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

Kuldeep MauryaUpdated on 24 Jan 2022, 02:21 PM IST

RD Sharma class 12th exercise RE has become the holy grail of students in India. Hundreds of students have already placed their trust in RD Sharma Solutions and have experienced the magic of the book. The RD Sharma class 12 chapter 19 exercise RE is a must-have from their series and should be availed by all aspiring students. RD Sharma Solutions The solutions in the book will be of immense help to students preparing for board exams.

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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:
$\begin{aligned} &=\int_{0}^{3} x^{2} d x+\int_{0}^{3} 1 d x \\ &=\left[\frac{x^{3}}{3}\right]_{0}^{3}+[x]_{0}^{3} \\ &=\left[\frac{27}{3}-\frac{0}{3}\right]+[3-0] \\ &=9+3 \\ &=12 \end{aligned}$

The 19th chapter in class 12 mathematics syllabus consist of many concepts that are challenging for the students to understand. This chapter consists of five exercises, ex 19.1 to ex 19.5. The concepts that these exercises cover is Evaluation of Definite Integrals, solving sums with Integrals and Riemann Integral formulas. There are 69 questions in total including its subparts given in the RD Sharma Class 12 Solutions Definite Integrals RE book. to help the students cover all these sums easily without cracking their heads, the Class 12 RD Sharma Chapter 19 Exercise RE Solutions helps a lot.

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