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RD Sharma Solutions Class 12 Mathematics Chapter 19 RE

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RD Sharma Solutions Class 12 Mathematics Chapter 19 RE

RD Sharma Solutions Class 12 Mathematics Chapter 19 RE

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Updated on 24 Jan 2022, 02:21 PM IST

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RD Sharma Class 12 Solutions Chapter19 RE Definite Integrals - Other Exercise
Definite Integrals Excercise:RE
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter19 RE Definite Integrals - Other Exercise

Definite Integrals Excercise:RE

Definite Integrals Exercise Revision Exercise Question 1

Answer:$\frac{128}{15}$
Given:$\int_{0}^{4} x \sqrt{4-x} d x$
Hint: Use the formula $\int_{0}^{a} f(x) d x$
Solution:
$\begin{aligned} &I=\int_{0}^{4} x \sqrt{4-x} d x \\ & \end{aligned}$
$I=\int_{0}^{4}(4-x) \sqrt{4-(4-x)} d x$
$\left(\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$
$\begin{aligned} &=\int_{0}^{4}(4-x) \sqrt{x} d x \\ & \end{aligned}$
$=\int_{0}^{4} 4 \sqrt{x}-x^{\frac{3}{2}}$
$\begin{aligned} &=4\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)_{0}^{4}-\left(\frac{-x^{\frac{5}{2}}}{\frac{5}{2}}\right)_{0}^{4} \\ & \end{aligned}$
$=\frac{8}{3}(8-0)-\frac{2}{5}(32-0)$
$\begin{aligned} &=\frac{64}{3}-\frac{64}{5} \\ & \end{aligned}$
$=\frac{320-192}{15} \\$
$=\frac{128}{15}$

Definite Integrals Exercise Revision Exercise Question 2

Answer:$\frac{326}{135}$
Given:$\int_{1}^{2} x \sqrt{3 x-2} d x$
Hint: Use the substitution method
Solution:$\int_{1}^{2} x \sqrt{3 x-2} d x$
Let $3 x-2=t$
$3dx = dt$ (differentiating w.r.t x)
$\begin{aligned} &\Rightarrow \int_{1}^{2} \frac{t+2}{3} \times \sqrt{t} \times \frac{d t}{3} \\ & \end{aligned}$
$=\frac{1}{9} \int_{1}^{2}(t+2) \sqrt{t} d t \\$
$=\frac{1}{9} \int_{1}^{2}(t)^{\frac{3}{2}}+2 \sqrt{t} d t$
$\begin{aligned} &=\frac{1}{9}\left[\left(\frac{t^{\frac{5}{2}}}{\frac{5}{2}}\right)_{1}^{2}+2\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right)_{1}^{2}\right] \\ & \end{aligned}$
$=\frac{1}{9}\left\{\frac{2}{5}\left[(3 x-2)^{\frac{5}{2}}\right]_{1}^{2}+\frac{4}{3}\left[(3 x-2)^{\frac{3}{2}}\right]_{1}^{2}\right\}$
$\begin{aligned} &=\frac{1}{9}\left[\frac{2}{5}(32-1)+\frac{4}{3}(8-1)\right] \\ & \end{aligned}$
$=\frac{1}{9}\left(\frac{2}{5} \times 31+\frac{4}{3} \times 7\right)=\frac{1}{9}\left(\frac{186+140}{15}\right) \\$
$=\frac{326}{135}$

Definite Integrals Exercise Revision Exercise Question 3

Answer:$\frac{16}{3}$
Given:$\int_{1}^{5} \frac{x}{\sqrt{2 x-1}} d x$
Hint: Let the denominator$(2x-1) = t$
Solution:
$\int_{1}^{5} \frac{x}{\sqrt{2 x-1}} d x$
Let $t=2 x-1=>x=\frac{t+1}{2}$
$dt = 2dx$ (differentiating w.r.t x)
$\begin{aligned} &I=\int_{1}^{5} \frac{t+1}{\frac{2}{\sqrt{t}}} \times \frac{d t}{2} \\ & \end{aligned}$
$I=\int_{1}^{5} \frac{t+1}{2 \sqrt{t}} d t=\frac{1}{4} \int_{1}^{5} \sqrt{t}+(t)^{\frac{-1}{2}} d t$
$=\frac{1}{4}\left(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)_{1}^{5}$
$=\frac{1}{4}\left(\frac{2}{3}(2 x-1)^{\frac{3}{2}}+2(2 x-1)^{\frac{1}{2}}\right)_{1}^{5}$
$\begin{aligned} &=\frac{1}{4}\left(\frac{2}{3}(27-1)+2(3-1)\right) \\ & \end{aligned}$
$=\frac{1}{4}\left(\frac{2}{3}(26)+2(2)\right)$
$\begin{aligned} &=\frac{1}{4}\left(\frac{52+12}{3}\right) \\ & \end{aligned}$
$=\frac{1}{4} \times \frac{64}{3} \\$
$=\frac{16}{3}$

Definite Integrals Exercise Revision Exercise Question 5

Answer:$\frac{\pi}{4}-\frac{1}{2} \log 2$
Given:$\int_{0}^{1} \tan ^{-1} x d x$
Hint: Use the integration by parts method
Solution:
$\int_{0}^{1} \tan ^{-1} x d x$
$\int_{0}^{1} \tan ^{-1} x \cdot 1 d x$
We know that $\int u v d x=u \int v d x-v \int u d x$
$\begin{aligned} &=\left(x\tan ^{-1} x \right)_{0}^{1}-\int_{0}^{1} \frac{1}{1+x^{2}} \times x d x \\ & \end{aligned}$
$=\frac{\pi}{4}-\frac{1}{2} \int_{0}^{1} \frac{1}{t} d t \\$
$=\frac{\pi}{4}-\frac{1}{2}[\log (t)]_{0}^{1}=\frac{\pi}{4}-\frac{1}{2}\left[\log \left(1+x^{2}\right)\right]_{0}^{1}$
$\begin{aligned} &=\frac{\pi}{4}-\frac{1}{2}(\log 2-\log 1) \\ & \end{aligned}$ $(\because log 1= 0)$
$=\frac{\pi}{4}-\frac{1}{2} \log 2$

Definite Integrals Excercise Revision Exercise Question 6

Answer:$\frac{\pi}{2}-\log 2$
Given:$\int_{0}^{1} \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x$
Hint: Use the substitution method and trigonometric identities
Solution:
$\int_{0}^{1} \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) d x$
Put $\begin{aligned} &x=\tan \theta \\ & \end{aligned}$
$\Rightarrow d x=\sec ^{2} \theta d \theta$
$\begin{aligned} &=\int_{0}^{1} \cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \times \sec ^{2} \theta d \theta \\ & \end{aligned}$
$=\int_{0}^{1} \cos ^{-1}(\cos 2 \theta) \times \sec ^{2} \theta d \theta$ $\quad\left(\because \cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$
$\begin{aligned} &=2 \int_{0}^{1} \theta \sec ^{2} \theta d \theta \\ & \end{aligned}$
$=2\left[\theta \int_{0}^{1} \sec ^{2} \theta d \theta-\int_{0}^{1} \frac{d}{d \theta} \theta\left[\sec ^{2} \theta d \theta\right]_{0}^{1}\right] \\$
$=2\left[\theta \tan \theta-\int \tan \theta d \theta\right]_{0}^{1}$
$\begin{aligned} &=2(\theta \tan \theta)_{0}^{1}-2 \int_{0}^{1} \frac{\sin \theta}{\cos \theta} d \theta \\ & \end{aligned}$
$=2\left[\tan ^{-1} x \times \tan \left(\tan ^{-1} x\right)\right]_{0}^{1}-2 \int_{0}^{1} \frac{\sin \theta}{\cos \theta} d \theta$
Let $\begin{aligned} &\cos \theta=t \\ & \end{aligned}$
$\Rightarrow-\sin \theta d \theta=\mathrm{dt}$ (differentiate w.r.t $\theta$)
$\begin{aligned} =& 2\left(\tan ^{-1} 1 \times 1\right)+2 \int_{0}^{1} \frac{1}{t} d t \\ \end{aligned}$
$= 2 \times \frac{\pi}{4}+2[\log |t|]_{0}^{1} \\$
$=\frac{\pi}{2}+2[\log |\cos \theta|]_{0}^{1}$
$\begin{aligned} &=\frac{\pi}{2}+2\left[\log \left|\cos \left(\tan ^{-1} x\right)\right|\right]_{0}^{1} \\ & \end{aligned}$
$=\frac{\pi}{2}+2\left(\log \frac{1}{\sqrt{2}}-\log 1\right) \\$
$=\frac{\pi}{2}-\log 2$

Definite Integrals Excercise Revision Exercise Question 7

Answer: $\frac{\pi}{2}-\log 2$
Given:$\int_{0}^{1} \tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$
Hint: Use substitution method and then apply the formula of $\tan 2\theta$
Solution:
$\int_{0}^{1} \tan ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$
Let $\mathrm{x}=\tan \theta$
$d x=\sec ^{2} \theta d \theta$ (differentiate w.r.t x )
Now, $0 < x <1$
$0<\tan \theta<1$
$0<\theta<\frac{\pi}{4}$
Now,
$\begin{aligned} &\int_{0}^{\frac{\pi}{4}} \tan ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) \sec ^{2} \theta d \theta \\ & \end{aligned}$
$=2 \int_{0}^{\frac{\pi}{4}} \theta \sec ^{2} \theta d \theta \\$
$=2\left[\theta \int \sec ^{2} \theta d \theta-\int \frac{d}{d \theta} \theta \int \sec ^{2} \theta d \theta\right]_{0}^{\frac{\pi}{4}}$
$\begin{aligned} &{\left[\because \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x\right]} \\ & \end{aligned}$
$=2\left[(\theta \tan \theta)_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} \tan \theta d \theta\right]$
$\begin{aligned} &=2\left(\frac{\pi}{4}-0\right)-2 \int_{0}^{\frac{\pi}{4}} \tan \theta d \theta \\ & \end{aligned}$
$=\frac{\pi}{2}-2 \int_{0}^{\frac{\pi}{4}} \frac{\sin \theta}{\cos \theta} d \theta$
Let $\cos \theta=t$
$-\sin \theta d \theta=\mathrm{dt}$ (differentiate w.r.t

)
$\begin{aligned} &=\frac{\pi}{2}+2 \int_{0}^{\frac{\pi}{4}} \frac{1}{t} d t \\ & \end{aligned}$
$=\frac{\pi}{2}+2(\log |t|)_{0}^{\frac{\pi}{4}}$
$\begin{aligned} &=\frac{\pi}{2}+2(\log |\cos \theta|)_{0}^{\frac{\pi}{4}}\\ & \end{aligned}$
$=\frac{\pi}{2}+2\left(\log \frac{1}{\sqrt{2}}-\log 1\right)\\$
$=\frac{\pi}{2}+2 \log \frac{1}{\sqrt{2}}\\$ $(\because \log 1=0)\\$
$=\frac{\pi}{2}-\log 2$

Definite Integrals Excercise Revision Exercise Question 8

Answer:$\frac{\pi}{2 \sqrt{3}}-\frac{3}{2} \log \frac{4}{3}$
Given: $\int_{0}^{\frac{1}{\sqrt{3}}} \tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right) d x$
Hint: You must know about trigonometric identities and formula of $\int u v d x$
Solution:
$\int_{0}^{\frac{1}{\sqrt{3}}} \tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right) d x$
Let $\tan x=\theta$
$d x=\sec ^{2} \theta d \theta$ (differentiate w.r.t x )
We know ,
$\begin{aligned} &0<\tan \theta<\frac{1}{\sqrt{3}} \\ & \end{aligned}$
$0<\theta<\frac{\pi}{6}$
$\therefore \int_{0}^{\frac{\pi}{6}} \tan ^{-1}\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right) \times \sec ^{2} \theta d \theta$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{6}} \tan ^{-1}(\tan 3 \theta) \times \sec ^{2} \theta d \theta \\ & \end{aligned}$
$=3 \int_{0}^{\frac{\pi}{6}} \theta \sec ^{2} \theta d \theta$
We know that $\left[\because \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x\right]$
$\begin{aligned} &=3(\theta \tan \theta)_{0}^{\frac{\pi}{6}}-3 \int_{0}^{\frac{\pi}{6}} \tan \theta d \theta \\ & \end{aligned}$
$=3\left(\frac{\pi}{6} \times \frac{1}{\sqrt{3}}-0\right)-3 \int_{0}^{\frac{\pi}{6}} \tan \theta d \theta$
Let $\cos \theta=t$
$-\sin \theta d \theta=\mathrm{dt}$ (differentiate w.r.t $\theta$)
$\begin{aligned} &=\frac{\pi}{2 \sqrt{3}}+3 \int_{0}^{\frac{\pi}{6}} \frac{1}{t} d t \\ & \end{aligned}$
$=\frac{\pi}{2 \sqrt{3}}+3(\log |t|)_{0}^{\frac{\pi}{6}}=\frac{\pi}{2 \sqrt{3}}+3(\log |\cos \theta|)_{0}^{\frac{\pi}{6}}$
$\begin{aligned} &=\frac{\pi}{2 \sqrt{3}}+3\left(\log \frac{\sqrt{3}}{2}-\log 1\right)_{0}^{\frac{\pi}{6}} \\ & \end{aligned}$
$\frac{\pi}{2 \sqrt{3}}-\frac{3}{2} \log \frac{4}{3}$

Definite Integrals Excercise Revision Exercise Question 9

Answer: $2\log2 -1$
Given:$\int_{0}^{1} \frac{1-x}{1+x} d x$
Hint: Apply integration by parts method
Solution:
$\int_{0}^{1} \frac{1-x}{1+x} d x$
$\begin{aligned} &=\int_{0}^{1} \frac{1-x-1+1}{1+x} d x=\int_{0}^{1} \frac{2-(x+1)}{1+x} d x \\ & \end{aligned}$
$=\int_{0}^{1} \frac{2}{1+x} d x-\int_{0}^{1} d x$
$\begin{aligned} &=2[\log (1+x)]_{0}^{1}-(x)_{0}^{1} \\ & \end{aligned}$
$=2(\log 2-\log 1)-1 \\$
$=2 \log 2-1$

Definite Integrals Excercise Revision Exercise Question 10

Answer:$\frac{1}{4} \log \left(\frac{2 \sqrt{3}+3}{3}\right)$
Given:$\int_{0}^{\frac{\pi}{3}} \frac{\cos x}{3+4 \sin x} d x$
Hint: Apply Substitution method
Solution:
$\int_{0}^{\frac{\pi}{3}} \frac{\cos x}{3+4 \sin x} d x$
Let $3+4 \sin x=t$
$4 \cos d x=d t$ (diff w.r.t x)
$\begin{aligned} &\frac{1}{4} \int_{0}^{\frac{\pi}{3}} \frac{1}{t} d t=\frac{1}{4}(\log t)_{0}^{\frac{\pi}{3}} \\ & \end{aligned}$
$=\frac{1}{4}[\log |3+4 \sin x|]_{0}^{\frac{\pi}{3}}$
$\begin{aligned} &=\frac{1}{4}\left(\log \left(3+4 \times \frac{\sqrt{3}}{2}\right)-\log 3\right) \\ & \end{aligned}$
$=\frac{1}{4}(\log (3+2 \sqrt{3})-\log 3) \\$
$=\frac{1}{4} \log \left(\frac{3+2 \sqrt{3}}{3}\right)$

Definite Integrals Exercise Revision Exercise question 11

Answer:$2-\frac{\pi}{2}$
Given:$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{(1+\cos x)^{2}} d x$
Hint:
Use the formula

and then apply integration rule by parts method.
Solution:
$\begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{(1+\cos x)^{2}} d x \\ & \end{aligned}$
$=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos ^{2} x}{(1+\cos x)^{2}} d x$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos x}{(1+\cos x)} d x \\ & \end{aligned}$
$=\int_{0}^{\frac{\pi}{2}} \frac{1-\cos x-1+1}{1+\cos x} d x$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{2-(1+\cos x)}{1+\cos x} d x \\ & \end{aligned}$
$=\int_{0}^{\frac{\pi}{2}} \frac{2}{1+\cos x} d x-\int_{0}^{\frac{\pi}{2}} \frac{1+\cos x}{1+\cos x} d x$
$\begin{aligned} &=2 \int_{0}^{\frac{\pi}{2}} \frac{(1-\cos x)}{\left(1-\cos ^{2} x\right)} d x-\frac{\pi}{2} \\ & \end{aligned}$
$=2 \int_{0}^{\frac{\pi}{2}} \frac{(1-\cos x)}{\sin ^{2} x} d x-\frac{\pi}{2}$
$\begin{aligned} &=2 \int_{0}^{\frac{\pi}{2}} \cos e c^{2} x-\operatorname{cosec} x \cdot \cot x d x-\frac{\pi}{2} \\ & \end{aligned}$
$=2(-\cot x+\cos e c x)_{0}^{\frac{\pi}{2}}-\frac{\pi}{2} \\$
$=2(-0+1+0)-\frac{\pi}{2} \\$
$=2-\frac{\pi}{2}$

Definite Integrals Exercise Revision Exercise question 12

Answer: $2(\sqrt{2-1})$
Given: $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{1+\cos x}} d x$
Hint: Use substitution Method
Solution:$\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sqrt{1+\cos x}} d x$
Let $1+\cos x=t$
$-\sin x dx = dt$ (diff w.r.t x)
Now,
$\begin{aligned} &\int_{0}^{\frac{\pi}{2}}-\frac{1}{\sqrt{t}} d t \\ & \end{aligned}$
$=-\left(\frac{\sqrt{t}}{\frac{1}{2}}\right)_{0}^{\frac{\pi}{2}}=-2(\sqrt{1+\cos x})_{0}^{\frac{\pi}{2}}$
$\begin{aligned} &=-2(1-\sqrt{2}) \\ & \end{aligned}$
$=2(\sqrt{2}-1)$

Definite Integrals Exercise Revision Exercise question 13

Answer:$\frac{\pi}{4}$
Given: $\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1+\sin ^{2} x} d x$
Hint: You must know about the integration of $\frac{1}{1+x^{2}}$
Solution:$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{1+\sin ^{2} x} d x$
Let $\sin x=t$
$\cos xdx=dt$ (Differentiate w.r.t to x)
Now $\int_{0}^{\frac{\pi}{2}} \frac{1}{1+t^{2}} d t$
$\begin{aligned} &=\left(\tan ^{-1} t\right)_{0}^{\frac{\pi}{2}}=\left[\tan ^{-1}(\sin x)\right]_{0}^{\frac{\pi}{2}} \\ & \end{aligned}$
$=\tan ^{-1} 1-\tan ^{-1} 0 \\$
$=\frac{\pi}{4}-0\left(\because \tan \frac{\pi}{4}=1\right)$

Definite Integrals Exercise Revision Exercise question 14
Answer: $\frac{8}{3}$

Given:$\int_{0}^{\pi} \sin ^{3} x(1+2 \operatorname{cox})(1+\cos x)^{2} d x$
Hint: Use trigonometric identity and then apply substitution method
Solution:$\int_{0}^{\pi} \sin ^{3} x(1+2 \operatorname{cox})(1+\cos x)^{2} d x$
$\begin{aligned} &\int_{0}^{\pi} \sin x\left(1-\cos ^{2} x\right)(1+2 \operatorname{cox})(1+\cos x)^{2} d x \\ & \end{aligned}$
${\left[\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \sin ^{2} x=1-\cos ^{2} x \end{array}\right]}$
Now,
Let
$\begin{aligned} &\cos x=t \\ & \end{aligned}$
$-\sin x d x=d t$ (Diff w.r.t to x)
$\begin{aligned} &\Rightarrow-\int_{0}^{\pi}\left(1-t^{2}\right)(1+2 t)(1+t)^{2} d x \\ & \end{aligned}$
$\Rightarrow-\int_{0}^{\pi}\left(1-t^{2}\right)(1+2 t)(1+t)^{2} d t \\$
$\Rightarrow-\int_{0}^{\pi}\left(1+t^{2}+2 t+2 t+2 t^{3}+4 t^{2}-2 t^{2}-t^{4}-2 t^{3}-2 t^{5}-4 t^{4}\right) d t$
$\begin{aligned} &\Rightarrow-\int_{0}^{\pi}\left(1+4 t+4 t^{2}-2 t^{3}-5 t^{4}-2 t^{5}\right) d t \\ &\end{aligned}$
$\Rightarrow-\left(t+2 t^{2}+\frac{4 t^{3}}{3}-\frac{t^{4}}{2}-t^{5}-\frac{t^{6}}{3}\right)_{-1}^{1} \\$
$\left\{\begin{array}{l} \therefore 0<x<\pi \\ 1<\cos x<-1 \\ 1<t<-1 \end{array}\right\}$
$\begin{aligned} &\Rightarrow\left(t+2 t^{2}+\frac{4 t^{3}}{3}-\frac{t^{4}}{2}-t^{5}-\frac{t^{6}}{3}\right)_{-1}^{1} \\ & \end{aligned}$
$\Rightarrow\left(1+2+\frac{4}{3}-\frac{1}{2}-1-\frac{1}{3}\right)_{-1}^{1}-\left((-1)+2-\frac{4}{3}-\frac{1}{2}+1-\frac{1}{3}\right)_{-1}^{1}$
$\begin{aligned} &\Rightarrow 2-\frac{1}{2}+1-2+\frac{5}{3}+\frac{1}{2} \\ & \end{aligned}$
$=1+\frac{5}{3}=\frac{8}{3}$

Definite Integrals Exercise Revision Exercise question 15

Answer:$\frac{\pi}{4}$
Given:$\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x$
Hint: Use Partial fraction method
Solution:
$\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x$
Using Partial fraction
$\begin{aligned} &\frac{x}{(1+x)\left(1+x^{2}\right)}=\frac{A}{1+x}+\frac{B x+c}{1+x^{2}} \\ & \end{aligned}$
$\Rightarrow x=A\left(1+x^{2}\right)+(B x+C)(1+x) \\$
$\Rightarrow x=A+A x^{2}+B x+B x^{2}+C+C x \\$
$\Rightarrow x=(A+C)+(A+B) x^{2}+(B+C) x$
On comparing
$\begin{aligned} &A+C=0, B+C=1, A+B=0 \\ & \end{aligned}$
$\therefore A=\frac{-1}{2}, B=\frac{1}{2}, C=\frac{1}{2}$
Hence,
$\begin{aligned} &\int_{0}^{\infty} \frac{x}{(1+x)\left(1+x^{2}\right)} d x=\int_{0}^{\infty}\left(\frac{-1}{2(1+x)}+\frac{1}{2}\left(\frac{x+1}{1+x^{2}}\right)\right) d x \\ & \end{aligned}$
$=-\frac{1}{2}[\log |1+x|]_{0}^{\infty}+\frac{1}{2} \int_{0}^{\infty} \frac{x}{1+x^{2}}+\frac{1}{1+x^{2}} d x$
$\begin{aligned} &=-\frac{1}{2}\left[\log \frac{\sqrt{1+x^{2}}}{x+1}\right]_{0}^{\infty}+\frac{1}{2}\left(\tan ^{-1} x\right)_{0}^{\infty} \\ & \end{aligned}$
$=\frac{1}{2} \times 0+\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}$

Definite Integrals Exercise Revision Exercise Question 16

Answer:$\frac{3}{5\sqrt{2}}$
Given:$\int_{0}^{\frac{\pi}{4}} \sin 2 x \sin 3 x d x$
Hint: You must know the identity 2 sin x sin y
Solution: $\int_{0}^{\frac{\pi}{4}} \sin 2 x \sin 3 x d x$
$\begin{aligned} &{[\therefore 2 \sin x \sin y=\cos (x-y)-\cos (x+y)]} \\ & \end{aligned}$
$=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \cos x-\cos 5 x d x \\$
$(\therefore \cos (-x)=\cos x)$
$\begin{aligned} &=\frac{1}{2}\left(\sin x-\frac{\sin 5 x}{5}\right)_{0}^{\frac{\pi}{4}}=\frac{1}{2}\left(\sin \frac{\pi}{4}+\frac{\sin \frac{\pi}{4}}{5}\right) \\ & \end{aligned}$
$=\frac{1}{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{5 \sqrt{2}}\right)=\frac{1}{2} \times\left(\frac{5+1}{5 \sqrt{2}}\right) \\$
$=\frac{1}{2} \times \frac{6}{5 \sqrt{2}}=\frac{3}{5 \sqrt{2}}$

Definite Integrals Exercise Revision Exercise Question 17

Answer:$\frac{\pi}{2}-1$
Given:$\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x$
Hint: Do rationalization and then integrate by parts
Solution:
$\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x$
$\begin{aligned} &\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} d x \\ & \end{aligned}$
$=\int_{0}^{1} \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x=\int_{0}^{1} \sqrt{\frac{(1-x)^{2}}{1-x^{2}}} d x$
$\begin{aligned} &=\int_{0}^{1} \frac{1-x}{\sqrt{1-x^{2}}} d x=\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}} d x-\int_{0}^{1} \frac{x}{\sqrt{1-x^{2}}} d x \\ & \end{aligned}$
$=\left(\sin ^{-1} x\right)_{0}^{1}-\left(-\frac{1}{2} \sqrt{\frac{1-x^{2}}{\frac{1}{2}}}\right)_{0}^{1}$
$\begin{aligned} &=\left(\sin ^{-1} x\right)_{0}^{1}+\left[\sqrt{1-x^{2}}\right]_{0}^{1} \\ & \end{aligned}$
$=\left(\frac{\pi}{2}-1\right)$

Definite Integrals Exercise Revision Exercise Question 18

Answer:$\frac{\sqrt{e}-1}{e}$
Given: $\int_{1}^{2} \frac{1}{x^{2}} e^{-\frac{1}{x}} d x$
Hint: Use Substitution method
Solution:
Let
$\begin{aligned} &\frac{-1}{x}=t \\ & \end{aligned}$ (Differentiating w.r.t to x)
$\frac{1}{x^{2}} d x=d t$
$\begin{aligned} &\int_{1}^{2} \frac{1}{x^{2}} e^{-\frac{1}{x}} d x\\ & \end{aligned}$
$\int_{1}^{2} e^{t} d t=\left(e^{t}\right)_{1}^{2}=\left(e^{\frac{-1}{x}}\right)_{1}^{2}\\$
$\begin{aligned} &=e^{\frac{-1}{2}}-e^{-1} \\ \end{aligned}$
$=e^{-\frac{1}{2}}-\frac{1}{e} \\$
$=\frac{\sqrt{e}-1}{e}$

Definite Integrals Exercise Revision Exercise Question 19

Answer:$\frac{2}{35}-\frac{9}{280 \sqrt{2}}$
Given:$\int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin ^{3} x d x$
Hint: Use trigonometric identities.
Solution: $\int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin ^{3} x d x$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \cos ^{4} x \sin x\left(1-\cos ^{2} x\right) d x \\ & \end{aligned}$
$\left(\begin{array}{l} \because \sin ^{2} x+\cos ^{2} x=1 \\ \sin ^{2} x=1-\cos ^{2} x \end{array}\right)$
$\begin{aligned} &\text { Let } \\ & \end{aligned}$
$\cos x=t \\$ (Differentiate w.r.t to x)
$-\sin x d x=d t$
$\begin{aligned} &=-\int_{0}^{\frac{\pi}{4}} t^{4}\left(1-t^{2}\right) d t \\ & \end{aligned}$
$=\int_{0}^{\frac{\pi}{4}} t^{4}\left(t^{2}-1\right) d t \\$
$=\int_{0}^{\frac{\pi}{4}}\left(t^{6}-t^{4}\right) d t=\left(\frac{t^{7}}{7}-\frac{t^{5}}{5}\right)_{0}^{\frac{\pi}{4}}$
$\begin{aligned} &=\left(\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}\right)_{0}^{\frac{\pi}{4}}=\frac{1}{(\sqrt{2})^{7}} \times \frac{1}{7}-\frac{1}{(\sqrt{2})^{5}} \times \frac{1}{5}-\frac{1}{7}+\frac{1}{5} \\ & \end{aligned}$
$=\frac{1}{56 \sqrt{2}}-\frac{1}{20 \sqrt{2}}+\frac{2}{35}$
$\begin{aligned} &=\frac{1}{4 \sqrt{2}}\left(\frac{1}{14}-\frac{1}{5}\right)+\frac{2}{35} \\ & \end{aligned}$
$=\frac{2}{35}+\frac{1}{4 \sqrt{2}}\left(\frac{5-14}{70}\right) \\$
$=\frac{2}{35}-\frac{9}{280 \sqrt{2}}$

Definite Integrals Exercise Revision Exercise Question 20

Answer: $\frac{3}{2}$
Given: $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x$
Hint: Do rationalization and then apply substitution method
Solution:
$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} d x$
$\begin{aligned} &=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}} \times \frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} d x\\ & \end{aligned}$
$=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sqrt{1-\cos ^{2} x}}{(1-\cos x)^{2}} d x=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sin x}{(1-\cos x)^{2}} d x$
$\begin{aligned} &\text { Let } \\ & \end{aligned}$
$1-\cos x=t \\$ (differentiate w.r.t to x)
$\sin x d x=d t$
$\begin{aligned} &=\int_{\frac{1}{2}}^{1} \frac{1}{t^{2}} d t=-\frac{1}{2}\left(t^{-1}\right)_{\frac{1}{2}}^{1}=-\frac{1}{4} \\ & \end{aligned}$

Definite Integrals Excercise Revision Exercise Question 21

Answer: $\frac{-\pi}{4}$
Given:$\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x$
Hint: Apply integration by parts method
Solution:$\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x$
$\begin{aligned} &{\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x \frac{\sin 2 x}{2} d x} \\ & \end{aligned}$
${\left[\because \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x\right]}$
$\begin{aligned} &=\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\left[\left(-x \frac{\cos 2 x}{2}\right)+\int_{0}^{\frac{\pi}{2}} \frac{\cos 2 x}{2} d x\right] \\ \end{aligned}$
$=\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}+\left(x \frac{\cos 2 x}{2}\right)_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left(\frac{\sin 2 x}{2}\right)_{0}^{\frac{\pi}{2}} \\$
$=0-\frac{\pi}{2} \times \frac{1}{2}-0=-\frac{\pi}{4}$

Definite Integrals Excercise Revision Exercise Question 22

Answer:$\log\left ( \frac{4}{e} \right )$
Given:$\int_{0}^{1} \log (1+x) d x$
Hint: Apply the formula of $\int u v d x$
Solution: $\int_{0}^{1} \log (1+x) d x$
$\begin{aligned} &=\int_{0}^{1} \log (1+x) \cdot 1 d x \\ & \end{aligned}$
$=[\log (1+x) x]_{0}^{1}-\int_{0}^{1} \frac{x}{1+x} d x \\$
${\left[\because \int u v d x=u \int v d v-\int \frac{d}{d x} u \int v d x\right]}$
$\begin{aligned} &=[\log (1+x) x]_{0}^{1}-\int_{0}^{1} \frac{x+1-1}{1+x} d x \\ & \end{aligned}$
$=[\log (1+x) x]_{0}^{1}-\int_{0}^{1} 1-\frac{1}{1+x} d x$
$\begin{aligned} &=[\log (1+x) x]_{0}^{1}-[x-\log (1+x)]_{0}^{1} \\ & \end{aligned}$
$=\log 2-\log 0-[(1-\log 2)-(0-\log 1)]$
We know $\log1 = 0$
$\begin{aligned} &\therefore \log 2-1+\log 2=2 \log 2-1 \\ & \end{aligned}$
$=\log 4-\log e$
$\begin{aligned} &\left(\begin{array}{l} \therefore \log e=1, a \log b=\log b^{a} \\ \therefore \log a-\log b=\log \frac{a}{b} \end{array}\right) \\ & \end{aligned}$
$=\log \left(\frac{4}{e}\right)$

Definite Integrals Excercise Revision Exercise Question 23

Answer:$\frac{57}{5}-\sqrt{5}$
Given:$\int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x$
Hint: Use substitution method
Solution: $\int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x$
Let
$\begin{aligned} &2 x+1=t^{2} \\ & \end{aligned}$
$2 d x=2 t d t \\$ (Differentiate w.r.t to x)
$d x=t d t$
Now, $\int_{2}^{4} \frac{x^{2}+x}{\sqrt{2 x+x}} d x$
$\begin{aligned} &=\int_{2}^{4} \frac{\left(\frac{t^{2}-1}{2}\right)^{2}+\left(\frac{t^{2}-1}{2}\right)}{t} t d t \\ & \end{aligned}$
$=\int_{2}^{4} \frac{\left(t^{2}-1\right)^{2}+2\left(t^{2}-1\right)}{4} d t$
$\begin{aligned} &=\frac{1}{4} \int_{2}^{4}\left(t^{4}+1-2 t^{2}+2 t^{2}-2\right) d t \\ & \end{aligned}$
$=\frac{1}{4} \int_{2}^{4} t^{4}-1 d t \\$
$=\frac{1}{4}\left(\frac{t^{5}}{5}-t\right)^{4}$
$\begin{aligned} &=\frac{1}{4}\left[\frac{(2 x+1)^{2} \sqrt{2 x+1}}{5}-\sqrt{2 x+1}\right]_{2}^{4} \\ & \end{aligned}$
$=\frac{1}{4}\left[\left(\frac{81 \times 3}{5}-3\right)-\left(\frac{25 \sqrt{5}}{5}-\sqrt{5}\right)\right]$
$\begin{aligned} &=\frac{1}{4}\left[\left(\frac{243-15}{5}\right)-\left(\frac{25 \sqrt{5}-5 \sqrt{5}}{5}\right)\right] \\ & \end{aligned}$
$=\frac{1}{4} \times \frac{228}{5}-\frac{1}{4} \times \frac{20 \sqrt{5}}{5} \\$
$=\frac{57}{5}-\sqrt{5}$

Definite Integrals Excercise Revision Exercise Question 24

Answer:$\frac{\pi^{2}}{16}-\frac{\pi}{4}+\frac{1}{2} \log 2$
Given:$\int_{0}^{1} x\left(\tan ^{-1} x\right)^{2} d x$
Hint: Apply the formula of $\int uv dx$
Solution:
$\int_{0}^{1} x\left(\tan ^{-1} x\right)^{2} d x$
Let
$\begin{aligned} &\tan ^{-1} x=t \\ & \end{aligned}$
$x=\tan t \\$ (Differentiate w.r.t to x)
$d x=\sec ^{2} t d t$
$\begin{aligned} &=\int_{0}^{1}(\tan t) t^{2} \sec ^{2} t d t \\ & \end{aligned}$
$=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-\int_{0}^{1} 2 t \times \frac{\tan ^{2} t}{2} d t \\$
$=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-\int_{0}^{1} t\left(\sec ^{2} t-1\right) d t$
$\begin{aligned} &=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-\int_{\ell}^{1} t \sec ^{2} t d t+\int_{0}^{1} t d t \\ & \end{aligned}$
$=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-(t \tan t)_{0}^{1}+\int_{0}^{1} \tan t d t+\left(\frac{t^{2}}{2}\right)_{0}^{1} \\$
$=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{1}-(t \tan t)_{0}^{1}+[\log |\sec t|]_{0}^{1}+\left(\frac{t^{2}}{2}\right)_{0}^{1}$
Now $0< x< 1$
$\begin{aligned} &\Rightarrow 0<\tan ^{-1} x<\frac{\pi}{4} \\ & \end{aligned}$
$\Rightarrow 0<t<\frac{\pi}{4}$
$\begin{aligned} &=\left[t^{2} \frac{\tan ^{2} t}{2}\right]_{0}^{\frac{\pi}{4}}-(t \tan t)_{0}^{\frac{\pi}{4}}+[\log |\sec t|]_{0}^{\frac{\pi}{4}}+\left(\frac{t^{2}}{2}\right)_{0}^{\frac{\pi}{4}} \\ & \end{aligned}$
$=\left(\frac{\pi}{4}\right)^{2} \times \frac{1}{2}-\frac{\pi}{4}+\log \sqrt{2}+\left(\frac{\pi}{4}\right)^{2} \times \frac{1}{2}$
$\begin{aligned} &=\frac{\pi^{2}}{16} \times \frac{1}{2}-\frac{\pi}{4}+\log \sqrt{2}+\frac{\pi^{2}}{16} \times \frac{1}{2} \\ & \end{aligned}$
$=2 \times \frac{\pi^{2}}{16} \times \frac{1}{2}-\frac{\pi}{4}+\log \sqrt{2} \\$
$=\frac{\pi^{2}}{16}-\frac{\pi}{4}+\log \sqrt{2}$

Definite Integrals Excercise Revision Exercise Question 25
Answer: $\pi-2$

Given: $\int_{0}^{1}\left(\cos ^{-1} x\right)^{2}$
Hint: Using substitution method and formula of $\int uv dx$
Solution:$\int_{0}^{1}\left(\cos ^{-1} x\right)^{2}$
Let
$\begin{aligned} &\cos ^{-1} x=\theta \Rightarrow x=\cos \theta \\ & \end{aligned}$ (Differentiate w.r.t to x)
$d x=-\sin \theta d \theta \\$
$=\int_{0}^{1} \theta^{2}(-\sin \theta d \theta)$
$\begin{aligned} &=-\int_{0}^{1} \theta^{2} \sin \theta d \theta \\ & \end{aligned}$
$=-\left[\theta^{2}(\cos \theta)\right]_{0}^{1}+\int_{0}^{1} 2 \theta(-\cos \theta) d \theta \\$
$=\theta^{2}(\cos \theta)_{0}^{1}-2\left[(\theta \sin \theta)_{0}^{1}-\int_{0}^{1} \sin \theta d \theta\right]$
$\begin{aligned} &=(\theta^{2}\cos \theta)_{0}^{1}-2(\theta \sin \theta)_{0}^{1}+2(-\cos \theta)_{0}^{1} \\ \end{aligned}$
$=\left[\left(\cos ^{-1} x\right)^{2} x\right]_{0}^{1}-2\left[\cos ^{-1} x \sin \left(\cos ^{-1} x\right)\right]_{0}^{1}-2(x)_{0}^{1} \\$
$= 0-2\left(0-\frac{\pi}{2}\right)-2=\frac{2 \pi}{2}-2 \\$
$= \pi-2$

Definite Integrals Exercise Revision Exercise Question 26

Answer:$\frac{1}{2}\log 6$
Given:$\int_{1}^{2} \frac{(x+3)}{x(x+2)} d x$
Hint: Do integration by parts
Solution:$\int_{1}^{2} \frac{(x+3)}{x(x+2)} d x$
$\begin{aligned} &=\int_{1}^{2} \frac{(x+2+1)}{x(x+2)} d x \\ & \end{aligned}$
$=\int_{1}^{2} \frac{1}{x} d x+\int_{1}^{2}\left(\frac{1}{x(x+2)}\right) d x \\$
$=\int_{1}^{2} \frac{1}{x} d x+\frac{1}{2} \int_{1}^{2}\left(\frac{x+2-x}{x(x+2)}\right) d x$
$\begin{aligned} &=\int_{1}^{2} \frac{1}{x} d x+\frac{1}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} d x \\ & \end{aligned}$
$=\frac{3}{2} \int_{1}^{2} \frac{1}{x} d x-\frac{1}{2} \int_{1}^{2} \frac{1}{x+2} d x$
$\begin{aligned} &=\frac{3}{2}(\log x)_{1}^{2}-\frac{1}{2}[\log (x+2)]_{1}^{2} \\ & \end{aligned}$
$=\frac{3}{2} \log 2-\frac{1}{2} \log 4+\frac{1}{2} \log 3 \\$
$=\frac{3}{2} \log 2-\log 2+\frac{1}{2} \log 3$
$\begin{aligned} &=\frac{1}{2} \log 2+\frac{1}{2} \log 3=\frac{1}{2}(\log 2+\log 3) \\ & \end{aligned}$
$=\frac{1}{2} \log 6(\therefore \log a+\log b=\log a b)$

Definite Integrals Exercise Revision Exercise Question 27

Answer:$\frac{1}{2}$
Given:$\int_{0}^{\frac{\pi}{4}} e^{x} \sin x d x$
Hint: Apply the formula of $\int uvdx$
Solution: $I=\int_{0}^{\frac{\pi}{4}} e^{x} \sin x d x$ ……… (1)
$\begin{aligned} &I=\left(-e^{x} \cos x\right)_{0}^{\frac{\pi}{4}}+\int_{0}^{\frac{\pi}{4}} e^{x} \cos x d x \\ & \end{aligned}$
${\left[\because \int u v d x=u \int v d v-\int \frac{d}{d x} u \int v d x\right]}$
$\begin{aligned} &\Rightarrow I=\left(-e^{x} \cos x\right)_{0}^{\frac{\pi}{4}}+\left(e^{x} \sin x\right)_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} e^{x} \sin x d x \\ & \end{aligned}$
$\Rightarrow I=\left(-e^{x} \cos x\right)_{0}^{\frac{\pi}{4}}+\left(e^{x} \sin x\right)_{0}^{\frac{\pi}{4}}-I[\therefore \text { from }(1)]$
$\begin{aligned} &\Rightarrow 2 I=-e^{\frac{\pi}{4}} \times \frac{1}{\sqrt{2}}-(-1)+e^{\frac{\pi}{4}} \times \frac{1}{\sqrt{2}}-0 \\ & \end{aligned}$
$\Rightarrow 2 I=1 \Rightarrow I=\frac{1}{2}$

Definite Integrals Exercise Revision Exercise Question 28

Answer:

Answer:$\frac{\pi}{4}-\frac{2}{3}$
Given: $\int_{0}^{\frac{\pi}{4}} \tan ^{4} x d x$
Hint: Use trigonometric identities and then integrate by parts
Solution:
$\int_{0}^{\frac{\pi}{4}} \tan ^{4} x d x=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x \tan ^{2} x d x$
$=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x\left(\sec ^{2} x-1\right) d x \\$
$\begin{aligned} & &=\int_{0}^{\frac{\pi}{4}} \tan ^{2} x \sec ^{2} x d x-\int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x-1\right) d x \end{aligned}$
$\begin{aligned} &p u t \\ & \end{aligned}$
$\tan x=t \\$ (Differentiate w.r.t to x)
$\sec ^{2} x d x=d t$
$\begin{aligned} &=\left(\frac{t^{3}}{3}\right)_{0}^{\frac{\pi}{4}}-\left(1-\frac{\pi}{4}\right)=\frac{1}{3}-1+\frac{\pi}{4} \\ & \end{aligned}$
$=\frac{\pi}{4}-\frac{2}{3}$

Definite Integrals Exercise Revision Exercise Question 29

Answer:$\frac{1}{2}$
Given:$\int_{0}^{1}|2 x-1| d x$
Hint: You must know how to open mode
Solution:$\int_{0}^{1}|2 x-1| d x$
$\begin{aligned} &=\int_{0}^{\frac{1}{2}}-(2 x-1) d x+\int_{\frac{1}{2}}^{1}(2 x-1) d x \\ & \end{aligned}$
$=\left(\frac{-2 x^{2}}{2}+x\right)_{0}^{\frac{1}{2}}+\left(\frac{2 x^{2}}{2}-x\right)^{\frac{1}{2}}$
$\begin{aligned} &=-\left(\frac{1}{2}\right)^{2}+\frac{1}{2}-\left(\frac{1}{2}\right)^{2}+\frac{1}{2} \\ & \end{aligned}$
$=-\frac{1}{4}+\frac{1}{2}-\frac{1}{4}+\frac{1}{2} \\$
$=\frac{-1}{2}+1$$=\frac{-1}{2}+1$
$= \frac{1}{2}$

Definite Integrals Exercise Revision Exercise Question 30

Answer:$2$
Given: $\int_{1}^{2}\left|x^{2}-2 x\right| d x$
Hint: Open the mode and then do integration by parts
Solution:
$\int_{1}^{2}\left|x^{2}-2 x\right| d x$
$\begin{aligned} &=\int_{1}^{2}-\left(x^{2}-2 x\right) d x+\int_{2}^{3}\left(x^{2}-2 x\right) d x \\ & \end{aligned}$
$=\left(\frac{-x^{3}}{3}+x^{2}\right)_{1}^{2}+\left(\frac{x^{3}}{3}-x^{2}\right)_{2}^{3}$
$\begin{aligned} &=\frac{-8}{3}+4+\frac{1}{3}-1+9-9-\frac{8}{3}+4 \\ & \end{aligned}$
$=7-\frac{15}{3} \\$
$=\frac{21-15}{3} \\$
$=\frac{6}{3}$
$= 2$

Definite Integrals Exercise Revision Exercise question 31

Answer:$2\sqrt{2}-2$
Hint: To solve this question we will split $\sin x.\cos x$ in two forms. $\left[\begin{array}{l} \because \sin x=\cos x \\ \tan x=1 \\ x=\frac{\pi}{4} \end{array}\right]$
Given:$\int_{0}^{\frac{\pi}{2}}|\sin x-\cos x| d x$
Solution:
$\begin{aligned} &\int_{0}^{\frac{\pi}{4}}(\sin x-\cos x) d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) d x \\ & \end{aligned}$
$=-[-\cos x-\sin x]_{0}^{\frac{\pi}{4}}+[-\cos x-\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$ $\left[\begin{array}{l} |x|=-x, x<0 \\ =x, x>0 \end{array}\right]$
$\begin{aligned} &=-[\cos x-\sin x]_{0}^{\frac{\pi}{4}}-[\cos x+\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \\ & \end{aligned}$
$=\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(1+0)-\left[(0+1)-\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)\right] \\$
$=2 \sqrt{2}-2$

Definite Integrals Exercise Revision Exercise question 36

Answer:$0$
Hint: To solve the equation, the value of negative is always zero
Given:$\int_{-a}^{a} \frac{x e^{x^{2}}}{1+x^{2}} d x$
Solution:
$\begin{aligned} &\int_{-a}^{a} f(x) d x=\int_{-a}^{a} f(a-a-x) d x \\ & \end{aligned}$
$=\int_{-a}^{a} f(-x) d x$
$\begin{aligned} &=-\int_{-a}^{a} f(x) d x \\ & \end{aligned}$
$I=\int_{-a}^{a} \frac{(-x) e^{x^{2}}}{1+x^{2}} d x \\$
$2 I=0 \\$
$I=0$

Definite Integrals Exercise Revision Exercise question 32

Answer:$\frac{2}{\pi}$
Hint: To solve this question we will do integration by separation
Given:$\int_{0}^{1}|\sin 2 \pi x| d x$
Solution:
$\int_{0}^{1}|\sin 2 \pi x| d x$
$\begin{aligned} &=\int_{0}^{\frac{1}{2}}(\sin 2 \pi x) d x+\int_{\frac{1}{2}}^{1}-\sin 2 \pi x d x \\ & \end{aligned}$
$=\left[\frac{-\cos 2 \pi x}{2 \pi}\right]_{0}^{\frac{1}{2}}+\left[\frac{\cos 2 \pi x}{2 \pi}\right]_{\frac{1}{2}}^{1}$
$\begin{aligned} &=\frac{-1}{2 \pi}[\cos \pi-\cos 0]+\frac{1}{2 \pi}[\cos 2 \pi-\cos \pi] \\ & \end{aligned}$
$=\frac{-1}{2 \pi}[-1-1]+\frac{1}{2 \pi}[1-(-1)] \\$
$=\frac{-1}{2 \pi} \times-2+\frac{1}{2 \pi} \times 2 \\$
$=\frac{1+1}{\pi} \\$
$=\frac{2}{\pi}$

Definite Integrals Exercise Revision Exercise question 33

Answer: $\frac{28}{3}$
Hint: To solve this question we need to use $\int x^{n}$ formula
Given:$\int_{1}^{3}\left|x^{2}-4\right| d x$ $\left\{\begin{array}{l} x^{2}-4, x^{2} \geq 4 \\\\ -\left(x^{2}-4\right), x^{2} \leq 4 \end{array}\right.$
$\begin{aligned} &I=-\int_{1}^{2}\left(x^{2}-4\right) d x+\int_{2}^{3}\left(x^{2}-4\right) d x \\ & \end{aligned}$
$I=-\left(\frac{x^{2}}{3}-4 x\right)_{1}^{2}+\left(\frac{x^{3}}{3}-4 x\right)_{2}^{3}$
$\begin{aligned} &I=-\left(\frac{8}{3}-8-\frac{1}{3}+4\right)+\left(\frac{27}{3}-12-\frac{8}{3}+8\right) \\ & \end{aligned}$
$I=-\frac{8}{3}+8+\frac{1}{3}-4+\frac{27}{3}-12-\frac{8}{3}+8 \\$
$I=\frac{28}{3}$

Definite Integrals Exercise Revision Exercise question 34

Answer: $0$
Hint: When the value of $f\left ( x \right )$ is odd then answer is zero
Given:$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{9} x d x$
Solution:$\begin{aligned} &f(x)=\sin ^{9} x \\ & \end{aligned}$
$x \rightarrow-x$ $\left[\int_{-a}^{a} f(x) d x=0, f(x)->o d d\right]$
$f(-x)=\sin ^{9}(-x)=>(-\sin x)^{9}$
$=-\sin ^{9} x$
$f(-x)=-f(x) \Rightarrow \mathrm{f}(\mathrm{x}) \text { is odd }$
$\int_{-a}^{a} f(x) d x=0, \text { when } \mathrm{f}(\mathrm{x}) \text { is odd }$
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{9} x d x=0$

Definite Integrals Exercise Revision Exercise question 35

Answer:$0$
Hint: When the value of $f\left ( x \right )$ is odd then answer is zero
Given:$\int_{\frac{1}{2}}^{\frac{1}{2}} \cos x \log \left(\frac{1+x}{1-x}\right) d x$
Solution:
$\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$ $\left[\int_{-a}^{a} f(x) d x=0, f(x)->o d d\right]$
$\begin{aligned} &f(x)=\cos x \log \left(\frac{1+x}{1-x}\right) \\ & \end{aligned}$
$f(-x)=\cos (-x) \log \left(\frac{1+(-x)}{1-(-x)}\right)$
$\begin{aligned} &=\cos x \log \left(\frac{1-x}{1+x}\right) \\ & \end{aligned}$
$f(-x)=-f(x)$
F(x) is odd function
$\int_{\frac{-1}{2}}^{\frac{1}{2}} \cos x \log \left(\frac{1+x}{1-x}\right) d x=0$

Definite Integrals Exercise Revision Exercise Question 37

Answer:$\frac{\pi}{4}$
Hint: To solve this equation we have to change cot into tan
Given:$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot ^{7} x} d x$
Solution:
$\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot ^{7} x} d x=-\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\frac{1}{\cot ^{7} x}} d x$ $I=\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{7} x}{1+\tan ^{7} x} d x$ ………….. (1)
$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot \left(0+\frac{\pi}{2}-x\right)^{3}} d x$
$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{7} x} d x$ …….. (2)
Adding equation (1) and (2)
$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{7} x}{1+\tan ^{7} x} d x+\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{7} x} d x$
$\begin{gathered} 2 I=\int_{0}^{\frac{\pi}{2}} d x \\ \end{gathered}$
$2 I=[x]_{0}^{\frac{\pi}{2}}$
$\begin{aligned} &2 I=\frac{\pi}{2}-0 \\ & \end{aligned}$
$I=\frac{\pi}{2 \times 2} \\$
$I=\frac{\pi}{4}$

Definite Integrals Exercise Revision Exercise Question 38

Answer: $0$
Hint: To solve this equation, we have to use f(x) formula
Given:$\int_{0}^{2 \pi} \cos ^{7} x d x$
Solution:
$\begin{aligned} &I=\int_{0}^{2 a} f(x) d x=\int_{0}^{a}[f(x)+f(2a-x)] d x \\ & \end{aligned}$
$I=\int_{0}^{\frac{\pi}{2}} \cos ^{7} x+\cos ^{7}(2 \pi-x) d x$
$\begin{aligned} &I=2 \int_{0}^{\frac{\pi}{2}} \cos ^{7} x d x \\ & \end{aligned}$

$I=2 \int_{0}^{\frac{\pi}{2}}\left(\cos ^{7} x-\cos ^{7} x\right) d x$
$I= 0$

Definite Integrals Exercise Revision Exercise Question 39

Answer: $\frac{a}{2}$
Hint: To solve this equation use formula $\int_{0}^{a} f(x) d x$
Given:
$\int_{0}^{a} f(x) d x$
Solution:
$\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\\ & \end{aligned}$
$I=\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}} d x\\$ ........(1)
$I=\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x$ ........(2)
Adding (1) and (2)
$I+I=\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}} d x+\int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x$
$\begin{aligned} &2 I=\int_{0}^{a} \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{x}} d x \\ & \end{aligned}$
$2 I=\int_{0}^{a} d x \\$
$2 I=(x)_{0}^{a}$
$\begin{aligned} &\quad \Rightarrow a-0=a \\ & \end{aligned}$
$2 I=a \\$
$I=\frac{a}{2}$

Definite Integrals Exercise Revision Exercise Question 40

Answer:$\frac{\pi}{6}$
Hint: To solve this equation we will use $\int f\left ( x \right )dx$ formula
Given: $\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{3} x} d x$
Solution:
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{3}\left(\frac{\pi}{2}-x\right)} d x \\ & \end{aligned}$ $\left[\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$\because \tan \left(\frac{\pi}{2}-x\right)=\cot x=\frac{1}{\tan x}$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{3} x}{\tan ^{3} x+1} d x$ …………. (1)
$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{\tan ^{3} x+1} d x$ ……………. (2)
$2 I=\int_{0}^{\frac{\pi}{2}}\left(\frac{1+\tan ^{3} x}{\tan ^{3} x+1}\right) d x$

$2 I=[x]_{0}^{\frac{\pi}{2}} \\$

$I=\frac{\pi}{4}$

Definite Integrals Exercise Revision Exercise Question 41

Answer:$\frac{\pi}{4}$
Hint: In this Statement, we will use $\int_{0}^{a} f(x) d x$ formula
Given:$\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$
Solution:
$\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$
$\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}$
$I=\int_{0}^{\pi} \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^{2}(\pi-x)} d x \\$
$=\int_{0}^{\pi} \frac{(\pi-x) \sin x}{1+\cos ^{2} x} d x$
$\begin{aligned} &=\int_{0}^{\pi} \frac{\pi \sin x}{1+\cos ^{2} x} d x-\int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x \\ & \end{aligned}$
$I=\int_{0}^{\pi} \frac{\pi \sin x}{1+\cos ^{2} x} d x-I$
$\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x \\ & \end{aligned}$
$2 I=\pi \int_{1}^{-1} \frac{-d t}{1+t^{2}}$
$\begin{aligned} &I=\frac{\pi}{2} \int_{1}^{-1} \frac{d t}{1+t^{2}} \\ & \end{aligned}$
$I=\frac{\pi}{2}\left[\tan ^{-1} t\right]_{-1}^{1} \\$
$I=\frac{\pi}{2}\left(\tan ^{-1}-\tan ^{-1}(-1)\right. \\$
$I=\frac{\pi}{2}\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]$
$\begin{aligned} &I=\frac{\pi}{2}\left[\frac{\pi}{2}\right] \\ & \end{aligned}$
$I=\frac{\pi}{4}$

Definite Integrals Exercise Revision Exercise Question 42

Answer:$I= \frac{\pi}{5}$
Hint: In this equation we have $\int_{0}^{a} f(x)$ formula
Given: $\int_{0}^{\pi} x \sin x \cos ^{4} x d x$
Solution:
$\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}$
$I=\int_{0}^{\pi}(\pi-x) \sin (\pi-x) \cos ^{4}(\pi-x) d x \\$
$=\int_{0}^{\pi}(\pi-x) \sin x \cdot \cos ^{4} x d x$
$\begin{aligned} &=\pi \int_{0}^{\pi} \sin x \cdot \cos ^{4} x d x-\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{4} x d x \\ & \end{aligned}$
$I=\frac{\pi}{2} \int_{0}^{\pi} \sin x \cdot \cos ^{4} x d x-I$
$I=-\frac{\pi}{2} \int_{1}^{-1} t^{4} d t$ $\left[\begin{array}{l} \cos x=t \\ \sin x d x=d t \\ x \rightarrow 0 \\ t \cos x=1 \end{array}\right]$
$\begin{aligned} &I=\frac{\pi}{2} \int_{-1}^{1} t^{4} d t \\ & \end{aligned}$
$I=\frac{\pi}{10}\left[t^{5}\right]_{-1}^{1} \\$
$I=\frac{\pi}{10}\left[1^{5}-(-1)^{5}\right] \\$
$I=\frac{\pi}{10}(1+1) \\$
$I=\frac{2 \pi}{10} \\$
$I=\frac{\pi}{5}$

Definite Integrals Exercise Revision Exercise Question 43

Answer:$\frac{\pi^{2}}{2ab}$
Hint: To solve this equation we will convert statement in terms of tan and sec.
Given:$\int_{0}^{\pi} \frac{x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x$
Solution:
$\begin{aligned} & \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}$
$I=\int_{0}^{\pi} \frac{x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x$
$\begin{aligned} I &=\int_{0}^{\pi} \frac{\pi-x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x \\ \end{aligned}$
$I =\int_{0}^{\pi} \frac{\pi}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x-\int_{0}^{\pi} \frac{\pi}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x$
$\begin{aligned} &2 I=\int_{0}^{\pi} \frac{\pi}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x \\ & \end{aligned}$
$2 I=\pi \int_{0}^{\pi} \frac{1}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x} d x \\$
$I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec ^{2} x}{a^{2}+b^{2} \tan ^{2} x} d x$
$\begin{aligned} &I=\frac{\pi}{2} \times 2 \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{a^{2}+b^{2} \tan ^{2} x} d x \\ & \end{aligned}$
$I=\frac{\pi}{b} \int_{0}^{\infty} \frac{1}{a^{2}+t^{2} d t}$ $\left[\begin{array}{l} b \tan x=t \\ b \sec ^{2} d x=d t \end{array}\right]$
$I=\frac{\pi}{b}\left[\frac{1}{a} \tan ^{-} \frac{t}{a}\right]_{0}^{\infty}$
$\begin{aligned} &I=\frac{\pi}{a b}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \\ & \end{aligned}$
$I=\frac{\pi}{a b}\left[\frac{\pi}{2}-0\right] \\$
$I=\frac{\pi^{2}}{2 a b}$

Definite Integrals Exercise Revision Exercise Question 44

Answer:$\log 2$
Hint: To solve this equation we convert tan in form of sec
Given: $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}|\tan x| d x$
Solution:
$\begin{aligned} &I=\int_{-\frac{\pi}{4}}^{0}-\tan x d x+\int_{0}^{\frac{\pi}{4}} \tan x d x \\ & \end{aligned}$
$I=\int_{0}^{-\frac{\pi}{4}} \tan x d x+\int_{0}^{\frac{\pi}{4}} \tan x d x$
$\begin{aligned} I &=[\log |\sec x|]_{0}^{-\frac{\pi}{4}}+[\log |\sec x|]_{0}^{\frac{\pi}{4}} \\ & \end{aligned}$
$=\log \sqrt{2}+\log \sqrt{2} \\$
$=2 \log \sqrt{2} \\$
$=2 \log (2)^{\frac{1}{2}} \\$
$=\log 2$

Definite Integrals Exercise Revision Exercise Question 45

Answer:$2-\sqrt{2}$
Hint: To solve this equation we will split the x
Given:$\int_{0}^{15}\left[x^{2}\right] d x$
Solution:$I= \int_{0}^{15}\left[x^{2}\right] d x$
$\begin{aligned} &I=\int_{0}^{\frac{3}{2}}(x)^{2} d x \\ & \end{aligned}$
$I=\int_{0}^{1} 0 d x+\int_{1}^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{\frac{3}{2}} 2 d x$
$\begin{aligned} &I=0+[x]_{1}^{\sqrt{2}}+[2 x]_{\sqrt{2}}^{\frac{3}{2}} \\ & \end{aligned}$
$I=(\sqrt{2}-1)+(3-2 \sqrt{2}) \\$
$I=2-\sqrt{2}$

Definite Integrals Excercise Revision Exercise Question 46

Answer: $\frac{\pi}{\sin \alpha}[\pi+\alpha]$
Hint: To solve this equation we convert cos and sin into tan
Given: $\frac{\pi}{\sin \alpha}[\pi+\alpha]$
Solution: $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$
$\begin{aligned} I &=\int_{0}^{\pi} \frac{\pi-x}{1+\cos \alpha \sin (\pi-x)} d x \\ \end{aligned}$
$I =\int_{0}^{\pi} \frac{\pi-x}{1+\cos \alpha \sin x} d x \\$
$I =\int_{0}^{\pi} \frac{\pi}{1+\cos \alpha \sin x} d x-\int_{0}^{\pi} \frac{x}{1+\cos \alpha \sin x} d x$
Let
$\begin{aligned} 2 I &=\int_{0}^{\pi} \frac{\pi}{1+\cos \alpha \sin x} d x \\ \end{aligned}$
$2 I =\int_{0}^{\pi} \frac{\pi}{1+\cos x \cdot \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$
$\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}-2 \cos \alpha \tan \frac{x}{2}} d x \\ & \end{aligned}$
$I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}-2 \cos \alpha \tan \frac{x}{2}+1} d x$
Let
$\begin{aligned} &\tan \frac{x}{2}=t \\ & \end{aligned}$
$\Rightarrow>\sec ^{2} \frac{x}{2} d x=2 d t \\$
$x=0 \rightarrow t=\tan 0=0 \\$
$x=x \rightarrow t=\tan \frac{\pi}{2}=\infty$
$\begin{aligned} &I=\frac{\pi}{2} \int_{0}^{\infty} \frac{2 d t}{t^{2}-2 \cos \alpha t+1} \\ & \end{aligned}$
$I=\pi \int_{0}^{\infty} \frac{d t}{(t-\cos \alpha)^{2}+\left(1-\cos ^{2} x\right)} \\$
$I=\pi \int_{0}^{\infty} \frac{d t}{\sin ^{2} \alpha+(t-\cos x)^{2}}$
$\begin{aligned} &I=\pi-\frac{1}{\sin \alpha} \cdot \tan ^{-1}\left(\frac{t-\cos \alpha}{\sin \alpha}\right)_{0}^{\infty} \\ & \end{aligned}$
$I=\frac{\pi}{\sin \alpha}\left(\tan ^{-1} \infty-\tan ^{-1}\left(\frac{-\cos \alpha}{\sin \alpha}\right)\right) \\$
$I=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}-\tan ^{-1}(-\cot \alpha)\right)$
$\begin{aligned} &=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\tan ^{-1}(\cot \alpha)\right) \\ & \end{aligned}$
$=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\tan ^{-1} \tan \left(\frac{\pi}{2}+\alpha\right)\right) \\$
$=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\left(\frac{\pi}{2}+\alpha\right)\right)$
$= \frac{\pi}{\sin \alpha}[\pi+\alpha]$

Definite Integrals Excercise Revision Exercise Question 47

Answer:$\frac{\pi^{2}}{16}$
Hint: To solve this we use $a^{4}+b^{4}, \int_{0}^{a} f(x)$ form
Given:$\int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$
Solution:
$I= \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$ ………. (1)
$\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x$
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \cos x \sin x}{\cos ^{4} x+\sin ^{4} x} d x \\ & \end{aligned}$ …….. (2)
$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x \sin x}{\cos ^{4} x+\sin ^{4} x} d x \\$
${\left[a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}\right]}$
$\begin{aligned} &I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\cos x \sin x}{\cos ^{2} x+\sin ^{2} x-2 \cos ^{2} x \cdot \sin ^{2} x} d x \\ & \end{aligned}$
$I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\cos x \sin x}{1-(2 \cos x \cdot \sin x)^{2}} d x$
$\begin{aligned} &\sin 2 x=2 \sin x \cdot \cos x \\ & \end{aligned}$
$\sin x \cos x=\frac{\sin 2 x}{2}$
$\begin{aligned} &I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin \frac{2 x}{2}}{1-\left(\frac{\sin 2 x}{2}\right)^{2}} d x \\ & \end{aligned}$
$I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin \frac{2 x}{2}}{\left(\frac{2-\sin ^{2} 2 x}{2}\right)} d x$
$\begin{aligned} &I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 x}{1+1-\sin ^{2} x} d x \\ & \end{aligned}$
$I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^{2} x} d x$
$\begin{aligned} &\cos 2 x=t \\ & \end{aligned}$
$-2 \sin 2 x=\frac{d t}{d x} \\$
$\sin 2 x d x=\frac{-d t}{2}$
$\begin{aligned} &I=\frac{\pi}{4} \times \frac{-1}{2} \int \frac{d t}{1+t^{2}} \\ & \end{aligned}$ $\left[\begin{array}{l} x=0, t=1 \\ x=\frac{\pi}{2}, t=-1 \end{array}\right]$
$I=\frac{-\pi}{8} \int_{+1}^{-1} \frac{d t}{1+t^{2}}$
$\int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x \\$
$I=\frac{\pi}{8} \int_{-1}^{1} \frac{d t}{1+t^{2}}$
$\begin{aligned} &I=\frac{\pi}{8}\left[\tan ^{-1} t\right]_{-1}^{1} \\ \end{aligned}$
$I=\frac{\pi}{8}\left[\tan ^{-1} 1-\tan ^{-1}(-1)\right] \\$
$I=\frac{\pi}{8}\left[\frac{\pi}{4}+\frac{\pi}{4}\right] \\$
$I=\frac{\pi}{8}\left(\frac{2 \pi}{4}\right) \\$
$I=\frac{\pi^{2}}{16}$

Definite Integrals Excercise Revision Exercise Question 48

Answer:$I=\frac{1}{\sqrt{2}} \ln |\sqrt{2}+1|$
Hint: To solve this question we use f(x) form
Given:$\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\sin x+\cos x} d x$
Solution:$I= \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\sin x+\cos x} d x$
$\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}$
$I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2}\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x$
$\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{\cos x+\sin x} d x \\ \end{aligned}$
$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\sin x+\cos x} d x+\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{\sin x+\cos x} d x \\$
$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x+\cos ^{2} x}{\sin x+\cos x} d x$ #check the steps and contents again
$\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2}(\sin x+\cos x)} d x\\ & \end{aligned}$
$2 I=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos x \frac{1}{\sqrt{2}}+\sin x \frac{1}{\sqrt{2}}} d x\\$
$2 I=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos x \cos \frac{\pi}{4}+\sin x \sin \frac{\pi}{4}} d x$
$\begin{aligned} &2 I=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos \left(x-\frac{\pi}{4}\right)} d x \\ & \end{aligned}$
$2 I=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \sec \left(x-\frac{\pi}{4}\right) d x$
$\begin{aligned} &\left.2 I=\frac{1}{\sqrt{2}}\left[\ln \mid \sec \left(x-\frac{\pi}{4}\right)+\tan \left(x-\frac{\pi}{4}\right)\right]\right]_{0}^{\frac{\pi}{2}} \\ & \end{aligned}$
$2 I=\frac{1}{\sqrt{2}}\left[\ln \mid \sec \left(\frac{\pi}{2}-\frac{\pi}{4}\right)+\tan \left(\frac{\pi}{2}-\frac{\pi}{4}\right)\right]-\ln \left[\sec \left(0-\frac{\pi}{4}\right)+\tan \left(0-\frac{\pi}{4}\right)\right]$
$\begin{aligned} &2 I=\frac{1}{\sqrt{2}}\left[\ln \left|\frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1) \sqrt{2}-1}\right|\right] \\ & \end{aligned}$
$2 I=\frac{1}{\sqrt{2}}\left[\ln \left|\frac{(\sqrt{2}+1)^{2}}{(\sqrt{2})^{2}-(1)^{2}}\right|\right.$
$\begin{aligned} &2 I=\frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{2}} \ln |\sqrt{2}+1| \\ & \end{aligned}$
$I=\frac{1}{\sqrt{2}} \ln |\sqrt{2}+1|$

Definite Integrals Excercise Revision Exercise Question 49

Answer:$-\frac{\pi}{2}$
Hint: To solve this equation we use By Part
Given:$\int_{0}^{\pi} \cos 2 x \log \sin x d x$
Solution:$\int f(x) g(x) d x\\$
$\downarrow$
By Parts
$\begin{aligned} &f(x) g(x) d x=\int \frac{d(f x)}{d x} \int g(x) d x(d x) \\ & \end{aligned}$
$f(x)=\log \sin x \\$
$g(x)=\cos 2 x$
$\begin{aligned} &\Rightarrow\left[\log (\sin x) \int \cos 2 x d x-\int \frac{d(\log \sin x)}{d x} \int \cos 2 x d x d x\right]_{0}^{\pi} \\ & \end{aligned}$
${\left[\log (\sin x) \frac{\sin 2 x}{2}-\int \frac{1}{\sin x} \cos x \cdot \frac{\sin 2 x}{x} d x\right]_{0}^{\pi}}$
$\begin{aligned} &\Rightarrow\left[\frac{1}{2} \sin 2 x \cdot \log \sin x-\frac{1}{2} \int 2 \cos ^{2} x d x\right]_{2}^{\pi} \\ & \end{aligned}$
$\Rightarrow\left[\frac{1}{2} \sin 2 x \cdot \log \sin x-\frac{1}{2} \int(\cos 2 x-1)\right]_{2}^{\pi} \\$
$\Rightarrow\left[\left[\frac{1}{2} \sin 2 x \cdot \log (\sin x)-\frac{1}{2}\left(\frac{\sin 2 x}{2}+x\right)\right]_{0}^{\pi}\right]$
$\begin{aligned} &I=\left[\frac{1}{2} \sin 2 x \cdot \log (\sin x)-\frac{1}{2} \frac{\sin 2 x}{2}-\frac{x}{2}\right]_{0}^{\pi} \\ & \end{aligned}$
$\frac{1}{2} \sin 2 \pi \cdot \log (\sin \pi)-\frac{1}{2} \frac{\sin 2 \pi}{2}-\frac{\pi}{2}-\frac{1}{2} \sin 0 \log \sin 0+\frac{1}{2} \sin 0+0 \\$
$=0-0-\frac{\pi}{2}-0+0+0 \\$
$=-\frac{\pi}{2}$

Definite Integrals Excercise Revision Exercise Question 50

Answer:$\frac{\pi^{2}}{2 a \sqrt{a^{2}-1}}$
Hint: To solve this equation we convert cos
Given:$\int_{0}^{\pi} \frac{x}{a^{2}-\cos ^{2} x} d x, a>1$
Solution:
Let $I=\int_{0}^{\pi} \frac{x}{a^{2}-\cos ^{2} x} d x$
$\begin{aligned} &I=\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}$
$I=\int_{0}^{\pi} \frac{\pi-x}{a^{2}-\cos ^{2}(\pi-x)} d x=\int_{0}^{\pi} \frac{\pi-x}{a^{2}-\cos ^{2} x} d x$
$\begin{aligned} &2 I=\int_{0}^{\pi} \frac{\pi}{a^{2}-\cos ^{2} x} d x \\ & \end{aligned}$
$2 I=\pi \int_{0}^{\pi} \frac{1}{2 a}\left[\frac{1}{1+a \cos x}+\frac{1}{1-\cos x}\right] d x \\$
$2 I=\frac{\pi}{2 a} \int_{0}^{\pi} \frac{1}{a+\cos x} d x+\frac{\pi}{2 a} \int_{0}^{\pi} \frac{1}{1-a \cos x} d x$
$\begin{aligned} &2 I=\frac{\pi}{a} \int_{0}^{\pi} \frac{1}{a+\cos x} d x \\ & \end{aligned}$
$\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
$\begin{aligned} &2 I=\frac{\pi}{a} \int_{0}^{\pi} \frac{1}{a+\frac{1-\tan ^{2} \frac{x}{2}}{\sec ^{2} \frac{x}{2}}} d x\\ & \end{aligned}$
$2 I=\frac{\pi}{a} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{(a-1) \tan ^{2} \frac{x}{2}+a+1} d x$
$\begin{aligned} &2 I=\frac{\pi}{a} \int_{0}^{\pi} \frac{\frac{d}{d x} \tan \frac{x}{2} d x}{(a-1) \tan ^{2} \frac{x}{2}+(a+1)} \\ & \end{aligned}$
$I=\frac{\pi}{a(a-1)} \int_{0}^{\pi} \frac{d \tan \frac{x}{2}}{\tan ^{2} \frac{x}{2}+\frac{a+1}{a-1}}$
$\begin{aligned} &I=\frac{\pi}{a(a-1)} \times \sqrt{\frac{a+1}{a+1}}\left|\tan ^{-1} \frac{\tan \frac{x}{2}}{\sqrt{\frac{a+1}{a-1}}}\right|_{0}^{\pi} \\ & \end{aligned}$
$I=\frac{\pi}{a \sqrt{a^{2}-1}}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)$
$\begin{aligned} &=\frac{\pi}{a} \times \frac{1}{a \sqrt{a^{2}-1}} \times \frac{\pi}{2} \\ & \end{aligned}$
$=\frac{\pi^{2}}{2 a \sqrt{a^{2}-1}}$

Definite Integrals Exercise Revision Exercise question 51
Answer: $I=\pi\left(\frac{\pi}{2}-1\right)$

Hint: To solve this equation we split tan x and sec x
Given:$\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$
Solution:$I=\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$
$I=\int_{0}^{\pi} \frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} d x$ $\left[\begin{array}{l} \because \tan x=\frac{\sin x}{\cos x} \\ \sec =\frac{1}{\cos x} \end{array}\right]$
$I=\int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x$
$\begin{aligned} &I=\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}$
$I=\int_{0}^{\pi} \frac{(\pi-x) \sin (\pi-x)}{1+\sin (\pi-x)} d x \\$
$I=\int_{0}^{\pi} \frac{(\pi-x) \sin x}{1+\sin x} d x$
$\begin{aligned} &I=\int_{0}^{\pi} \frac{\pi \sin x-x \sin x}{1+\sin x} d x \\ & \end{aligned}$
$I=\int_{0}^{\pi} \frac{\sin x}{1+\sin x} d x-\int_{0}^{\pi} \frac{x \sin x}{1+\sin x} d x \\$
$I=\int_{0}^{\pi} \frac{\sin x}{1+\sin x} d x-I$
$2 I=\pi \int_{0}^{\pi} \frac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)} d x$
$\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{\sin x-\sin ^{2} x}{\cos ^{2} x} d x \\ & \end{aligned}$
$I=\pi \int_{0}^{\pi}\left(\frac{\sin x}{\cos ^{2} x}-\frac{\sin x}{\cos ^{2} x}\right) d x$
$\begin{aligned} &I=\pi \int_{0}^{\pi}\left(\frac{1}{\cos x} \frac{\sin x}{\cos x}-\tan ^{2} x\right) d x \\ & \end{aligned}$
$I=\pi \int_{0}^{\pi} \sec x \tan x d x-\pi \int_{0}^{\pi} \tan ^{2} x d x$
$\begin{aligned} &I=\pi \int_{0}^{\pi} \sec x \tan x d x-\pi \int_{0}^{\pi} \sec ^{2} x d x+\pi \int_{0}^{\pi} 1 d x \\ & \end{aligned}$
$2 I=\pi[\sec x]_{0}^{\pi}-\pi[\tan x]_{0}^{\pi}+\pi[x]_{0}^{\pi}$
$\begin{aligned} &2 I=\pi(\sec \pi-\sec 0)-\pi(\tan \pi-\tan 0)+\pi(\pi-0) \\ & \end{aligned}$
$2 I=\pi(-1-1)-\pi(-0)+\pi^{2} \\$
$I=\frac{-2 \pi+\pi^{2}}{2} \\$
$I=\pi\left(\frac{-2}{2}+\frac{\pi}{2}\right) \\$
$I=\pi\left(\frac{\pi}{2}-1\right)$

Definite Integrals Exercise Revision Exercise question 52

Answer: $\frac{}{}$$\frac{1}{2}$
Hint: To use this equation, we use $\int_{0}^{a} f(x) d x$ formula
Given:$\int_{2}^{3} \frac{\sqrt{x}}{\sqrt{5-x}+\sqrt{x}} d x$
Solution:
$I=\int_{2}^{3} \frac{\sqrt{x}}{\sqrt{5-x}+\sqrt{x}} d x$ ……………. (1)
$\begin{aligned} &I=\int_{2}^{3} \frac{\sqrt{5-x}}{\sqrt{5-(5-x)}+\sqrt{5-x}} d x \\ & \end{aligned}$
$I=\int_{2}^{3} \frac{\sqrt{5-x}}{\sqrt{x}+\sqrt{5-x}} d x$ ……………… (2)
By adding (1) and (2)
$\begin{aligned} &2 I=\int_{2}^{3} \frac{\sqrt{x}+\sqrt{5-x}}{\sqrt{x}+\sqrt{5-x}} d x \\ & \end{aligned}$
$2 I=\int_{2}^{3} d x \Rightarrow 2 I=[x]_{3}^{2} \\$
$2 I=[3-2] \\$
$I=\frac{1}{2}$

Definite Integrals Exercise Revision Exercise question 53

Answer:$\frac{2}{\sqrt{2}} \log (\sqrt{2}+1)$
Hint: To solve this equation we use $\int_{0}^{a} f\left ( x \right )$ formula
Given:$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{\sin x+\cos x} d x$
Solution:
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2}\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x \\ & \end{aligned}$
$=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos x+\sin x} d x$
$\begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{\cos x+\sin x} d x \\ & \end{aligned}$
$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x+\sin ^{2} x}{\cos x+\sin x} d x$
$\begin{aligned} I &=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\sqrt{2} \cos x+\frac{1}{\sqrt{2}} \sin x} d x \\ & \end{aligned}$
$=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos \left(x-\frac{\pi}{4}\right)} d x \\$
$=\frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \sec \left(x-\frac{\pi}{4}\right) d x$
$\begin{aligned} &=\frac{1}{2}\left\{\log \sec \left|x-\frac{\pi}{9}\right|+\tan \left|x-\frac{\pi}{9}\right|\right\}_{0}^{\frac{\pi}{2}} \\ & \end{aligned}$
$=\frac{1}{\sqrt{2}} \log \left|\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right|-\log \left|\sec \left(-\frac{\pi}{4}\right)+\tan \left(-\frac{\pi}{4}\right)\right| \\$
$=\frac{1}{\sqrt{2}}\{\log (\sqrt{2}+1)-\log (\sqrt{2}-1)\}$
$\begin{aligned} &=\frac{1}{\sqrt{2}} \log \left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) \\ & \end{aligned}$
$=\frac{1}{\sqrt{2}} \log \left(\frac{(\sqrt{2}+1)^{2}}{2-1}\right) \\$
$=\frac{2}{\sqrt{2}} \log (\sqrt{2}+1)$

Definite Integrals Exercise Revision Exercise question 54

Answer: $\frac{\pi^{2}}{8}$
Hint: In this equation, we use $\int x^{n}dx$ formula
Given:$\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin ^{2} x+\cos ^{2} x} d x$
Solution:
$I=\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin ^{2} x+\cos ^{2} x} d x$
$I=\int_{0}^{\frac{\pi}{2}} x d x$ $\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]$
$\begin{aligned} &I=\left[\frac{x^{2}}{2}\right]_{0}^{\frac{\pi}{2}} \\ & \end{aligned}$
$I=\frac{\pi^{2}}{8}-0 \\$
$I=\frac{\pi^{2}}{8}$

Definite Integrals Exercise Revision Exercise question 55

Answer:$0$
Hint: In this equation we use $f\left ( x \right )= -f\left ( x \right )$
Given:$\int_{-\pi}^{\pi} x^{10} \sin ^{7} x d x$
Solution:
$\begin{aligned} &I=f(x) \Rightarrow-f(x) \\ & \end{aligned}$
$\int_{-a}^{a} f(x) d x=0 \\$
$I=\int_{-\pi}^{\pi} x^{10} \sin ^{7} x d x$
$\begin{aligned} &I=\int_{-\pi}^{\pi} x^{10} \sin ^{7} x d x \\ & \end{aligned}$
$f(x)=x^{10} \sin ^{7} x d x \\$
$\begin{aligned} &f(-x)=(-x)^{10} \sin ^{7}(-x) d x \\ & \end{aligned}$
$f(-x)=-x^{10} \sin ^{7} x \\$
$f(-x)=-f(x) \\$
$\int_{-a}^{a} f(x) d x=0 \\$
$\int_{-\pi}^{\pi} x^{10} \sin ^{7} x d x=0$

Definite Integrals Exercise Revision Exercise question 56

Answer:$I=\frac{\pi}{2}-2 \log \sqrt{2}$
Hint: To this equation convert cot into tan
Given:$\int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right) d x$
Solution:
$\begin{aligned} &\int_{0}^{1} \tan ^{-1}\left(\frac{1}{1-x+x^{2}}\right) d x \\ & \end{aligned}$
$\int_{0}^{1} \tan ^{-1}\left(\frac{1}{(1-x)(1-x)}\right) d x \\$
$=\int_{0}^{1} \tan ^{-1}\left(\frac{x+1-x}{(1-x)(1-x)}\right)$
$\begin{aligned} &=\int_{0}^{1}\left(\tan ^{-1} x+\tan ^{-1}(1-x)\right) d x \\ & \end{aligned}$
$=\int_{0}^{1} \tan ^{-1} x d x+\int_{0}^{1} \tan ^{-1}(1-x) d x$
$\tan^{-1}x=t$
Put $x=\tan t$
$\begin{aligned} &1=\sec ^{2} \frac{d t}{d x} \\ & \end{aligned}$
$d x=\sec ^{2} d t \\$
$I=\int \tan ^{-1} x d x \\$
$=\int t \sec ^{2} t d t$
$\begin{aligned} &=t \int \sec ^{2} t d t-\int \tan t d t \\ & \end{aligned}$
$=t \tan t-\int \tan t d t \\$
$=t \tan t-\log |\sec t|$
$\begin{aligned} &I_{2}=\int \tan ^{-1}(1-x) d x\\ & \end{aligned}$
$\text { Put }$ $1-x=m\\$
$-1=\frac{d m}{d x}\\$
$d x=-d m\\$
$\text { Put }$
$\begin{aligned} &1=\sec ^{2} \frac{v d v}{d m} \\ & \end{aligned}$
$m=\tan v \\$
$d m=\sec ^{2} v d v \\$
$I_{2}=-\int \tan ^{-1} m d n \\$
$=-\int v \cdot \sec ^{2} v d v$
$\begin{aligned} &=(v \tan v-\log |\sec v|) \\ & \end{aligned}$
$\int \cot ^{-1}\left(1-x+x^{2}\right) d x=I_{1}+I_{2} \\$
$=t \tan t-\log \sec t|-v \tan v-\log | \sec v \mid \\$
$=t \tan t-\log \log |\sec t|-v \tan v+\log |\sec v| \\$
$t=\tan ^{-1} x \\$
$\tan t=x \\$
$\tan v=m=1-x$
$\begin{aligned} &=\left[x \tan ^{-1}-\log \sqrt{1+x^{2}}-(1-x) \tan ^{-1}(1-x)+\log \left(\sqrt{1-x^{2}+1}\right]_{0}^{1}\right.\\ & \end{aligned}$
$=\left(\frac{\pi}{4}-\log \sqrt{2}\right)-\left(\frac{-\pi}{4}+\log \sqrt{2}\right)$
$\begin{aligned} &=\frac{\pi}{4}-\log \sqrt{2}+\frac{\pi}{4}-\log \sqrt{2} \\ & \end{aligned}$
$=\frac{\pi}{2}-2 \log \sqrt{2}$

Definite Integrals Exercise Revision Exercise Question 57

Answer: $\frac{\pi}{\sqrt{35}}$
Hint: To solve this equation, we convert cos x into tan x
Given: $\int_{0}^{\pi} \frac{d x}{6-\cos x}$
Solution
$I= \int _{0}^{\pi}\frac{dx}{6-\left ( \frac{1-\tan ^{2}\frac{x}{2}}{1+\tan\frac{x}{2}} \right )}$
$=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{6+6 \tan ^{2} \frac{x}{2}-1+\tan ^{2} \frac{x}{2}} d x$
$\begin{aligned} &=\frac{1}{7} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{\frac{5}{7}+\tan ^{2} \frac{x}{2}} d x \\ & \end{aligned}$
$\text { Let } \tan \frac{x}{2}=t \\$
$\frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t$
$\begin{aligned} &\sec ^{2} \frac{x}{2} d x=2 d t \\ & \end{aligned}$
$I=\frac{2}{7} \int_{0}^{\pi} \frac{d t}{\sqrt{\frac{5}{7}}^{2}+t^{2}}$
$\begin{aligned} &I=\frac{2}{7} \frac{1}{\sqrt{\frac{5}{7}}}\left(\tan ^{-1} \frac{t}{\sqrt{\frac{5}{7}}}\right)_{0}^{\infty} \\ & \end{aligned}$
$=\frac{2}{\sqrt{35}}\left(\frac{\pi}{2}-0\right) \\$
$I=\frac{\pi}{\sqrt{35}}$

Definite Integrals Exercise Revision Exercise Question 58

Answer:$\frac{-1}{2 \sqrt{5}} \log \left[\frac{2 \sqrt{5}-\sqrt{5}-3}{2 \sqrt{5}+\sqrt{5}-3}\right]$
Hint: To solve this we convert sin and cos into tan
Given:
$\int_{0}^{\frac{\pi}{2}} \frac{1}{2 \cos x+4 \sin x} d x$
Solution:
Let
$\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \frac{1}{2 \cos x+4 \sin x} d x \\ & \end{aligned}$
$I=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos x+2 \sin x} d x$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2}}{1-\tan ^{2} \frac{x}{2}+4 \tan ^{2} \frac{x}{2}} d x$
Let
$\begin{aligned} &t=\tan ^{2} \frac{x}{2} \\ & \end{aligned}$
$\frac{d t}{d x}=\sec ^{2} \frac{x}{2} \times \frac{1}{2}$
$2 d t=\sec ^{2} \frac{x}{2} d x$
Now
$\begin{aligned} &\frac{1}{2} \times 2 \int_{0}^{1} \frac{d t}{1-t^{2}+4 t} \\ & \end{aligned}$
$I=-\int_{0}^{1} \frac{d t}{t^{2}-4 t-1} \\$
$I=-\int_{0}^{1} \frac{d t}{t^{2}-4 t-1+2^{2}-2^{2}}$
$\begin{aligned} &I=-\int_{0}^{1} \frac{d t}{(t-2)^{2}-(\sqrt{5})^{2}} \\ & \end{aligned}$
$I=\frac{-1}{2 \sqrt{5}} \log \left|\frac{t-2-\sqrt{5}}{t-2+\sqrt{5}}\right|_{0}^{1} \\$
$I=\frac{-1}{2 \sqrt{5}} \log \left|\frac{-1-\sqrt{5}}{-1+\sqrt{5}}\right|-\log \left|\frac{-2-\sqrt{5}}{-2+\sqrt{5}}\right|$
$\begin{aligned} &=\frac{-1}{2 \sqrt{5}} \log \left[\left|\frac{-1-\sqrt{5}}{-1+\sqrt{5}}\right| \times\left|\frac{-2-\sqrt{5}}{-2-\sqrt{5}}\right|\right] \\ & \end{aligned}$
$=\frac{-1}{2 \sqrt{5}} \log \left(\frac{2-\sqrt{5}+2 \sqrt{5}-5}{2+\sqrt{5}-2 \sqrt{5}-\sqrt{5}}\right) \\$
$=\frac{-1}{2 \sqrt{5}} \log \left(\frac{2 \sqrt{5}-\sqrt{5}-3}{-2 \sqrt{5}+\sqrt{5}-3}\right)$

Definite Integrals Exercise Revision Exercise Question 59

Answer: $\tan ^{-1}2-\frac{\pi}{4}$
Hint: To solve this we assume cosec x and cot x in t
Given:$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\cos e c x \cot x}{1+\operatorname{cosec}^{2} x} d x$
Solution:
$\begin{aligned} &\cos e c x=t \\ & \end{aligned}$
$\operatorname{cosec} x \cdot \cot x d x=d t \\$
$\int_{2}^{1} \frac{-d t}{1+t^{2}}=-\left[\tan ^{-1} t\right]_{0}^{1} \\$
$=-\left(\frac{\pi}{4}-\tan ^{-1} 2\right) \\$
$=\tan ^{-1} 2-\frac{\pi}{4}$

Definite Integrals Exercise Revision Exercise Question 60

Answer:

Answer:$\frac{1}{2 \sqrt{5}} \log \left|\frac{1-\frac{1}{2}-\frac{\sqrt{5}}{2}}{1-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|-\log \left|\frac{1+\sqrt{5}}{1-\sqrt{5}}\right|$
Hint: To this equation we convert cos and sin in terms of tan
Given:$\int_{0}^{\frac{\pi}{2}} \frac{d x}{4 \cos x+2 \sin x}$
Solution:$I= \int_{0}^{\frac{\pi}{2}} \frac{d x}{4 \cos x+2 \sin x}$
$\left[\begin{array}{l} \sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \\ \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}} \end{array}\right]$
$\begin{aligned} I &=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(\frac{4-4 \tan ^{2} \frac{x}{2}}{4+4 \tan ^{2} \frac{x}{2}}\right)+\frac{4 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} \\ \end{aligned}$
$I =\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} \frac{x}{2} d x}{4-4 \tan ^{2} \frac{x}{2}+4 \tan \frac{x}{2}}$
$\begin{aligned} &{\left[\begin{array}{l} \tan \frac{x}{2}=t \\ \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t \\ \sec ^{2} \frac{x}{2} d x=2 d t \end{array}\right]} \\ & \end{aligned}$
$I=\int \frac{2 d t}{4-4 t^{2}-4 t} \\$
$I=-\frac{1}{2} \int \frac{d t}{t^{2}-2 t-1}$
$\begin{aligned} I &=-\frac{1}{2} \int \frac{d t}{t^{2}+2 t-1} \\ \end{aligned}$
$I =-\frac{1}{2} \int \frac{d t}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}}$
$\begin{aligned} &I=-\frac{1}{2} \frac{1}{2 \times \frac{\sqrt{5}}{2}} \log \left|\frac{t-\frac{1}{2}-\frac{\sqrt{5}}{2}}{t-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right| \\ & \end{aligned}$
$I=-\frac{1}{2 \sqrt{5}} \log \left|\frac{\tan \frac{x}{2}-\frac{1+\sqrt{5}}{2}}{\tan \frac{x}{2}-\frac{1-\sqrt{5}}{2}}\right|_{0}^{\frac{\pi}{2}}$
$\begin{aligned} &\frac{1}{2 \sqrt{5}} \log \left|\frac{1-\frac{1}{2}-\frac{\sqrt{5}}{2}}{1-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|-\log \left|\frac{1+\sqrt{5}}{1-\sqrt{5}}\right| \\ & \end{aligned}$
$I=\frac{1}{2 \sqrt{5}} \log \left|\frac{1-\frac{1}{2}-\frac{\sqrt{5}}{2}}{1-\frac{1}{2}+\frac{\sqrt{5}}{2}}\right|-\log \left|\frac{1+\sqrt{5}}{1-\sqrt{5}}\right|$

Definite Integrals Excercise Revision Exercise Question 61

Answer:$1$
Hint: To solve the integration we have to just integrate x and put the limits.
Given:$\int_{0}^{1} x d x$
Solution:$\int_{0}^{1} x d x$
$\begin{aligned} &=\left[\frac{x^{2}}{2}\right]_{0}^{1} \\ & \end{aligned}$
$=\frac{1-0}{2} \\$
$=\frac{1}{2}$

Definite Integrals Excercise Revision Exercise Question 62

Answer:$\frac{34}{3}$
Hint: To solve the given statement we have to integrate them individually
Given:$\int_{0}^{2}\left(2 x^{2}+3\right) d x$
Solution:
$\int_{0}^{2}\left(2 x^{2}+3\right) d x$
$\begin{aligned} &=\int_{0}^{2} 2 x^{2} d x+\int_{0}^{2} 3 d x \\ & \end{aligned}$
$=2\left[\frac{x^{3}}{3}\right]_{0}^{2}+3[x]_{0}^{2} \\$
$=\frac{163}{3}+6 \\$
$=\frac{34}{3}$

Definite Integrals Excercise Revision Exercise Question 63

Answer:$\frac{57}{2}$
Hint: To solve the given question we have to integrate them individually.
Given:$\int_{1}^{4}\left(x^{2}+x\right) d x$
Solution: $\int_{1}^{4}\left(x^{2}+x\right) d x$
$\begin{aligned} &=\int_{1}^{4} x^{2} d x+\int_{1}^{4} x d x \\ & \end{aligned}$
$=\left[\frac{x^{3}}{3}\right]_{1}^{4}+\left[\frac{x^{2}}{2}\right]_{1}^{4} \\$
$=\frac{64-1}{3}+\frac{16-1}{2} \\$
$=\frac{63}{3}+\frac{15}{2} \\$
$=\frac{42+15}{2} \\$
$=\frac{57}{2}$

Definite Integrals Exercise Revision Exercise Question 64

Answer:$\frac{1}{2} \frac{e^{4}-1}{e^{2}}$
Hint: To solve the given statement we have to use the formula.
Given:$\int_{-1}^{1} e^{2 x} d x$
Solution:
$\begin{aligned} &\int_{a}^{b} f(x) d x=\frac{b-a}{n} \lim _{x \rightarrow \infty}[f(a)+f(a+h)+\ldots f(a+(h-1)] \\ & \end{aligned}$
$f(x)=e^{2 x} \\$
$a=-1, b=1, h=\frac{1+1}{n}=\frac{2}{n}$
$\begin{aligned} &\int_{1}^{1} e^{2 x} d x=\frac{2}{n^{2}} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2(a+h)}+\ldots .+e^{2(a+(n-1) h)}\right] \\ &\end{aligned}$
$=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2 a} e^{2 h}+\ldots .+e^{2 a} e^{2(n-1) h}\right] \\$
$=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}\left(1+e^{2 h}+\ldots .+e^{2(n-1) h}\right]\right. \\$
$=\frac{2}{n} e^{-2} \lim _{h \rightarrow \infty}\left[1+e^{2 h}+\ldots .+e^{2(n-1) h}\right]$
$\begin{aligned} &a=1, r=e^{2 h} \\ & \end{aligned}$
$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1} \\$
$\int_{1}^{1} e^{2 x} d x=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2 a+2 h}+\ldots .+e^{2 a+2(n-1) h}\right]$
$\begin{aligned} &=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2 a} e^{2 h}+\ldots .+e^{2 a} e^{2(n-1) h}\right] \\ & \end{aligned}$
$=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}\left(1+e^{2 h}+\ldots .+e^{2(n-1) h}\right]\right. \\$
$=\frac{2}{n e^{2}} \lim _{h \rightarrow \infty}\left[\frac{1 \cdot\left(e^{2 / n}-1\right)}{e^{2 h}-1}\right] \\$
$=\frac{2}{e^{2}} \lim _{h \rightarrow \infty}\left[\frac{1 \cdot\left(e^{2 m}-1\right)}{n e^{2 h}-1}\right]$
$=\frac{2}{e^{2}} \lim _{h \rightarrow \infty}\left[\frac{\left.e^{2 n} e^{2 / h}-1\right)}{\left(e^{2 n}-1\right) \frac{2}{h}}\right]$
$\begin{aligned} &h=\frac{2}{n} \Rightarrow n=\frac{2}{h} \\ & \end{aligned}$
$n \rightarrow \infty \\$
$n \rightarrow 0$
$\begin{aligned} &\int_{1}^{1} e^{2 x} d x=\frac{2}{n} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2 a+2 h}+\ldots .+e^{2 a+2(n-1) h}\right] \\ & \end{aligned}$
$=\frac{2}{n^{2}} \lim _{h \rightarrow \infty}\left[e^{2 a}+e^{2 a} e^{2 h}+\ldots .+e^{2 a} e^{2(n-1) h}\right] \\$
$=\frac{2}{n^{2}} \lim _{h \rightarrow \infty}\left[e^{2 a}\left(1+e^{2 h}+\ldots .+e^{2(n-1) h}\right]\right. \\$
$=\frac{2}{n e^{2} \lim _{h \rightarrow \infty}}\left[\frac{1 \cdot\left(e^{2 m n}-1\right)}{e^{2 h}-1}\right] \\$
$=\frac{2}{e^{2}} \lim _{h \rightarrow \infty}\left[\frac{1 \cdot\left(e^{2 n n}-1\right)}{n e^{2 h}-1}\right]$
$\begin{aligned} &=\frac{2}{e^{2}} \times\left(e^{4}-1\right) \lim _{n \rightarrow 0}\left[\frac{1}{\left(\frac{e^{2 n}-1}{h}\right) \times 2}\right] \\ & \end{aligned}$
$=\frac{2}{e^{2}} \times\left(e^{4}-1\right) \lim _{n \rightarrow 0}\left[\frac{1}{\left(\frac{e^{2 n}-1}{h}\right) \times 4}\right]$
$\begin{aligned} &=\frac{2\left(e^{4}-1\right)}{4 e^{2}} \lim _{n \rightarrow 0}\left[\frac{1}{\left(\frac{e^{2 n}-1}{h}\right)}\right] \\ & \end{aligned}$
$=\frac{\left(e^{4}-1\right)}{2 e^{2}} \\$
$\int_{-1}^{1} e^{2 x} d x=\frac{1}{2} \frac{e^{4}-1}{e^{2}}$

Definite Integrals Exercise Revision Exercise Question 65

Answer:$\frac{1}{e^{2}}\left[1-\frac{1}{e}\right]$
Hint: To solve the given statement use the formula of $e^{-x}$
Given:$\int_{2}^{3} e^{-x} d x$
Solution:$\int_{2}^{3} e^{-x} d x$
$\begin{aligned} &=\left[-e^{-x}\right]_{2}^{3} \\ & \end{aligned}$
$=\left(-e^{-3}\right)-\left(-e^{-2}\right) \\$
$=-e^{-3}+e^{-2} \\$
$=-\frac{1}{e^{3}}+\frac{1}{e^{2}} \\$
$=\frac{1}{e^{2}}\left[1-\frac{1}{e}\right]$

Definite Integrals Exercise Revision Exercise Question 66

Answer:$\frac{112}{3}$
Hint: To solve the given statement we have to integrate them individually.
Given: $\int_{1}^{3}\left(2 x^{2}+5 x\right) d x$
Solution:
$\begin{aligned} &\int_{1}^{3} 2 x^{2} d x+\int_{1}^{3} 5 x d x \\ &=2\left[\frac{x^{3}}{3}\right]_{1}^{3}+5\left[\frac{x^{2}}{2}\right]^{3} \\ & \end{aligned}$
$=2\left[\frac{27-1}{3}\right]+5\left[\frac{9-1}{2}\right] \\$
$=2\left[\frac{26}{3}\right]+5\left[\frac{8}{2}\right] \\$
$=\frac{52}{3}+20 \\$
$=\frac{52+60}{3} \\$
$=\frac{112}{3}$

Definite Integrals Exercise Revision Exercise question 67

Answer:$\frac{62}{3}$
Hint: To solve the given statement we have to integrate them individually.
Given:$\int_{1}^{3}\left(x^{2}+3 x\right) d x$
Solution:
$\begin{aligned} &\int_{1}^{3}\left(x^{2}+3 x\right) d x \\ &\int_{1}^{3}\left(x^{2}+3 x\right) d x \\ &=\int_{1}^{3} x^{2} d x+\int_{1}^{3} 3 x d x \end{aligned}$
$\begin{aligned} &=\left[\frac{x^{3}}{3}\right]_{1}^{3}+3\left[\frac{x^{2}}{2}\right]_{1}^{3} \\ &=\frac{27-1}{3}+3\left[\frac{9-1}{2}\right] \\ &=\frac{26}{3}+\frac{24}{2} \\ &=\frac{26+36}{3} \\ &=\frac{62}{3} \end{aligned}$

Definite Integrals Exercise Revision Exercise question 68

Answer: $\frac{20}{3}$
Hint: To solve the given statement we have to integrate them individually
Given:$\int_{0}^{2}\left(x^{2}+2\right) d x$
Solution:$\int_{0}^{2}\left(x^{2}+2\right) d x$
$\begin{aligned} &=\int_{0}^{2} x^{2} d x+\int_{0}^{2} 2 d x \\ &=\left[\frac{x^{3}}{3}\right]_{0}^{2}+[2 x]_{0}^{2} \\ &=\frac{8}{3}+4 \\ &=\frac{8+12}{3} \\ &=\frac{20}{3} \end{aligned}$

Definite Integrals Exercise Revision Exercise question 69

Answer:$12$
Hint: To solve the given statement we have to integrate them individually
Given:$\int_{0}^{3}\left(x^{2}+1\right) d x$
Solution:
$\begin{aligned} &=\int_{0}^{3} x^{2} d x+\int_{0}^{3} 1 d x \\ &=\left[\frac{x^{3}}{3}\right]_{0}^{3}+[x]_{0}^{3} \\ &=\left[\frac{27}{3}-\frac{0}{3}\right]+[3-0] \\ &=9+3 \\ &=12 \end{aligned}$

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