Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 52.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 52.
 

Answers (1)

Answer :   a\left [ \frac{\pi}{2} -1\right ]

Hint: Use indefinite formula and the given limits to solve this integral

Given:

\int_{a}^{0}\sin^{-1}\sqrt{\frac{x}{a+x}}dx

Solution:

\int_{a}^{0}\sin^{-1}\sqrt{\frac{x}{a+x}}dx

Putting x=a\tan ^{2}\theta

\Rightarrow dx=2a \tan \theta \cdot\sec^ {2}\theta d\theta
 When x=0\Rightarrow \theta=0  and

x=a\Rightarrow \theta =\frac{\pi}{4}
\begin{aligned} &\therefore \int_{0}^{a} \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=\int_{0}^{\pi / 4} \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a+a \tan ^{2} \theta}} 2 a \tan \theta \cdot \sec ^{2} \theta d \theta \\ &=\int_{0}^{\pi / 4} \sin ^{-1} \sqrt{\frac{a \tan ^{2} \theta}{a\left(1+\tan ^{2} \theta\right)}} 2 a \cdot \tan \theta \cdot \sec ^{2} \theta d \theta \\ &=\int_{0}^{\pi / 4} \sin ^{-1} \sqrt{\frac{\operatorname{an}^{2} \theta}{\sec ^{2} \theta}} 2 a \tan \theta \cdot \sec ^{2} \theta d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because 1+\tan ^{2} \theta=\sec ^{2} \theta\right] \end{aligned}

\begin{aligned} &=\int_{0}^{\pi / 4} \sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right) \cdot 2 a \tan \theta \cdot \sec ^{2} \theta d \theta \\ &=\int_{0}^{\pi / 4} \sin ^{-1}\left(\frac{\sin \theta}{\cos \theta \cdot \frac{1}{\cos \theta}}\right) 2 a \tan \theta \cdot \sec ^{2} \theta d \theta \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \therefore \frac{\sin A}{\cos A}=\tan A \\ \frac{1}{\cos A}=\sec A \end{array}\right] \\ &=\int_{0}^{\pi / 4} \theta \cdot 2 a \tan \theta \cdot \sec ^{2} \theta d \theta\; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \because \sin ^{-1}(\sin A)=A \end{aligned}

Again putting \tan \theta=u

\Rightarrow \sec^2\theta d\theta =du
 When \theta =0 then u=0

and \theta =\frac{\pi}{4} then u=1

\begin{aligned} &\therefore \int_{0}^{a} \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=2 a \int_{0}^{1} \tan ^{-1} u \cdot u \cdot d u\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \tan \theta=u \Rightarrow \theta=\tan ^{-1} u\right] \\ &=2 a \int_{0}^{1} u \tan ^{-1} u d u \end{aligned}

Applying integration by parts, method then

\begin{aligned} &=2 a\left\{\left[\tan ^{-1} u \int u d u\right]_{0}^{1}-\int_{0}^{1}\left[\frac{d}{d u}\left(\tan ^{-1} u\right) \int u d u\right] d u\right\} \\ &=2 a\left\{\left[\tan ^{-1} u \frac{u^{2}}{2}\right]_{0}^{1}-\int_{0}^{1}\left[\frac{1}{1+u^{2}} \cdot \frac{u^{2}}{2}\right] d u\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\begin{array}{l} \because \frac{d\left(\tan ^{-1} x\right)}{d x}=\frac{1}{1+x^{2}} \\ \left.\& \int x d x=\frac{x^{2}+1}{x+1}-\frac{x^{2}}{2}\right] \end{array}\right. \end{aligned}

\begin{aligned} &=2 a\left\{\left[\frac{u^{2}}{2} \tan ^{-1} u\right]_{0}^{1}-\frac{1}{2} \int_{0}^{1} \frac{u^{2}+1-1}{1+u^{2}} d u\right\} \\ &=2 a\left\{\left[\frac{1}{2} \tan ^{-1}(1)-\frac{0}{2} \cdot \tan ^{-1} 0\right]-\frac{1}{2} \int_{0}^{1}\left(\frac{1+u^{2}}{1+u^{2}}-\frac{1}{1+u^{2}}\right) d u\right\} \\ &=2 a\left\{\left[\frac{1}{2} \cdot \frac{\pi}{4}-0\right]-\frac{1}{2} \int_{0}^{1}\left(1-\frac{1}{1+u^{2}}\right) d u\right\} \quad\left[\begin{array}{l} \tan ^{-1} 1=\frac{\pi}{4} \\ \tan ^{-1} 0=0 \end{array}\right] \end{aligned}

\begin{aligned} &=2 a \cdot \frac{1}{2} \cdot \frac{\pi}{4}-a \int_{0}^{1} 1 d u+a \int_{0}^{1} \frac{1}{1+\mathrm{u}^{2}} d u \\ &=\frac{a \pi}{4}-a\left[\frac{u^{0+1}}{0+1}\right]_{0}^{1}+a\left[\tan ^{-1} u\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int 1 d x=x \& \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \\ &=\frac{\pi a}{4}-a[u]_{0}^{1}+a\left[\tan ^{-1}(1)-\tan ^{-1} 0\right] \end{aligned}

\begin{aligned} &=\frac{\pi a}{4}-a[1-0]+a \cdot\left[\frac{\pi}{4}-0\right] \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan ^{n} 1=\frac{\pi}{4}, \tan ^{-1} \theta=0\right] \\ &=\frac{\pi a}{4}-a+\frac{\pi a}{4} \\ &=\frac{2 \pi a}{4}-a \\ &=a\left(\frac{\pi}{2}-1\right) \end{aligned}

 

Posted by

infoexpert24

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 result verification 2024 application out at cbse.gov.in; last date, fees
anam-khan

CBSE results 2024 later for 102 students over fake certificates; board issues show cause notice
anam-khan

What after CBSE Class 12 results 2024? From CUET UG to JEE Advanced, all you need to know

Latest Question