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RD Sharma Class 12 Exercise 19.1 Definite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 19.1 Definite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 02:21 PM IST

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: 2
Hint: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 2
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{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 thent=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:1
Hint: 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: 2
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 }{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}
Whenx=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: 2
Hint: 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: 2
Hint: 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}}

Hint: Use indefinite formula then put the limit to solve this integral
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: 2
Hint: 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: 2
Hint: 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=duin the 2nd integral, then when x=1

\Rightarrow u=2and 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 thent=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, then0=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:

\frac{57-\sqrt{3}}{5}
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=4then
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:0
Hint: 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: LetI=\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}
SoA=-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
Putt=\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: 0
Hint: 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: 8
Hint: 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.

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Frequently Asked Question (FAQs)

1. Which solution books are the most prescribed ones for the class 12 students to clarify their doubts on the mathematics chapter 19?

The RD Sharma Class 12th Exercise 19.1 material is the most prescribed solution book for the class 12 students to clarify their doubts on mathematics chapter 19.

2. Where can I find the RD Sharma books for free?

The RD Sharma solution books are available at the Career 360 website that can be accessed for free of cost.

3. How many questions are answered for exercise 18.1 in the RD Sharma book?

There are around sixty-eight questions, including its subparts in exercise 19.1, and the solutions are provided for every question in the RD Sharma Class 12th Exercise 19.1 book.

4. How to download the RD Sharma solution book from the career 360 websites?

Visit the Career 360 website and search for the name of the solution book you require. You’ll soon be able to access the book, click on the Download option to download it.

5. Which pattern do the RD Sharma books follow?

The RD Sharma Solution books follow the NCERT pattern. This makes it useful for the CBSE board students.

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