#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 113 Maths Textbook Solution.

Answer:$a\left[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}\right]+C$

Hint: to solve this statement we will assume sin as tan

Given:   $\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$

Solution:

$\mathrm{I}=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$

$\text { Put } x=a \tan ^{2} t=>\tan t=\sqrt{\frac{x}{a}}$

$d x=a\left(2 \tan (t) \sec ^{2} t d t\right.$

$I=\int \sin ^{-1} \sqrt{\frac{\operatorname{atan}^{2} t}{a+\operatorname{atan}^{2} t}} a\left(2 \tan (t) \sec ^{2} t d t\right.$

$I=2 a \int \operatorname{ttan}(t) \sec ^{2} t d t$

$\text { Let t be the first function and } \tan (t) \sec ^{2} t d t \text { be the} e \text { second, applying Byparts }$

$I=a\left(\operatorname{ttan}^{2} t-\tan t+t\right)+C$

$I=a\left(\tan ^{2} t+t-\tan t\right)+C$

$I=a\left[\frac{x}{a} \tan ^{-1} \sqrt{\frac{x}{a}}+\tan ^{-1} \sqrt{\frac{x}{a}}-\sqrt{\frac{x}{a}}\right]+C$

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