RD Sharma Solutions Class 12 Mathematics Chapter 19 RE

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.

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

• Chapter 19 - Definite Integrals - Ex-19.1

• Chapter 19 - Definite Integrals - Ex-19.2

• Chapter 19 - Definite Integrals - Ex-19.3

• Chapter 19 - Indefinite Integrals - Ex-19.4(a)

• Chapter 19 - Definite Integrals - Ex-19.4(b)

• Chapter 19 - Definite Integrals - Ex-19.5

• Chapter 19 - Definite Integrals - Ex-FBQ

• Chapter 19 - Definite Integrals - Ex-MCQ

• Chapter 19 - Definite Integrals - Ex-VSA

Definite Integrals Excercise:RE

Definite Integrals Excercise Revision Exercise Question 7

Answer: $\frac{\pi }{2}-\log 2$
Given:${\int }_0^1{\tan }^{-1}\left(\frac{2x}{1+x^2}\right)dx$
Hint: Use substitution method and then apply the formula of $\tan 2\theta$
Solution:
${\int }_0^1{\tan }^{-1}\left(\frac{2x}{1+x^2}\right)dx$
Let $\mathrm{x}=\tan \theta$
$dx={\sec }^2\theta d\theta$ (differentiate w.r.t x )
Now, $0<x<1$
$0<\tan \theta <1$
$0<\theta <\frac{\pi }{4}$
Now,
$\begin{array}{rl}& {\int }_0^{\frac{\pi }{4}}{\tan }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right){\sec }^2\theta d\theta \\ & \end{array}$
$=2{\int }_0^{\frac{\pi }{4}}\theta {\sec }^2\theta d\theta$
$=2{[\theta \int {\sec }^2\theta d\theta -\int \frac{d}{d\theta }\theta \int {\sec }^2\theta d\theta ]}_0^{\frac{\pi }{4}}$
$\begin{array}{rl}& [\because \int uvdx=u\int vdx-\int \frac{d}{dx}u\int vdx]\\ & \end{array}$
$=2[(\theta \tan \theta {)}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}\tan \theta d\theta ]$
$\begin{array}{rl}& =2(\frac{\pi }{4}-0)-2{\int }_0^{\frac{\pi }{4}}\tan \theta d\theta \\ & \end{array}$
$=\frac{\pi }{2}-2{\int }_0^{\frac{\pi }{4}}\frac{\sin \theta }{\cos \theta }d\theta$
Let $\cos \theta =t$
$-\sin \theta d\theta =\mathrm{dt}$ (differentiate w.r.t )
$\begin{array}{rl}& =\frac{\pi }{2}+2{\int }_0^{\frac{\pi }{4}}\frac{1}{t}dt\\ & \end{array}$
$=\frac{\pi }{2}+2(\log |t|{)}_0^{\frac{\pi }{4}}$
$\begin{array}{rl}& =\frac{\pi }{2}+2(\log |\cos \theta |{)}_0^{\frac{\pi }{4}}\\ & \end{array}$
$=\frac{\pi }{2}+2(\log \frac{1}{\sqrt{2}}-\log 1)$
$=\frac{\pi }{2}+2\log \frac{1}{\sqrt{2}}$ $(\because \log 1=0)$
$=\frac{\pi }{2}-\log 2$

Definite Integrals Exercise Revision Exercise question 11

Answer:$2-\frac{\pi }{2}$
Given:${\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{(1+\cos x{)}^2}dx$
Hint:
Use the formula and then apply integration rule by parts method.
Solution:
$\begin{array}{rl}& {\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{(1+\cos x{)}^2}dx\\ & \end{array}$
$={\int }_0^{\frac{\pi }{2}}\frac{1-{\cos }^{2}x}{(1+\cos x{)}^2}dx$
$\begin{array}{rl}& ={\int }_0^{\frac{\pi }{2}}\frac{1-\cos x}{(1+\cos x)}dx\\ & \end{array}$
$={\int }_0^{\frac{\pi }{2}}\frac{1-\cos x-1+1}{1+\cos x}dx$
$\begin{array}{rl}& ={\int }_0^{\frac{\pi }{2}}\frac{2-(1+\cos x)}{1+\cos x}dx\\ & \end{array}$
$={\int }_0^{\frac{\pi }{2}}\frac{2}{1+\cos x}dx-{\int }_0^{\frac{\pi }{2}}\frac{1+\cos x}{1+\cos x}dx$
$\begin{array}{rl}& =2{\int }_0^{\frac{\pi }{2}}\frac{(1-\cos x)}{(1-{\cos }^{2}x)}dx-\frac{\pi }{2}\\ & \end{array}$
$=2{\int }_0^{\frac{\pi }{2}}\frac{(1-\cos x)}{{\sin }^{2}x}dx-\frac{\pi }{2}$
$\begin{array}{rl}& =2{\int }_0^{\frac{\pi }{2}}\cos e{c}^{2}x-\mathrm{cosec}x\cdot \cot xdx-\frac{\pi }{2}\\ & \end{array}$
$=2(-\cot x+\cos ecx{)}_0^{\frac{\pi }{2}}-\frac{\pi }{2}$
$=2(-0+1+0)-\frac{\pi }{2}$
$=2-\frac{\pi }{2}$