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  • 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}

 

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