{\sqrt{17}+3}{\sqrt{17}-3} \times \frac{\sqrt{17}+1}{\sqrt{17}-1}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{17+3 \sqrt{17}+\sqrt{17}+3}{17-3 \sqrt{17}-\sqrt{17}+3}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{4(5+\sqrt{17})}{4(5-\sqrt{17})}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{5+\sqrt{17}}{5-\sqrt{17}}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{5+\sqrt{17}}{5-\sqrt{17}} \times \frac{5+\sqrt{17}}{5+\sqrt{17}}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{(5+\sqrt{17})^{2}}{(5-\sqrt{17})(5+\sqrt{17})}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{5^{2}+(\sqrt{17})^{2}+2.5 \cdot \sqrt{17}}{(5)^{2}-(\sqrt{17})^{2}}\right)$
$\left[\begin{array}{l}\because(a+b)^{2}=a^{2}+b^{2}+2 a b \\ \left(a+b(a-b)=\left(a^{2}-b^{2}\right)\right.\end{array}\right]$
$=\frac{1}{\sqrt{17}} \log \left(\frac{25+17+10 \sqrt{17}}{25-17}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{2(21+5 \sqrt{17})}{8}\right)$
$=\frac{1}{\sqrt{17}} \log \left(\frac{21+5 \sqrt{17}}{4}\right)$
Definite Integrals Exercise 19.1 Question 40
Answer:$\frac{2}{\sqrt{7}}\left [ \tan ^{-1}\frac{5}{\sqrt{7}}-\tan ^{-1}\frac{1}{\sqrt{7}} \right ]$
Hint: Use indefinite formula and then put limits to solve this integral
Given:
$\int_{0}^{1}\frac{1}{2x^{2}+x+1}dx$
Solution:
$\int_{0}^{1} \frac{1}{2 x^{2}+x+1} d x=\int_{0}^{1} \frac{1}{2\left(x^{2}+\frac{x}{2}+\frac{1}{2}\right)} d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x^{2}+\frac{x}{2}+\frac{1}{2}\right)} d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{x^{2}+2 \cdot x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}+\frac{1}{2}} d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{1}{4}\right)^{2}+\frac{1}{2}} d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\frac{1}{16}+\frac{1}{2}} d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{1-8}{16}\right)} d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}-\left(\frac{-7}{16}\right)} d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(x+\frac{1}{4}\right)^{2}+\left(\frac{\sqrt{7}}{4}\right)^{2}} d x$
Put
$x+\frac{1}{4}=u\Rightarrow dx=du$
When $x=0$ then
$u=\frac{1}{4}$
when $x=1$ then
$u=1+\frac{1}{4}=\frac{4+1}{4}=\frac{5}{4}$
Then
$\begin{aligned} &\frac{1}{2} \int_{0}^{1} \frac{1}{2 x^{2}+x+1} d x=\frac{1}{2} \int_{\frac{1}{4}}^{\frac{5}{4}} \frac{1}{u^{2}+\left(\frac{\sqrt{7}}{4}\right)^{2}} d u \\ &=\frac{1}{2}\left[\frac{\frac{1}{\sqrt{7}}}{4} \tan ^{-1} \frac{u}{\frac{\sqrt{7}}{4}}\right]_{\frac{1}{4}}^{\frac{5}{4}} \\ &=\frac{1}{2} \times \frac{4}{\sqrt{7}}\left[\tan ^{-1} \frac{4 u}{\sqrt{7}}\right]_{\frac{1}{4}}^{\frac{5}{4}} \end{aligned}$
$\begin{aligned} &=\frac{2}{\sqrt{7}}\left[\tan ^{-1} \frac{4 \times \frac{5}{4}}{\sqrt{7}}-\tan ^{-1} \frac{4 \times \frac{1}{4}}{\sqrt{7}}\right] \\ &=\frac{2}{\sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1} \frac{1}{\sqrt{7}}\right] \end{aligned}$
Definite Integrals Exercise 19.1 Question 41
Answer:$\frac{\pi }{8}$
Hint: Use indefinite integral formula then put limits to solve this integral
Given:
$\int_{0}^{1}\sqrt{x\left (1-x \right )}dx$
Solution:
$\int_{0}^{1}\sqrt{x\left (1-x \right )}dx$
Put
$x=\sin ^{2}\theta \Rightarrow dx=2\sin \theta .\cos \theta d\theta$
When $x=0$ then
$\theta =0$
When $x=1$ then
$\theta =\frac{\pi }{2}$
Then
$\int_{0}^{1} \sqrt{x(1-x)} d x=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)} \cdot 2 \sin \theta \cdot \cos \theta d \theta$
$\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right]$
$=\int_{0}^{\frac{\pi}{2}} \sin \theta \sqrt{\cos ^{2} \theta} \cdot 2 \sin \theta \cdot \cos \theta d \theta$
$=\int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta \cdot 2 \sin \theta \cdot \cos \theta d \theta$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 4 \sin ^{2} \theta \cdot \cos ^{2} \theta d \theta$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}(2 \sin \theta \cdot \cos \theta)^{2} d \theta$
$\left [ \because \sin 2\theta =2\sin \theta \cos \theta \right ]$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}(\sin 2 \theta)^{2} d \theta$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2} 2 \theta d \theta$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta$
$\left[\begin{array}{l}\because 2 \sin ^{2} \theta=1-\cos 2 \theta \\ \Rightarrow \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}\end{array}\right]$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}-\frac{\cos 2 \theta}{2}\right) d \theta$
$=\frac{1}{2 \times 2} \int_{0}^{\frac{\pi}{2}} 1 d \theta-\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \cos 2 \theta d \theta$
$\left[\because \int 1 d x=x\right]$
$\left[\because \int \cos a x d x=\frac{\sin a x}{a}\right]$
$=\frac{1}{4}[\theta]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}}$
$=\frac{1}{4}\left[\frac{\pi}{2}-0\right]-\frac{1}{8}[\sin 2 \theta]_{0}^{\frac{\pi}{2}}$
$=\frac{1}{4} \frac{\pi}{2}-\frac{1}{8}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right]$
$=\frac{\pi}{8}-\frac{1}{8}[\sin \pi-\sin 0]$
$=\frac{\pi}{8}-\frac{1}{8}[0-0]\; \; \; \; \; \; \; \; \; \; \quad[\because \sin \pi=\sin 0=0]$
$=\frac{\pi }{8}$
Definite Integrals Exercise 19.1 Question 42
Answer:$\frac{\pi }{3}$
Hint: Use indefinite formula and put the limits to solve this integral
Given:
$\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^{2}}}dx$
Solution:
$\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^{2}}}dx=\int_{0}^{2}\frac{1}{\sqrt-{\left ( x^{2}-2x-3 \right )}}dx$
$=\int_{0}^{2} \frac{1}{\sqrt{-\left\{x^{2}-2 x \cdot 1+1-1-3\right\}}} d x$
$=\int_{0}^{2} \frac{1}{\sqrt{-\left\{(x-1)^{2}-1-3\right\}}} d x$
$\left [ \because \left ( a-b \right )^{2}=a^{2}-2ab+b^{2} \right ]$
$=\int_{0}^{2} \frac{1}{\sqrt{-\left\{(x-1)^{2}-4\right\}}} d x$
$=\int_{0}^{2} \frac{1}{\sqrt{4-(x-1)^{2}}} d x$
$=\int_{0}^{2} \frac{1}{\sqrt{2^{2}-(x-1)^{2}}} d x$
$=\left[\sin ^{-1}\left(\frac{x-1}{2}\right)\right]_{0}^{2}$
$\left [ \because \int \frac{1}{\sqrt{a^{2}-x^{2}}}dx=\sin ^{-1}\frac{x}{a} \right ]$
$=\left[\sin ^{-1}\left(\frac{2-1}{2}\right)-\sin ^{-1}\left(\frac{0-1}{2}\right)\right]$
$=\left[\sin ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}\left(\frac{-1}{2}\right)\right]$
$\left [ \because \sin ^{-1}\left ( -\theta \right )=-\sin \theta \right ]$
$=\left[\sin ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)\right]$
$=2 \sin ^{-1}\left(\frac{1}{2}\right)$
$\left [ \because \sin ^{-1}\frac{1}{2}=\frac{\pi }{6} \right ]$
$=2.\frac{\pi }{6}$
$=\frac{\pi }{3}$
Definite Integrals Exercise 19.1 Question 43
Answer:$\pi$
Hint: Use indefinite formula and put the limits to solve this integral
Given:
$\int_{0}^{4}\frac{1}{\sqrt{4x-x^{2}}}dx$
Solution:
$\int_{0}^{4}\frac{1}{\sqrt{4x-x^{2}}}dx=\int_{0}^{4}\frac{1}{\sqrt-{\left ( x^{2}-2.2x+\left ( 2 \right )^{2}-\left ( 2 \right )^{2} \right )}}dx$
$=\int_{0}^{4}\frac{1}{\sqrt-{\left \{ \left ( x-2 \right )^{2}-4 \right \}}}dx$
$\left [ \because \left ( a-b \right )^{2}=a^{2}-2ab+b^{2} \right ]$
$=\int_{0}^{4} \frac{1}{\sqrt{4-(x-2)^{2}}} d x$
$=\int_{0}^{4} \frac{1}{\sqrt{2^{2}-(x-2)^{2}}} d x\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)\right]$
$=\left[\sin ^{-1}\left(\frac{x-2}{2}\right)\right]_{0}^{4}$
$=\left[\sin ^{-1}\left(\frac{4-2}{2}\right)-\sin ^{-1}\left(\frac{0-2}{2}\right)\right]$
$=\left[\sin ^{-1}\left(\frac{2}{2}\right)-\sin ^{-1}\left(\frac{-2}{2}\right)\right]$
$=\left[\sin ^{-1}(1)-\sin ^{-1}(-1)\right]$
$=\left[\sin ^{-1}(1)+\sin ^{-1}(1)\right]\left[\because \sin ^{-1}(-\theta)=-\sin \theta\right]$
$=2 \sin ^{-1}(1)$
$=2.\frac{\pi }{2}$
$=\pi$
Definite Integrals Exercise 19.1 Question 44
Answer:$\frac{\pi }{8}$
Hint: Use indefinite formula and put the limits to solve this integral
Given:
$\int_{-1}^{1}\frac{1}{x^{2}+2x+5}dx$
Solution:
$\int_{-1}^{1}\frac{1}{x^{2}+2x+5}dx=\int_{-1}^{1}\frac{1}{x^{2}+2x+4+1}dx$
$=\int_{-1}^{1} \frac{1}{\left(x^{2}+2 x+1\right)+4} d x$
$=\int_{-1}^{1} \frac{1}{(x+1)^{2}+2^{2}} d x$
$=\frac{1}{2}\left[\tan ^{-1}\left(\frac{x+1}{2}\right)\right]_{-1}^{1}$
$=\frac{1}{2}\left[\tan ^{-1}\left(\frac{1+1}{2}\right)-\tan ^{-1}\left(\frac{-1+1}{2}\right)\right]$
$=\frac{1}{2}\left[\tan ^{-1}\left(\frac{2}{2}\right)-\tan ^{-1}\left(\frac{0}{2}\right)\right]$
$=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right]$
$=\frac{1}{2}\left[\frac{\pi}{4}-0\right]$
$\left [ \because \tan ^{-1}\left ( 1 \right )=\frac{\pi }{4},\tan ^{-1}\left ( 0 \right ) =0\right ]$
$=\frac{\pi }{8}$
Definite Integrals Exercise 19.1 Question 45
Answer:
Hint: Use indefinite formula and put the limits to solve this integral
Given:
$\int_{1}^{4}\frac{x^{2}+x}{\sqrt{2x+1}}dx$
Sol:
$\int_{1}^{4}\frac{x^{2}+x}{\sqrt{2x+1}}dx$
Putting
$\Rightarrow 2x+1=t^{2}$
$\Rightarrow 2dx=2tdt$
$\Rightarrow dx=tdt$
and When $x=4$then
$t=3$
and when $x=4$ then
$t=3$
then
$\int_{\sqrt{3}}^{3} \frac{\left(\frac{t^{2}-1}{2}\right)^{2}+\left(\frac{t^{2}}{2}-\frac{1}{2}\right)}{\sqrt{t^{2}}} \cdot t d t=\int_{\sqrt{3}}^{3} \frac{\frac{1}{4}\left(t^{2}-1\right)^{2}+\frac{1}{2}\left(t^{2}-1\right)}{t} \cdot t d t$
$=\int_{\sqrt{3}}^{3}\left(\frac{1}{4}\left(t^{4}+1-2 t^{2}\right)+\frac{t^{2}-1}{2}\right) d t$
$=\int_{\sqrt{3}}^{3}\left(\frac{t^{4}+1-2 t^{2}+2 t^{2}-2}{4}\right) d t$
$=\int_{\sqrt{3}}^{3}\left(\frac{t^{4}-1}{4}\right) d t$
$=\int_{\sqrt{3}}^{3}\left(\frac{t^{4}}{4}-\frac{1}{4}\right) d t$
$=\frac{1}{4} \int_{\sqrt{3}}^{3} t^{4} d t-\frac{1}{4} \int_{\sqrt{3}}^{3} t^{0} d t$
$=\frac{1}{4}\left[\frac{t^{4+1}}{4+1}\right]_{\sqrt{3}}^{3}-\frac{1}{4}\left[\frac{t^{0+1}}{0+1}\right]_{\sqrt{3}}^{3}$
$=\frac{1}{4}\left[\frac{t^{5}}{5}\right]_{\sqrt{3}}^{3}-\frac{1}{4}[t]_{\sqrt{3}}^{3}$
$=\frac{1}{20}\left[t^{5}\right]_{\sqrt{3}}^{3}-\frac{1}{4}[t]_{\sqrt{3}}^{3}$
$=\frac{\left[(3)^{5}-(\sqrt{3})^{5}\right]}{20}-\frac{1}{4}[3-\sqrt{3}]$
$=\frac{1}{4}\left[\frac{243-9 \sqrt{3}}{5}-3+\sqrt{3}\right]$
$=\frac{1}{4}\left[\frac{243-9 \sqrt{3}-3 \times 5+5 \sqrt{3}}{5}\right]$
$=\frac{1}{4}\left[\frac{243-9 \sqrt{3}-15+5 \sqrt{3}}{5}\right]$
$=\frac{1}{4}\left[\frac{228-4 \sqrt{3}}{5}\right]$
$=\frac{228}{20}-\frac{\sqrt{3}}{5}$
$=\frac{114}{10}-\frac{\sqrt{3}}{5}$
$=\frac{57}{5}-\frac{\sqrt{3}}{5}$
$=\frac{57-\sqrt{3}}{5}$
Definite Integrals Exercise 19.1 Question 46
Answer:$\frac{1}{42}$
Hint: Use indefinite integral formula then put limits to solve this integral
Given:
$\int_{0}^{1}x\left ( 1-x \right )^{5}dx$
Sol:
$\int_{0}^{1} x(1-x)^{5} d x=\int_{0}^{1} x\left(1-5 x+10 x^{2}-10 x^{3}+5 x^{4}-x^{5}\right) d x$
(By Binomial theorem)
$=\int_{0}^{1}\left(x-5 x^{2}+10 x^{3}-10 x^{4}+5 x^{5}-x^{6}\right) d x$
$=\int_{0}^{1} x d x-5 \int_{0}^{1} x^{2} d x+10 \int_{0}^{1} x^{3} d x-10 \int_{0}^{1} x^{4} d x+5 \int_{0}^{1} x^{5} d x-\int_{0}^{1} x^{6} d x$
$=\left[\frac{x^{1+1}}{1+1}\right]_{0}^{1}-5\left[\frac{x^{2+1}}{2+1}\right]_{0}^{1}+10\left[\frac{x^{3+1}}{3+1}\right]_{0}^{1}-10\left[\frac{x^{4+1}}{4+1}\right]_{0}^{1}+5\left[\frac{x^{5+1}}{5+1}\right]_{0}^{1}-\left[\frac{x^{6+1}}{6+1}\right]_{0}^{1}$
$=\left[\frac{x^{2}}{2}\right]_{0}^{1}-5\left[\frac{x^{3}}{3}\right]_{0}^{1}+10\left[\frac{x^{4}}{4}\right]_{0}^{1}-10\left[\frac{x^{5}}{5}\right]_{0}^{1}+5\left[\frac{x^{6}}{6}\right]_{0}^{1}-\left[\frac{x^{7}}{7}\right]_{0}^{1}$
$=\frac{1}{2}[1-0]-\frac{5}{3}\left[1^{3}-0^{3}\right]+\frac{10}{4}\left[1^{4}-0^{4}\right]-\frac{10}{5}\left[1^{5}-0^{5}\right]+\frac{5}{6}\left[1^{6}-0^{6}\right]-\frac{1}{7}\left[1^{7}-0^{7}\right]$
$=\frac{1}{2}-\frac{5}{3}+\frac{5}{2}-2+\frac{5}{6}-\frac{1}{7}$
$=\frac{3-10+15-12+5}{6}-\frac{1}{7}$
$=\frac{1}{6}(23-22)-\frac{1}{7}$
$=\frac{7-6}{42}$
$=\frac{1}{42}$
Definite Integrals Exercise 19.1 Question 47
Answer:$\frac{e^{2}}{2}-e$
Hint: Use indefinite integral formula and put the limits to solve this integral
Given:
$\int_{1}^{2}\left ( \frac{x-1}{x^{2}} \right )e^{x}dx$
Sol:
$\int_{1}^{2}\left(\frac{x-1}{x^{2}}\right) e^{x} d x=\int_{1}^{2}\left(\frac{x}{x^{2}}-\frac{1}{x^{2}}\right)e^{x}d x$
$=\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) e^{x} d x$
$=\int_{1}^{2}\left(\frac{1}{x} e^{x}-\frac{1}{x^{2}} e^{x}\right) d x$
$=\int_{1}^{2} \frac{1}{x} e^{x} d x-\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x$
Applying integration by parts method in Ist integral, then
$\int_{1}^{2}\left(\frac{x-1}{x^{2}}\right)^{x} d x=\left[\frac{1}{x} \int e^{x} d x\right]_{1}^{2}-\int_{1}^{2}\left(\frac{d}{d x}\left(\frac{1}{x}\right) \int e^{x} d x\right) d x-\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x$
$=\left[\frac{1}{x} e^{x}\right]_{1}^{2}-\int_{1}^{2} \frac{-1}{x^{2}} e^{x} d x-\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x$
$=\left[\frac{1}{x} e^{x}\right]_{1}^{2}+\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x-\int_{1}^{2} \frac{1}{x^{2}} e^{x} d x$
$=\left[\frac{1}{2} \cdot e^{2}-\frac{1}{1} \cdot e^{1}\right]$
$=\frac{e^{2}}{2}-e$
Definite Integrals Exercise 19.1 Question 48
Answer:$\frac{e^{2}}{4}+\frac{1}{4}+\frac{2}{\pi }$
Hint: Use indefinite formula then put the limits to solve this integral
Given:
$\int_{0}^{1}\left(x e^{2 x}+\sin \frac{\pi x}{2}\right) d x$
Sol:
$\int_{0}^{1}\left(x e^{2 x}+\sin \frac{\pi x}{2}\right) d x=\int_{0}^{1} x e^{2 x} d x+\int_{0}^{1} \sin \frac{\pi x}{2} d x$
Applying integration by parts method in 1st integral then
$\left.=\left[x \int e^{2 x} d x\right]_{0}^{1}-\int_{0}^{1}\left(\frac{d(x)}{d x} \int e^{2 x} d x\right) d x+\left[\frac{\cos \frac{\pi}{2} x}{\frac{\pi}{2}}\right]_{0}^{1}\right]_{0}^{1}$
$=\left[x \frac{e^{2 x}}{2}\right]_{0}^{1}-\int_{0}^{1} \frac{e^{2 x}}{2} d x+\frac{2}{\pi}\left[\cos \frac{\pi x}{2}\right]_{0}^{1}$
$=\left[\frac{1 \cdot e^{2.1}}{2}-\frac{0 . e^{20}}{2}\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1} \frac{e^{2 x}}{2} d x+\frac{2}{\pi}\left[\cos \frac{\pi \times 1}{2}-\cos \frac{\pi \times 0}{2}\right]$
$=\left[\frac{e^{2}}{2}-0\right]-\frac{1}{2}\left[\frac{e^{2 x}}{2}\right]_{0}^{1}+\frac{2}{\pi}\left[\cos \frac{\pi}{2}-\cos 0\right]$
$=\left[\frac{e^{2}}{2}\right]-\frac{1}{4}\left[e^{2 \cdot 1}-e^{2.0}\right]+\frac{2}{\pi}[0-1]$
$=\left[\frac{e^{2}}{2}\right]-\frac{1}{4}\left[e^{2}-1\right]+\frac{2}{\pi}[-1]$
$=\left[\frac{e^{2}}{2}\right]-\frac{e^{2}}{4}+\frac{1}{4}-\frac{2}{\pi}$
$=\frac{2 e^{2}-e^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$
$=\frac{e^{2}}{4}+\frac{1}{4}+\frac{2}{\pi}$
Definite Integrals Exercise 19.1 Question 49
Answer:
Answer:$\int_{0}^{1}\left ( xe^{x}+\cos \frac{\pi x}{4} \right )dx$
Sol:$\int_{0}^{1}\left ( xe^{x}+\cos \frac{\pi x}{4} \right )dx=\int_{0}^{1} xe^{x}dx+\int_{0}^{1}\cos \frac{\pi x}{4}dx$
Applying integration by parts method in !st integral then
$=\left[x \int e^{x} d x\right]_{0}^{1}-\int_{0}^{1}\left(\frac{d(x)}{d x} \int e^{x} d x\right) d x+\left[\frac{\sin \frac{\pi}{4} x}{\frac{\pi}{4}}\right]_{0}^{1}$
$=\left[x e^{x}\right]_{0}^{1}-\int_{0}^{1} 1 . e^{x} d x+\frac{4}{\pi}\left[\sin \frac{\pi x}{4}\right]_{0}^{1}$
$=\left[1 . e^{1}-0 . e^{0}\right]-\left[e^{x}\right]_{0}^{1}+\frac{4}{\pi}\left[\sin \frac{\pi}{4}-\sin 0\right]$
$=[e-0]-\left[e^{1}-e^{0}\right]+\frac{4}{\pi}\left[\sin \frac{\pi}{4}-\sin 0\right]$
$= \left [ e \right ]-[e-1]+\frac{4}{\pi}\left[\frac{1}{\sqrt{2}}-0\right]$
$= e-e+1+\frac{4}{\pi} \times \frac{1}{\sqrt{2}}$
$=\frac{2\sqrt{2}}{\pi }+1$
Definite Integrals Exercise 19.1 Question 50
Answer:
Answer:$e\frac{\pi }{2}$Hint: Use indefinite integral formula then put the limits to solve this integral
Given:
$\int_{\frac{\pi }{2}}^{\pi }e^{x}\left ( \frac{1-\sin x}{1-\cos x} \right )dx$
Sol:
$\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x=\int_{\frac{\pi}{2}}^{\pi} e^{x} \frac{\left(1-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}\right)}{2 \sin ^{2} \frac{x}{2}} d x$
$\left[\begin{array}{l}\sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos 2 \theta=2 \sin ^{2} \theta\end{array}\right]$
$=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1}{2 \sin ^{2} \frac{x}{2}}-\frac{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}}\right) d x$
$\left[\frac{1}{\sin \theta}=\operatorname{cosec} \theta, \cot \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{\cos e c^{2} \frac{x}{2}}{2}-\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) d x$
$=\int_{\frac{\pi}{2}}^{\pi} \frac{e^{x} \cos e c^{2} \frac{x}{2}}{2}-\int_{\frac{\pi}{2}}^{\pi} e^{x} \cot \frac{x}{2} d x$
$=\frac{1}{2}\left[\int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2 \int_{\frac{\pi}{2}}^{\pi} e^{x} \cot \frac{x}{2} d x\right]$
$=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{x}{2} \int e^{x} d x\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\left(\frac{d\left(\cot \frac{x}{2}\right)}{d x} \int e^{x} d x\right) d x\right\}$
Using integration by parts in second term,
$=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{x}{2} e^{x}\right]_{\frac{\pi}{2}}^{\pi}-\int_{\frac{\pi}{2}}^{\pi}\left(-\cos e c^{2} \frac{x}{2}\right) \frac{1}{2} e^{x} d x\right\}$
$\left[\frac{d}{d x} \cot a x=-\cos e c^{2} a x \cdot a, \int e^{x} d x=e^{x}\right]$
$=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-2\left\{\left[\cot \frac{\pi}{2} \cdot e^{\pi}-\cot \frac{\pi}{4} e^{\frac{\pi}{2}}\right]+\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \cos e c^{2} \frac{x}{2} e^{x} d x\right\}$
$=\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} e^{x} \cos e c^{2} \frac{x}{2} d x-1\left(0 . e^{\pi}-1 \cdot e^{\frac{\pi}{2}}\right)-\frac{1}{2} \int_{\frac{\pi}{2}}^{\pi} \cos e c^{2} \frac{x}{2} e^{x} d x$
$\left[\cot \frac{\pi}{2}=0, \cot \frac{\pi}{4}=1\right]$
$=-\left(0-e^{\frac{\pi}{2}}\right)$
$=e^{\frac{\pi}{2}}$
Definite Integrals Exercise 19.1 Question 51
Answer:0Hint: Use indefinite formula then put the limits to solve this integral
Given:
$\int_{0}^{2\pi }e^\frac{x}{2}\sin \left ( \frac{x}{2}+\frac{\pi }{4} \right )dx$
Sol:
$\int_{0}^{2\pi }e^\frac{x}{2}\sin \left ( \frac{x}{2}+\frac{\pi }{4} \right )dx$
$=\int_{0}^{2 \pi} e^{\frac{x}{2}}\left[\sin \frac{\pi}{4} \cos \frac{x}{2}+\cos \frac{\pi}{4} \sin \frac{x}{2}\right] d x$
$[\because \sin (A+B)=\sin A \cos B+\cos A \sin B]$
$=\int_{0}^{2 \pi} e^{\frac{x}{2}}\left[\frac{1}{\sqrt{2}} \cos \frac{x}{2}+\frac{1}{\sqrt{2}} \sin \frac{x}{2}\right] d x$
$=\frac{1}{\sqrt{2}} \int_{0}^{2 \pi}\left[\cos \frac{x}{2} e^{\frac{x}{2}}+\sin \frac{x}{2} e^{\frac{x}{2}}\right] d x$
$=\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} \cos \frac{x}{2} e^{\frac{x}{2}} d x+\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} \sin \frac{x}{2} e^{\frac{x}{2}} d x$
$=\frac{1}{\sqrt{2}}\left\{\int_{0}^{2 \pi} e^{\frac{x}{2}} \sin \frac{x}{2} d x\right\}+\frac{1}{\sqrt{2}}\left\{\int_{0}^{2 \pi} e^{\frac{x}{2}} \cos \frac{x}{2} d x\right\}$
$=\frac{1}{\sqrt{2}}\left\{\left[\sin \frac{x}{2} \frac{e^{\frac{x}{2}}}{\frac{1}{2}}\right]_{0}^{2 \pi}-\int_{0}^{2 \pi} \cos \frac{x}{2} \cdot \frac{1}{2} \frac{e^{\frac{x}{2}}}{\frac{1}{2}} d x\right\}+\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} e^{\frac{x}{2}} \cos \frac{x}{2} d x$
[using integration by parts]
$=\frac{1}{\sqrt{2}}\left[2 \sin \frac{x}{2} e^{\frac{x}{2}}\right]_{0}^{2 \pi}-\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} e^{\frac{x}{2}} \cos \frac{x}{2} d x+\frac{1}{\sqrt{2}} \int_{0}^{2 \pi} e^{\frac{x}{2}} \cos \frac{x}{2} d x$
$=\frac{1}{\sqrt{2}} \times 2\left[\sin \frac{2 \pi}{2} \cdot e^{\frac{2 \pi}{2}}-\sin \frac{0}{2} \cdot e^{\frac{0}{2}}\right]$
$=\sqrt{2}\left[\sin \pi \cdot e^{\pi}-\sin 0 \cdot e^{0}\right]$
$=\sqrt{2}\left [ 0-0 \right ]$
$[\because \sin \pi=\sin 0=0]$
$=0$
Definite Integrals Exercise 19.1 Question 52
Answer:
Answer:$\frac{-3\sqrt{2}}{5}\left ( e^{2\pi}+1 \right )$Hint: Use indefinite formula then put the limit to solve this integral
Given:
$\int_{0}^{2\pi }e^{x}\cos \left ( \frac{\pi }{4}+\frac{x}{2} \right )dx$
Solution:
$I=\int_{0}^{2\pi }e^{x}\cos \left ( \frac{\pi }{4}+\frac{x}{2} \right )dx$ ....................(1)
Apply integration by parts method, then
$\int_{0}^{2 \pi} e^{x} \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) d x=\left[\cos \left(\frac{\pi}{4}+\frac{x}{2}\right) \int e^{x} d x\right]_{0}^{2 \pi}-\int_{0}^{2 \pi}\left\{\frac{d}{d x} \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) \int e^{x} d x\right\} d x$
$\Rightarrow I=\left[\cos \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]_{0}^{2 \pi}-\int_{0}^{2 \pi}-\sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \frac{1}{2} e^{x} d x \quad\left[\int e^{x} d x=e^{x} \frac{d \cos (a x)}{d x}=-\sin a x \cdot a\right]$
$\Rightarrow I=\left[e^{2 \pi} \cos \left(\frac{\pi}{4}+\frac{2 \pi}{2}\right)-e^{0} \cos \left(\frac{\pi}{4}+\frac{0}{2}\right)\right]+\frac{1}{2} \int_{0}^{2 \pi} \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) e^{x} d x$
$\Rightarrow I=\left[e^{2 \pi} \cos \left(\frac{\pi}{4}+\pi\right)-\cos \frac{\pi}{4}\right]+\frac{1}{2}\left[\left\{\sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \int e^{x} d x\right\}_{0}^{2 \pi}-\int_{0}^{2 \pi}\left\{\frac{d}{d x} \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \int e^{x} d x\right\} d x\right]$
$\Rightarrow I=\left[e^{2 \pi}\left(-\cos \frac{\pi}{4}\right)-\cos \left(\frac{\pi}{4}\right)\right]+\frac{1}{2}\left\{\left[\sin \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]_{0}^{2 \pi}-\int_{0}^{2 \pi} \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) \frac{1}{2} e^{x} d x\right\}$
(Again using integrating by parts method)
$\left[\frac{d}{d x} \sin a x=\cos a x \cdot a, \int e^{x}=\frac{e^{x}}{2 \pi}\right]$
$\Rightarrow I=\left[e^{2 \pi}\left(\frac{-1}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}}\right]+\frac{1}{2}\left[\sin \left(\frac{\pi}{4}+\frac{2 \pi}{2}\right) e^{2 \pi} \sin \left(\frac{\pi}{4}+\frac{0}{2}\right) e^{0}\right]-\frac{1}{4} \int_{0}^{2 \pi} \cos \left(\frac{\pi}{4}+\frac{x}{2}\right) e^{x} d x$
$\left [ \cos \frac{\pi }{4}=\frac{1}{\sqrt{2}} \right ]$
$\Rightarrow I=\left[\frac{-e^{2 \pi}}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right]+\frac{1}{2}\left[\sin \left(\pi+\frac{\pi}{4}\right) e^{2 \pi}-\sin \frac{\pi}{4}\right]-\frac{1}{4} I$
$\Rightarrow I=\frac{-1}{\sqrt{2}}\left(e^{2 \pi}+1\right)+\frac{1}{2}\left(-\sin \frac{\pi}{4} e^{2 \pi}-\sin \frac{\pi}{4}\right)-\frac{1}{4} I$ $\left [ \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right ]$
$\Rightarrow I=\frac{-1}{\sqrt{2}}\left(e^{2 \pi}+1\right)-\frac{1}{2}\left[\frac{1}{\sqrt{2}} e^{2 \pi}+\frac{1}{\sqrt{2}}\right]-\frac{1}{4} I$
$\Rightarrow I+\frac{1}{4} I=\frac{-1}{\sqrt{2}}\left(e^{2 \pi}+1\right)-\frac{1}{2 \sqrt{2}} e^{2 \pi}-\frac{1}{2 \sqrt{2}}$
$\Rightarrow \frac{4 I+I}{4}=\frac{-1}{\sqrt{2}}\left(e^{2 \pi}+1\right)-\frac{1}{2 \sqrt{2}}\left(e^{2 \pi}+1\right)$
$\Rightarrow \frac{5 I}{4}=\left(e^{2 \pi}+1\right)\left[\frac{-1}{\sqrt{2}}-\frac{1}{2 \sqrt{2}}\right]=\left(e^{2 \pi}+1\right)\left[\frac{-2-1}{2 \sqrt{2}}\right]$
$\Rightarrow I=\frac{4}{5} \frac{-3}{2 \sqrt{2}}\left(e^{2 \pi}+1\right)=\frac{2 \times \sqrt{2} \times \sqrt{2}}{5} \frac{-3}{2 \sqrt{2}}\left(e^{2 \pi}+1\right)$
$\Rightarrow I=\frac{-3 \sqrt{2}}{5}\left(e^{2 \pi}+1\right)$
Definite Integrals Exercise 19.1 Question 53
Answer:
Answer: $\frac{-1}{5\sqrt{2}}\left ( e^{2\pi }+1 \right )$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\pi} e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x$
Solution: Let$I=\int_{0}^{\pi} e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x$ ...............................(1)
Apply integration by parts method
$\qquad I=\left[\sin \left(\frac{\pi}{4}+x\right) \int \frac{e^{2 x}}{2}\right]_{0}^{\pi}-\int_{0}^{\pi}\left\{\frac{d}{d x}\left(\sin \left(\frac{\pi}{4}+x\right)\right) \int e^{2 x} d x\right\} d x$
$\Rightarrow I=\left[\sin \left(\frac{\pi}{4}+x\right) \frac{e^{2 x}}{2}\right]_{0}^{\pi}-\int_{0}^{\pi} \cos \left(\frac{\pi}{4}+x\right) \frac{e^{2 x}}{2} d x \quad\left[\begin{array}{l} \frac{d}{d x} \sin a x=\cos a x \cdot a \\ \int e^{a x} d x=\frac{e^{a x}}{a} \end{array}\right]$
$\Rightarrow I=\frac{1}{2}\left[\sin \left(\frac{\pi}{4}+x\right) e^{2 x}\right]_{0}^{\pi}-\frac{1}{2} \int_{0}^{\pi} \cos \left(\frac{\pi}{4}+x\right) e^{2 x} d x$
$\Rightarrow I=\frac{1}{2}\left[\sin \left(\frac{\pi}{4}+\pi\right) e^{2 \pi}-\sin \left(\frac{\pi}{4}+0\right) e^{0}\right]$
$-\frac{1}{2}\left\{\left[\cos \left(\frac{\pi}{4}+x\right) \int e^{2 x} d x\right]_{0}^{\pi}-\int_{0}^{\pi}\left(\frac{d}{d x} \cos \left(\frac{\pi}{4}+x\right) \int e^{2 x} d x\right) d x\right\} \\$
$\Rightarrow I=\frac{1}{2}\left[-\sin \frac{\pi}{4} e^{2 \pi}-\sin \frac{\pi}{4}\right]-\frac{1}{2}\left\{\left[\cos \left(\frac{\pi}{4}+x\right) \frac{e^{2 x}}{2}\right]_{0}^{\pi}-\int_{0}^{\pi}-\sin \left(\frac{\pi}{4}+x\right) \frac{e^{2 x}}{2} d x\right\} \\$
$\left[\because \frac{d}{d x} \cos a x=-\sin a x . a, \int e^{a x} d x=\frac{e^{a x}}{a}, \sin (\pi+x)=-\sin x\right] \\$
$\Rightarrow I=\frac{1}{2}\left[-\frac{1}{\sqrt{2}} e^{2 \pi}-\frac{1}{\sqrt{2}}\right]-\frac{1}{2} \cdot \frac{1}{2}\left[\cos \left(\frac{\pi}{4}+\pi\right) e^{2 \pi}-\cos \left(\frac{\pi}{4}+0\right) e^{0}\right]-\frac{1}{4} \int_{0}^{\pi} \sin \left(\frac{\pi}{4}+x\right) e^{2 x} d x$
$\left [ \because \sin \frac{\pi }{4}=\frac{1}{\sqrt{2}} \right ]$
$\begin{aligned} &\Rightarrow I=\frac{1}{2} \cdot \frac{-1}{\sqrt{2}}\left[e^{2 \pi}+1\right]-\frac{1}{4}\left[-\cos \frac{\pi}{4} e^{2 \pi}-\cos \frac{\pi}{4}\right]-\frac{1}{4} I \quad[\because \cos (\pi+x)=-\cos x] \\ &\Rightarrow I+\frac{1}{4} I=\frac{-1}{2 \sqrt{2}}\left[e^{2 \pi}+1\right]+\frac{1}{4}\left[\frac{1}{\sqrt{2}} e^{2 \pi}+\frac{1}{\sqrt{2}}\right] \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \\ &\Rightarrow \frac{4 I+I}{4}=\frac{-1}{2 \sqrt{2}}\left[e^{2 \pi}+1\right]+\frac{1}{4} \frac{1}{\sqrt{2}}\left(e^{2 \pi}+1\right) \\ &\Rightarrow \frac{5 I}{4}=\left[e^{2 \pi}+1\right]\left(\frac{1}{4 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right)=\left[e^{2 \pi}+1\right]\left(\frac{1-2}{4 \sqrt{2}}\right) \\ &\Rightarrow I=\frac{4}{5}\left(\frac{-1}{4 \sqrt{2}}\right)\left(e^{2 \pi}+1\right)=\frac{-1}{5 \sqrt{2}}\left(e^{2 \pi}+1\right) \end{aligned}$
Definite Integrals Exercise 19.1 Question 54
Answer:$\frac{4\sqrt{2}}{3}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} d x$
Solution:
Let,$I=\int_{0}^{1} \frac{1}{\sqrt{1+x}-\sqrt{x}} d x$
$\begin{aligned} &=\int_{0}^{1}\left(\frac{1}{\sqrt{1+x}-\sqrt{x}} \times \frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}\right) d x \\ &=\int_{0}^{1}\left(\frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x}-\sqrt{x})(\sqrt{1+x}+\sqrt{x})}\right) d x \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right] \\ &=\int_{0}^{1}\left(\frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x})^{2}-(\sqrt{x})^{2}}\right) d x \\ &=\int_{0}^{1}\left(\frac{\sqrt{1+x}+\sqrt{x}}{1+x-x}\right) d x \\ &=\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1} d x \\ &=\int_{0}^{1} \sqrt{1+x} d x+\int_{0}^{1} \sqrt{x} d x \end{aligned}$
Put $1+x=t\Rightarrow dx=dt$
$\begin{aligned} &I=\int_{0}^{1} \sqrt{t} d t+\int_{0}^{1} \sqrt{x} d x \\ &=\int_{0}^{1} t^{\frac{1}{2}} d t+\int_{0}^{1} x^{\frac{1}{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=\left[\frac{t^{\frac{1}{2}}+1}{\frac{1}{2}+1}\right]_{0}^{1}+\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{1} \end{aligned}$
$\begin{aligned} &=\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}+\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[(1+x)^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{3}\left[x^{\frac{3}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[(1+1)^{\frac{3}{2}}-(1+0)^{\frac{3}{2}}\right]_{0}^{1}+\frac{2}{3}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]_{0}^{1} \\ &=\frac{2}{3}\left[2^{\frac{3}{2}}-1^{\frac{3}{2}}\right]+\frac{2}{3}\left[1^{\frac{3}{2}}\right] \\ &=\frac{2}{3}\left[\sqrt{2}^{2 \frac{3}{2}}-1+1\right] \end{aligned}$
$=\frac{2}{3}\left ( \sqrt{2} \right )^{2}$
$=\frac{2}{3}2\sqrt{2}$
$=\frac{4\sqrt{2}}{3}$
Definite Integrals Exercise 19.1 Question 55
Answer:
Answer: $log\left ( \frac{32}{27} \right )$Hint: Use indefinite formula then put the limit to solve this integral
Given:$\int_{1}^{2} \frac{x}{(x+1)(x+2)} d x$
Solution:
$\int_{1}^{2} \frac{x}{(x+1)(x+2)} d x$ ..................(1)
To solve this integral, first we will final its partial fraction then integrate it with the given limits.
So,$\frac{x}{\left ( x+1 \right )\left ( x+2 \right )}=\frac{A}{x+1}+\frac{B}{x+2}$ ............................(2)
$\begin{aligned} &x=\frac{A(x+1)(x+2)}{(x+1)}+\frac{B(x+1)(x+2)}{(x+2)} \\ &\Rightarrow x=A(x+2)+B(x+1) \\ &\Rightarrow x=A x+2 A+B x+B \\ &\Rightarrow x=(A+B) x+2 A+B \end{aligned}$
Equating coefficient of x from both sides,
$1=A+B$ ........................(a)
And again equating coefficients of constant term from both sides, then
$0=2A+B\Rightarrow 2A=-B\Rightarrow B=-2A$ ........................(b)
Put the value of A from (b) in (a) then
$\begin{gathered} 1=A+(-2 A)=A-2 A=-A \\ A=-1 \\ \Rightarrow B=-2 A=-2(-1)=2 \end{gathered}$
So$A=-1,B=2$
Then equation (2) becomes
$\frac{x}{\left ( x+1 \right )\left ( x+2 \right )}=\frac{-1}{x+1}+\frac{2}{x+2}$
Now equation (1)
$\Rightarrow \int_{1}^{2} \frac{x}{(x+1)(x+2)} d x=\int_{1}^{2} \frac{-1}{x+1} d x+2 \int_{1}^{2} \frac{1}{x+2} d x$ ..................(3)
$\int_{1}^{2}\frac{1}{x+1}dx$
Putting $x+1=t\Rightarrow dx=dt$
When $x=1$ then $t=1+1=2$
And When $x=2$ then $t=2+1=3$
Then $\left.\left.\int_{1}^{2} \frac{1}{x+1} d x=\int_{1}^{2} \frac{1}{t} d t=[\log \mid t]\right]_{2}^{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d x=\log \mid x\right]\right]$
$=\left [ log\left | 3 \right |-log\left | 2 \right | \right ]$
$=\left [ log3-log2 \right ]$ .......................(4)
and $\int_{1}^{2}\frac{1}{x+1}dx$
Putting $x+2=u\Rightarrow dx=du$
When $x=1$ then $u=2+1=3$
And $x=2$ then $u=2+2=4$
Then$\begin{aligned} \int_{1}^{2} \frac{1}{x+2} d x &=\int_{3}^{4} \frac{1}{u} d u=[\log |u|]_{3}^{4} \\ &=\log |4|-\log |3|=\log 4-\log 3 \end{aligned}$ ......(5)
$\begin{aligned} &\int_{1}^{2} \frac{x}{(x+1)(x+2)} d x=-\int_{1}^{2} \frac{1}{x+1} d x+2 \int_{1}^{2} \frac{1}{x+2} d x \\ &=-[\log 3-\log 2]+2[\log 4-\log 3] \\ &=-\log 3+\log 2+2 \log 2^{2}-2 \log 3 \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\log a^{m}=m \log a\right] \\ &=-3 \log 3+\log 2+4 \log 2 \end{aligned}$
$\begin{aligned} &=-3 \log 3+5 \log 2\\ &\begin{aligned} &=5 \log 2-3 \log 3 \\ &=\log 2^{5}-\log 3^{3} \end{aligned}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\log a-\log b=\log \frac{a}{b}\right]\\ &=\log 32-\log 27 \end{aligned}$
$=log\left ( \frac{32}{27} \right )$
Definite Integrals Exercise 19.1 Question 56
Answer:
Answer: $\frac{2}{3}$Hint: Use indefinite formula then put the limit to solve this integral
Given:$\int_{0}^{\frac{\pi }{2}}\sin ^{3}xdx$
Solution: Let,$I=\int_{0}^{\frac{\pi }{2}}\sin ^{3}xdx=\int_{0}^{\frac{\pi }{2}}\sin ^{2}x\sin xdx$
$=\int_{0}^{\frac{\pi}{2}}\left(1-\cos ^{2} x\right) \sin x d x$ $\left[\sin ^{2} x=1-\cos ^{2} x\right]$
$=\int_{0}^{\frac{\pi}{2}}\left(\sin x-\cos ^{2} x \sin x\right) d x$
$=\int_{0}^{\frac{\pi}{2}} \sin x d x-\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \sin x d x$
Put$t=\cos x \Rightarrow d t=-\sin x d x \Rightarrow d x=\frac{-d t}{\sin x}$ in the 2nd Integral term
$=\int_{0}^{\frac{\pi}{2}} \sin x d x-\int_{0}^{\frac{\pi}{2}} t^{2} \sin x \frac{d t}{-\sin x}$
$=\int_{0}^{\frac{\pi}{2}} \sin x d x+\int_{0}^{\frac{\pi}{2}} t^{2} d t$ $\quad\left[\int \sin x d x=-\cos x, \int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$
$=[-\cos x]_{0}^{\frac{\pi}{2}}+\left[\frac{t^{2+1}}{2+1}\right]_{0}^{\frac{\pi}{2}}$
$=-\left[\cos \frac{\pi}{2}-\cos 0\right]+\frac{1}{3}\left[t^{3}\right]_{0}^{\frac{\pi}{2}}$
$=-\left[\cos \frac{\pi}{2}-\cos 0\right]+\frac{1}{3}\left[\cos ^{3} x\right]_{0}^{\frac{\pi}{2}}$ $\left [ \cos \frac{\pi }{2}=0,\cos 0=1 \right ]$
$=-[0-1]+\frac{1}{3}\left[\cos ^{3} \frac{\pi}{2}-\cos ^{3} 0\right]_{0}^{\frac{\pi}{2}}$
$=1+\frac{1}{3}\left[0^{3}-1^{3}\right]=1+\frac{1}{3}[0-1]$
$=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$
Definite Integrals Exercise 19.1 Question 57
Answer: 0Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\pi }\left ( \sin ^{2}\frac{x}{2}-\cos ^{2}\frac{x}{2} \right )dx$
Solution: $\int_{0}^{\pi }\left ( \sin ^{2}\frac{x}{2}-\cos ^{2}\frac{x}{2} \right )dx=-\int_{0}^{\pi }\left ( \cos ^{2}\frac{x}{2}-\sin ^{2}\frac{x}{2}\right )dx$
$=-\int_{0}^{\pi} \cos 2 \times \frac{x}{2} d x$ $\quad\left[\cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta\right] \\$
$=-\int_{0}^{\pi} \cos x d x \\$
$=-[\sin x]_{0}^{\pi}$ $\quad\left[\int \cos x d x=\sin x\right] \\$
$=-[\sin \pi-\sin 0]$ $\quad[\sin \pi=\sin 0=0] \\$
$=-1[0-0]$
$=0$
Definite Integrals Exercise 19.1 Question 58
Answer:
Answer: $\frac{e^{4}-2e^{2}}{4}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{1}^{2}e^{2x}\left ( \frac{1}{x}-\frac{1}{2x^{2}} \right )dx$
Solution:
$\begin{aligned} &\int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x=2 \int_{1}^{2} e^{2 x}\left(\frac{1}{2 x}-\frac{1}{2 \times 2 x^{2}}\right) d x \\ &\text { Put } 2 x=t \Rightarrow 2 d x=d t \Rightarrow d x=\frac{d t}{2} \end{aligned}$
When $x=1$ then $t=2$ and
When $x=2$ then $t=4$
Then $\int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x=2 \int_{1}^{2} e^{2 x}\left(\frac{1}{2 x}-\frac{1}{4 x^{2}}\right) d x$$\begin{aligned} &=2 \int_{2}^{4} e^{t}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) \frac{d t}{2} \\ &=\int_{2}^{4} e^{t}\left(\frac{1}{t}-\frac{1}{t^{2}}\right) d t \\ &=\int_{2}^{4} e^{t} \frac{1}{t} d t-\int_{2}^{4} e^{t} \frac{1}{t^{2}} d t \end{aligned}$
Apply integration by parts in 1st integral, then
$\int_{1}^{2} e^{2 x}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) d x=\left[\frac{1}{t} \int e^{t} d t\right]_{2}^{4}-\int_{2}^{4}\left(\frac{d}{d t}\left(\frac{1}{t}\right) \int e^{t} d t\right) d t-\int_{2}^{4} e^{t} \frac{1}{t^{2}} d t$
$=\left[\frac{1}{t} e^{t}\right]_{2}^{4}-\int_{2}^{4}\left(\frac{-1}{t^{2}} e^{t}\right) d t-\int_{2}^{4} e^{t} \frac{1}{t^{2}} d t$ $\quad\left[\begin{array}{l} \frac{d}{d x}\left(\frac{1}{x}\right)=\frac{d\left(x^{-1}\right)}{d x}=-1 x^{-1-1}=-1 \cdot x^{-2}=\frac{-1}{x^{2}} \\ \int e^{x} d x=e^{x} \end{array}\right] \\$
$=\left[\frac{1}{4} e^{4}-\frac{1}{2} e^{2}\right]+\int_{2}^{4} \frac{1}{t^{2}} e^{t} d t-\int_{2}^{4} \frac{1}{t^{2}} e^{t} d t$
$=\frac{1}{4}\left(e^{4}-2 e^{2}\right)=\frac{e^{4}-2 e^{2}}{4}$
Definite Integrals Exercise 19.1 Question 59
Answer: $\pi$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{1}^{2}\frac{1}{\sqrt{\left ( x-1 \right )\left ( 2-x \right )}}dx$
Solution:
$\int_{1}^{2} \frac{1}{\sqrt{(x-1)(2-x)}} d x=\int_{1}^{2} \frac{1}{\sqrt{2 x-x^{2}-2+x}} d x=\int_{1}^{2} \frac{1}{\sqrt{3 x-x^{2}-2}} d x\\$
$=\int_{1}^{2} \frac{1}{\sqrt{-\left(x^{2}-3 x+2\right)}} d x$
$=\int_{1}^{2} \frac{1}{\sqrt{-\left(x^{2}-2 x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+2\right)}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[a^{2}+b^{2}-2 a b=(a-b)^{2}\right]$
$=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}+2\right)}} d x$
$\begin{aligned} &=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\left(\frac{9-8}{4}\right)\right)}} d x \\ &=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{4}\right)\right)}} d x \\ &=\int_{1}^{2} \frac{1}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}} d x \end{aligned}$
Putting $\left ( x-\frac{3}{2} \right )=t\Rightarrow dx=dt$
When $x=1$ then $t=1-\frac{3}{2}=\frac{2-3}{2}=\frac{-1}{2}$
And When $x=2$ then $t=2-\frac{3}{2}=\frac{4-3}{2}=\frac{1}{2}$
Then $\int_{1}^{2} \frac{1}{\sqrt{(x-1)(2-x)}} d x=\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{\sqrt{\left(\frac{1}{2}\right)^{2}-t^{2}}} d t$
$=\left[\sin ^{-1}\left(\frac{t}{\frac{1}{2}}\right)\right]_{\frac{-1}{2}}^{\frac{1}{2}}$ $\left[\int \frac{1}{a^{2}-x^{2}} d x=\sin ^{-1} \frac{x}{a}\right]$
$=\left[\sin ^{-1}(2 t)\right] \frac{-\frac{1}{2}}{2}$
$=\left[\sin ^{-1}\left(2 \times \frac{1}{2}\right)-\sin ^{-1}\left(2 \times \frac{-1}{2}\right)\right]$ $[\sin (-\theta)=-\sin \theta]$
$=\sin ^{-1}(1)-\sin ^{-1}(-1)$
$=\sin ^{-1}(1)+\sin ^{-1}(1) \\$
$=2 \sin ^{-1}(1)$ $\left [ \sin ^{-1}=\frac{\pi }{2} \right ]$
$=2\times \frac{\pi }{2}$
$=\pi$
Definite Integrals Exercise 19.1 Question 60
Answer: $k=\frac{1}{2}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{k} \frac{1}{2+8 x^{2}} d x=\frac{\pi}{16}$
Solution: $\int_{0}^{k} \frac{1}{2+8 x^{2}} d x=\frac{\pi}{16}$
$\begin{aligned} &\Rightarrow \int_{0}^{k} \frac{1}{8\left(x^{2}+\frac{2}{8}\right)} d x=\frac{\pi}{16} \\ &\Rightarrow \frac{1}{8} \int_{0}^{k} \frac{1}{x^{2}+\frac{1}{4}} d x=\frac{\pi}{16} \\ &\Rightarrow \frac{1}{8} \int_{0}^{k} \frac{1}{x^{2}+\left(\frac{1}{2}\right)^{2}} d x=\frac{\pi}{16} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \\ &\Rightarrow \frac{1}{8}\left[\frac{1}{\frac{1}{2}} \tan ^{-1}\left(\frac{x}{\frac{1}{2}}\right)\right]_{0}^{k}=\frac{\pi}{16} \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{1}{8} \times 2\left[\tan ^{-1} 2 x\right]_{0}^{k}=\frac{\pi}{16} \\ &\Rightarrow \frac{1}{4}\left[\tan ^{-1} 2 k-0\right]=\frac{\pi}{16} \\ &\Rightarrow \tan ^{-1} 2 k=\frac{4 \pi}{16} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan \frac{\pi}{4}=1\right] \\ &\Rightarrow \tan ^{-1} 2 k=\frac{\pi}{4} \\ &\Rightarrow 2 k=\tan \frac{\pi}{4}=1 \\ &k=\frac{1}{2} \end{aligned}$
Definite Integrals Exercise 19.1 Question 61
Answer: $a=2$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{a}3x^{2}dx=8$
Solution: $\int_{0}^{a}3x^{2}dx=8$
$\begin{aligned} &\Rightarrow 3 \int_{0}^{a} x^{2} d x=8 \Rightarrow 3\left[\frac{x^{2+1}}{2+1}\right]_{0}^{a}=8 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &\Rightarrow 3\left[\frac{x^{3}}{3}\right]_{0}^{a}=8 \\ &\Rightarrow\left[x^{3}\right]_{0}^{a}=8 \Rightarrow\left[a^{3}-0^{3}\right]=8 \\ &\Rightarrow a^{3}=8 \\ &\Rightarrow a=2^{3} \\ &\Rightarrow a=2 \end{aligned}$
Definite Integrals Exercise 19.1 Question 62
Answer: $-\sqrt{2}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{\pi }^{\frac{3\pi }{2}}\sqrt{1-\cos 2xdx}$
Solution: $\int_{\pi}^{\frac{3 \pi}{2}} \sqrt{1-\cos 2 x} d x=\int_{\pi}^{\frac{3 \pi}{2}} \sqrt{2 \sin ^{2} x} d x \; \; \; \; \; \; \; \: \: \: \: \quad\left[1-\cos 2 \theta=2 \sin ^{2} \theta\right]$
$\begin{aligned} &=\int_{\pi}^{\frac{3 \pi}{2}} \sqrt{2} \sin x d x=\sqrt{2} \int_{\pi}^{\frac{3 \pi}{2}} \sin x d x \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \sin x d x=-\cos x\right] \\ &=\sqrt{2}[-\cos x]_{\pi}^{\frac{3 \pi}{2}} \\ &=-\sqrt{2}\left[\cos \frac{3 \pi}{2}-\cos \pi\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\cos \left(\pi+\frac{\pi}{2}\right)=-\cos \frac{\pi}{2}\right] \\ &=-\sqrt{2}\left[\cos \left(\pi+\frac{\pi}{2}\right)-\cos \pi\right] \\ &=-\sqrt{2}\left[-\cos \frac{\pi}{2}-\cos \pi\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{c} \cos \frac{\pi}{2}=0 \\ \cos \pi=-1 \end{array}\right] \\ &=-\sqrt{2}[-0-1) \\ &=-\sqrt{2} \times 1 \\ &=-\sqrt{2} \end{aligned}$
Definite Integrals Exercise 19.1 Question 63
Answer: 8Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{2 \pi} \sqrt{1+\sin \frac{x}{2}} d x$
Solution: $\int_{0}^{2 \pi} \sqrt{1+\sin \frac{x}{2}} d x=\int_{0}^{2 \pi} \sqrt{\cos ^{2} \frac{x}{4}+\sin ^{2} \frac{x}{4}+2 \sin \frac{x}{4} \cos \frac{x}{4}} d x$
${\left[1=\cos ^{2} \theta+\sin ^{2} \theta, \sin 2 \theta=2 \sin \theta \cos \theta\right]}$
$=\int_{0}^{2 \pi} \sqrt{\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)^{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[a^{2}+b^{2}+2 a b=(a+b)^{2}\right] \\$
$=\int_{0}^{2 \pi}\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right) d x$ $\left[\begin{array}{l} \int \sin a x d x=\frac{-\cos a x}{a} \\ \int \cos a x d x=\frac{\sin a x}{a} \end{array}\right]$
$=\int_{0}^{2_{\pi}} \cos \frac{x}{4} d x+\int_{0}^{2 \pi} \sin \frac{x}{4} d x$
$=\left[\frac{\sin \frac{x}{4}}{\frac{1}{4}}\right]_{0}^{2 \pi}-\left[\frac{\cos \frac{x}{4}}{\frac{1}{4}}\right]_{0}^{2 \pi}$
$=4\left[\sin \frac{2 \pi}{4}-\sin 0\right]-4\left[\cos \frac{2 \pi}{4}-\cos 0\right]$ $\left[\begin{array}{l} \sin \frac{\pi}{2}=\cos 0=1 \\ \sin 0=\cos \frac{\pi}{2}=0 \end{array}\right]$
$=4\left[\sin \frac{\pi}{2}-\sin 0\right]-4\left[\cos \frac{\pi}{2}-\cos 0\right]$
$=4[1-0]-4[0-1]$
$=4+4$
$=8$
Definite Integrals Exercise 19.1 Question 64
Answer: $\frac{\pi }{32}$Hint: Use indefinite formula then put the limit to solve this integral
Given:$\int_{0}^{\frac{\pi}{4}}(\tan x+\cot x)^{-2} d x$
Solution:
$\begin{aligned} &\int_{0}^{\frac{\pi}{4}}(\tan x+\cot x)^{-2} d x=\int_{0}^{\frac{\pi}{4}} \frac{1}{(\tan x+\cot x)^{2}} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)^{2}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}, \cot \theta=\frac{\cos \theta}{\sin \theta}\right] \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right)^{2}} d x \end{aligned}$
$=\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin x \cos x}{\sin ^{2} x+\cos ^{2} x}\right)^{2} d x$ $\left[\sin ^{2} x+\cos ^{2} x=1\right] \\$
$=\int_{0}^{\frac{\pi}{4}} \frac{(\sin x \cos x)^{2}}{1} d x \\$
$=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x \cos ^{2} x d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \$ $\left[1-\sin ^{2} \theta =\cos ^{2} \theta \right] \\$
$=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x\left(1-\sin ^{2} x\right) d x \\$ $\left[\begin{array}{l} 2 \sin ^{2} \theta=1-\cos 2 \theta \\ \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2} \end{array}\right]$
$=\int_{0}^{\frac{\pi}{4}} \sin ^{2} x d x-\int_{0}^{\frac{\pi}{4}} \sin ^{4} x d x$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2} d x-\int_{0}^{\frac{\pi}{4}}\left(\sin ^{2} x\right)^{2} d x \\ &=\int_{0}^{\frac{\pi}{4}} \frac{1}{2}(1-\cos 2 x) d x-\int_{0}^{\frac{\pi}{4}}\left(\frac{1-\cos 2 x}{2}\right)^{2} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}}(1-\cos 2 x) d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}(1-\cos 2 x)^{2} d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}\left(1+\cos ^{2} 2 x-2 \cos 2 x\right) d x \\ &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}} \cos ^{2} 2 x d x-\frac{2}{4} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x \\ &=\left(\frac{1}{2}-\frac{1}{4}\right) \int_{0}^{\frac{\pi}{4}} 1 d x-\left(\frac{1}{2}+\frac{1}{2}\right) \int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{4} \int_{0}^{\frac{\pi}{4}}\left(\frac{1+\cos 4 x}{2}\right) d x \end{aligned}$
$\left [ 1+\cos 2\theta =2\cos ^{2}\theta \right ]$
$\begin{aligned} &=\left(\frac{2-1}{4}\right) \int_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\frac{1}{4} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\left(\frac{1}{4}-\frac{1}{8}\right)_{0}^{\frac{\pi}{4}} 1 d x-\int_{0}^{\frac{\pi}{4}} \cos 2 x d x-\frac{1}{8} \int_{0}^{\frac{\pi}{4}} \cos 4 x d x \\ &=\left(\frac{2-1}{8}\right)[x]_{0}^{\frac{\pi}{4}}-\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{4}}-\frac{1}{8}\left[\frac{\sin 4 x}{4}\right]_{0}^{\frac{\pi}{4}} \end{aligned}$
$\left [ \because \int 1dx=x,\int \cos a\: xdx=\frac{\sin ax}{a} \right ]$
$=\frac{1}{8}\left(\frac{\pi}{4}-0\right)-\frac{1}{2}\left[\sin \frac{2 \pi}{4}-\sin 0\right]-\frac{1}{8}\left[\sin \frac{4 \pi}{4}-\sin 0\right]$
$=\frac{1}{8} \frac{\pi}{4}-\frac{1}{2}\left[\sin \frac{\pi}{2}-0\right]-\frac{1}{8}[\sin \pi-\sin 0]$
$\left [ \because \sin 0=0,\sin \pi =0 \right ]$
$=\frac{\pi}{32}-\frac{1}{2}[0-0]-\frac{1}{8}[0-0]$
$=\frac{\pi}{32}$
Definite Integrals Exercise 19.1 Question 65
Answer: $\frac{3}{8}log 3$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{1}xlog\left ( 1+2x \right )dx$
Solution: $\int_{0}^{1}xlog\left ( 1+2x \right )dx$
Apply integration by parts, we get
$=\left[\log (1+2 x) \int x d x\right]_{0}^{1}-\int_{0}^{1}\left\{\frac{d}{d x} \log (1+2 x) \int x d x\right\} d x$ $\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \frac{d}{d x}(\log (a x+b))=\frac{1}{a x+b} a \end{array}\right]$
$=\left[\log (1+2 x) \frac{x^{1+1}}{1+1}\right]_{0}^{1}-\int_{0}^{1}\left(\frac{1}{1+2 x} 2 \frac{x^{1+1}}{1+1}\right) d x$
$\begin{aligned} &=\left[\log (1+2 x) \frac{x^{2}}{2}\right]_{0}^{1}-2 \int_{0}^{1} \frac{1}{2 x+1} \frac{x^{2}}{2} d x \\ &=\left[\log (1+2(1)) \frac{1^{2}}{2}-\log (1+2(0)) \frac{0^{2}}{2}\right]-\frac{1}{2} \int_{0}^{1} \frac{2 x^{2}}{2 x+1} d x \\ &=\left[\frac{\log 3}{2}-\log 1 \times 0\right]-\frac{1}{2} \int_{0}^{1} \frac{2 x^{2}+x-x}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} \frac{x(2 x+1)-x}{2 x+1} d x \end{aligned}$
$\begin{aligned} &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1}\left(\frac{x(2 x+1)}{2 x+1}-\frac{x}{2 x+1}\right) d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} x d x+\frac{1}{2} \int_{0}^{1} \frac{2 x+1-1}{2(2 x+1)} d x \\ &=\frac{\log 3}{2}-\frac{1}{2} \int_{0}^{1} x d x+\frac{1}{2} \int_{0}^{1}\left(\frac{2 x+1}{(2 x+1)}-\frac{1}{2 x+1}\right) d x \\ &=\frac{\log 3}{2}-\frac{1}{2}\left[\frac{x^{1+1}}{1+1}\right]_{0}^{1}+\frac{1}{4} \int_{0}^{1} 1 d x-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \end{aligned}$ $\left [ x^{n}dx=\frac{x^{n+1}}{n+1} \right ]$
$\begin{aligned} &=\frac{\log 3}{2}-\frac{1}{2}\left[\frac{x^{2}}{2}\right]_{0}^{1}+\frac{1}{4}\left[\frac{x^{0+1}}{0+1}\right]_{0}^{1}-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{4}\left[1^{2}-0^{2}\right]+\frac{1}{4}[1-0]-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \\ &=\frac{\log 3}{2}-\frac{1}{4} \int_{0}^{1} \frac{1}{2 x+1} d x \end{aligned}$
Put $2x+1=t\Rightarrow 2dx=dt\Rightarrow dx=\frac{dt}{2}$
When $x=0$ then $t=2$ and
When $x=1$ then $t=3$
$\begin{aligned} &=\frac{\log 3}{2}-\frac{1}{4} \int_{1}^{3} \frac{1}{t} \frac{d t}{2} \\ &=\frac{\log 3}{2}-\frac{1}{8}[\log |t|]_{1}^{3} \\ &=\frac{\log 3}{2}-\frac{1}{8}[\log 3-\log 1] \\ &=\frac{\log 3}{2}-\frac{1}{8}[\log 3]+0 \\ &=\log 3\left(\frac{1}{2}-\frac{1}{8}\right) \\ &=\log 3\left(\frac{4-1}{8}\right) \\ &=\frac{3}{8} \log 3 \end{aligned}$
Definite Integrals Exercise 19.1 Question 66
Answer: $\frac{4}{\sqrt{3}}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(\tan x+\cot x)^{2} d x$
Solution:$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}(\tan x+\cot x)^{2} d x=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\tan ^{2} x+\cot ^{2} x+2 \tan x \cot x\right) d x\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$\begin{aligned} &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\left(\sec ^{2} x-1\right)+\left(\cos e c^{2} x-1\right)+2 \tan x \frac{1}{\tan x}\right) d x \\ &\left.=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\sec ^{2} x-1\right)+\left(\cos e c^{2} x-1\right)+2 \tan x \frac{1}{\tan x}\right) d x \\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec ^{2} x d x-1 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos e c^{2} x d x-1 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x+2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x \\ &=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\pi}{3} \sec ^{2} x d x+\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos e c^{2} x d x-2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x+2 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} d x \end{aligned}$
$\left[\begin{array}{l} \int \sec ^{2} x d x=\tan x \\ \int \cos e c^{2} x d x=-\cot x \end{array}\right]$
$\begin{aligned} &=\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)-\left(\frac{1}{\sqrt{3}}-\sqrt{3}\right) \\ &=\left(\sqrt{3}-\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}+\sqrt{3}\right) \\ &=2 \sqrt{3}-\frac{2}{\sqrt{3}} \\ &=\frac{2 \times 3-2}{\sqrt{3}} \\ &=\frac{6-2}{\sqrt{3}} \\ &=\frac{4}{\sqrt{3}} \end{aligned}$
Definite Integrals Exercise 19.1 Question 67
Answer: $\frac{\left ( a^{2}+b^{2} \right )\pi }{8}+\frac{\left ( a^{2}-b^{2} \right )}{4}$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi}{4}}\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right) d x$
Solution:
$\begin{aligned} &\int_{0}^{\frac{\pi}{4}}\left(a^{2} \cos ^{2} x+b^{2} \sin ^{2} x\right) d x=\int_{0}^{\frac{\pi}{4}}\left[a^{2}\left(1-\sin ^{2} x\right)+b^{2} \sin ^{2} x\right] d x \\ &{\left[1=\sin ^{2} \theta+\cos ^{2} \theta \Rightarrow 1-\sin ^{2} \theta=\cos ^{2} \theta\right]} \\ &=\int_{0}^{\frac{\pi}{4}}\left(a^{2}-a^{2} \sin ^{2} x+b^{2} \sin ^{2} x\right) d x \\ &=\int_{0}^{\frac{\pi}{4}}\left[a^{2}+\sin ^{2} x\left(b^{2}-a^{2}\right)\right] d x \\ &=a^{2} \int_{0}^{\frac{\pi}{4}} 1 d x+\left(b^{2}-a^{2}\right) \int_{0}^{\frac{\pi}{4}} \sin ^{2} x d x \end{aligned}$
$=a^{\frac{\pi}{4}} \int_{0}^{\frac{\pi}{4}} 1 d x+\left(b^{2}-a^{2}\right) \int_{0}^{\frac{\pi}{4}}\left(\frac{1-\cos 2 x}{2}\right) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1-\cos 2 \theta=2 \sin ^{2} \theta\right] \$
$=a^{\frac{\pi}{4}} \int_{0}^{\frac{\pi}{4}} 1 d x+\frac{\left(b^{2}-a^{2}\right)}{2} \int_{0}^{\frac{\pi}{4}} 1 d x-\frac{\left(b^{2}-a^{2}\right)}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x \\$
$=\left(a^{2}+\frac{\left(b^{2}-a^{2}\right)}{2}\right)_{0}^{\frac{\pi}{4}} x d x-\frac{\left(b^{2}-a^{2}\right)}{2} \int_{0}^{\frac{\pi}{4}} \cos 2 x d x \\$
$=\left(\frac{2 a^{2}+\left(b^{2}-a^{2}\right)}{2}\right)\left[\frac{x^{0+1}}{0+1}\right]_{0}^{\frac{\pi}{4}}-\frac{\left(b^{2}-a^{2}\right)}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{4}} \\$
$=\left(\frac{a^{2}+b^{2}}{2}\right)[x]_{0}^{\frac{\pi}{4}}-\frac{\left(b^{2}-a^{2}\right)}{2 \times 2}\left[\sin 2 \times \frac{\pi}{4}-\sin 2 \times 0\right] \\$
$=\left(\frac{a^{2}+b^{2}}{2}\right)\left(\frac{\pi}{4}-0\right)-\frac{\left(b^{2}-a^{2}\right)}{4}\left[\sin \left(\frac{\pi}{2}\right)-\sin 0\right]$
$\left [ \sin \frac{\pi }{2}=1,\sin 0=0 \right ]$
$\begin{aligned} &=\left(\frac{a^{2}+b^{2}}{2}\right) \frac{\pi}{4}-\frac{\left(b^{2}-a^{2}\right)}{4} \sin (1-0) \\ &=\left(\frac{a^{2}+b^{2}}{2}\right) \frac{\pi}{4}+\left(\frac{a^{2}-b^{2}}{4}\right) \end{aligned}$
Definite Integrals Exercise 19.1 Question 68
Answer: $\frac{1}{4}log\left ( 2e \right )$Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x$
Solution: $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x$
$\begin{aligned} &=\int_{0}^{1} \frac{1}{x^{4}+2 x^{3}+2 x^{2}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{4}+x^{2}+x^{2}+2 x^{3}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{4}+x^{2}+2 x^{3}+x^{2}+2 x+1} d x \\ &=\int_{0}^{1} \frac{1}{x^{2}\left(x^{2}+1\right)+2 x\left(x^{2}+1\right)+\left(x^{2}+1\right)} d x \\ &=\int_{0}^{1} \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2 x+1\right)} d x \\ &=\int_{0}^{1} \frac{1}{\left(x^{2}+1\right)(x+1)^{2}} d x \end{aligned}$
$\left [ \left ( a+b \right )^{2}=a^{2}+2ab+b^{2} \right ]$
To solve this integral, first we have to find its partial fractions then integrate it by using indefinite integral formula then put the limits to get required answer.
So,
$\begin{aligned} &\frac{1}{\left(x^{2}+1\right)(x+1)^{2}}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C x+D}{x^{2}+1} \\ &\Rightarrow 1=\frac{A(x+1)^{2}\left(x^{2}+1\right)}{x+1}+\frac{B(x+1)^{2}\left(x^{2}+1\right)}{(x+1)^{2}}+\frac{(C x+D)(x+1)^{2}\left(x^{2}+1\right)}{\left(x^{2}+1\right)} \\ &\Rightarrow 1=A(x+1)^{2}\left(x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)(x+1)^{2} \\ &\Rightarrow 1=A\left(x^{3}+x+x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)\left(x^{2}+2 x+1\right) \end{aligned}$
$\left [ \left ( a+b \right )^{2}=a^{2}+b^{2}+2ab \right ]$$\Rightarrow 1=A x^{3}+A x+A x^{2}+A+B x^{2}+B+C x^{3}+2 C x^{2}+C x+D x^{2}+2 D x+D$
Equating coefficient of and constant term respectively
$0=A+C$
$0=A+B+2C+D$
$0=A+C+2D$
$1=A+B+D$
From (a) and (c)
$A+C+2D=0$
$\Rightarrow 0+2D=0$
$\Rightarrow D=0$
Put the value of D in (b) and (d)
$A+B+2C=0$
$A+B=1$
Subtracting (e) – (f) then
$A+B+2C=0$
$\frac{A+B=1}{2C=-1}$
$\Rightarrow C=-\frac{1}{2}$
Then $\left ( a \right )=>A=-C=-\left ( -\frac{1}{2} \right )=\frac{1}{2}$
And (d)
$\begin{aligned} &\Rightarrow A+B+D=1 \\ &\Rightarrow \frac{1}{2}+B+0=1 \\ &\Rightarrow B=1-\frac{1}{2}=\frac{1}{2} \\ &A=\frac{1}{2}, B=\frac{1}{2}, C=-\frac{1}{2} \end{aligned}$
$\begin{aligned} &\frac{1}{\left(x^{2}+1\right)(x+1)^{2}}=\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}+\frac{\frac{-1}{2} x+0}{x^{2}+1} \\ &=\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}-\frac{1}{2} \frac{x}{x^{2}+1} \end{aligned}$
Now $\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{2}+x} d x=\int_{0}^{1}\left(\frac{1}{2(x+1)}+\frac{1}{2(x+1)^{2}}-\frac{1}{2} \frac{x}{x^{2}+1}\right) d x$
$=\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)} d x+\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)^{2}} d x-\frac{1}{2} \int_{0}^{1} \frac{x}{x^{2}} d x$ ..........(1)
Put $x+1=t\Rightarrow dx=dt$
When $x=0$ then $t=1$
And When $x=1$ then $t=2$
Then $\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)} d x+\frac{1}{2} \int_{0}^{1} \frac{1}{(x+1)^{2}} d x$
$=\frac{1}{2} \int_{1}^{2} \frac{1}{t} d t+\frac{1}{2} \int_{1}^{2} \frac{1}{t^{2}} d t=\frac{1}{2} \int_{1}^{2} \frac{1}{t} d t+\frac{1}{2} \int_{1}^{2} t^{-2} d t \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}, \int \frac{1}{x} d x=\log |x|\right]$
$\begin{aligned} &=\frac{1}{2}[\log 2-\log 1]+\frac{1}{2}\left[\frac{t^{-1}}{-1}\right]_{1}^{2} \\ &=\frac{1}{2}[\log 2-\log 1]-\frac{1}{2}\left[\frac{1}{t}\right]_{1}^{2} \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{1}{2}-\frac{1}{1}\right] \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{1-2}{2}\right] \\ &=\frac{1}{2} \log 2-\frac{1}{2}\left[\frac{-1}{2}\right] \\ &=\frac{1}{2} \log 2+\frac{1}{4} \end{aligned}$
And
$\frac{1}{2}\int_{0}^{1}\frac{x}{x^{2}+1}dx$
When $x=0$ then $p=1$ and
When $x=1$ then $p=2$
$\begin{aligned} &\frac{1}{2} \int_{0}^{1} \frac{x}{x^{2}+1} d x=\frac{1}{2} \int_{1}^{2} \frac{1}{p} \frac{d p}{2} \\ &=\frac{1}{4} \int_{1}^{2} \frac{1}{p} d p \\ &=\frac{1}{4}[\log |p|]_{1}^{2} \\ &=\frac{1}{4}[\log 2-\log 1] \\ &=\frac{1}{4}[\log 2] \end{aligned} \quad\left[\begin{array}{l} \left.\frac{1}{x} d x=\log |x|\right] \\ \end{array}\right.$$\left [ log1=0 \right ]$
Then From (1)
$\begin{aligned} &\int_{0}^{1} \frac{1}{1+2 x+2 x^{2}+2 x^{3}+x^{4}} d x=\frac{1}{2} \log 2+\frac{1}{4}-\frac{1}{4} \log 2 \\ &=\log 2\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{1}{4} \\ &=\log 2\left(\frac{2-1}{4}\right)+\frac{1}{4} \\ &=\frac{1}{4} \log 2+\frac{1}{4} \\ &=\frac{1}{4} \log 2+\frac{1}{4} \text { loge }[\because \text { loge }=1] \\ &=\frac{1}{4} l(\operatorname{og} 2+\log e) \\ &=\frac{1}{4} \log (2 e) \quad[\because \log m+\log n=\log (m n)] \end{aligned}$
Class 12, mathematics, chapter 19, Definite Integrals, is one of the challenging portions in the syllabus. There are five exercises in this chapter, ex 19.1 to ex 19.5. The first exercise of this chapter, ex 19.1, consists of 68 questions. These questions revolve around the concept of evaluating the Definite Integrals. There are two levels of questions in this exercise, Level 1 and Level 2. If the students face difficulties in any of the questions, they can very well refer to the RD Sharma Class 12 Chapter 19 Exercise 19.1 material to clarify their doubts.
As ex 19.1 is the first exercise in the Definite Integrals chapter, it consists of pure basics. A strong foundation for this chapter can be laid only through this exercise. Hence, the student needs to concentrate well while working with it. This book follows the NCERT pattern giving a valid reason for the CBSE students to follow it. The RD Sharma Class 12th Exercise 19.1 solution book consists of various practice questions to make the students understand this concept correctly.
If the students find this exercise challenging, they can solve exercise 19.1 with the help of Class 12 RD Sharma Chapter 19 Exercise 19.1 Solution material. All the solutions are provided by the staff members who are experts in this domain. With little practice, you can start solving Definite Integrations effortlessly. This exercise is very significant because the following exercises are based on the basics discussed here. Otherwise, it would be a huge burden to solve the rest of the four exercises.
The main benefit for the students who use the RD Sharma Solutions Definite Integrals Ex 19.1 is that they need not purchase it. These solution books are freely available at the Career 360 website. No payment or monetary charge is required to be paid by them. Therefore, an abundance of students has benefitted by learning the concepts through the RD Sharma Class 12th Exercise 19.1 material.
These books are used to prepare questionnaires by the staff who prepare questions for the public exams. Hence using the RD Sharma Class 12 Solutions Chapter 19 Ex 19.1 will help you achieve better marks in the exam.
RD Sharma Chapter wise Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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