$\int {{{\sin }^{ - 1}}} \sqrt {\frac{x}{{a + x}}} dx$

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

Let's put $x = a{\tan ^2}\theta$

$\Rightarrow$ $dx = 2a\tan \theta {\sec ^2}\theta d\theta$

therefore,$I = \int {{{\sin }^{ - 1}}} \sqrt {\frac{{a{{\tan }^2}\theta }}{{a + a{{\tan }^2}\theta }}} \left( {2a\tan \theta \cdot {{\sec }^2}\theta } \right)d\theta$

$= 2a\int {{{\sin }^{ - 1}}} \left( {\frac{{\tan \theta }}{{\sec \theta }}} \right)\tan \theta \cdot {\sec ^2}\theta d\theta$

$= 2a\int {{{\sin }^{ - 1}}} (\sin \theta )\tan \theta \cdot {\sec ^2}\theta d\theta$

$= \mathop {2a\int \theta \cdot \tan \theta {{\sec }^2}\theta d\theta }\limits_{{\rm{I}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}}$

$\quad$ I $\quad$ II
$= 2a\left[ {\theta \cdot \int {\tan } \theta \cdot {{\sec }^2}\theta d\theta - \int {\left( {\frac{d}{{d\theta }}\theta \cdot \int {\tan } \theta \cdot {{\sec }^2}\theta d\theta } \right)} d\theta } \right]$

$\left[ {\begin{array}{llllllllllllllllllll}{{\rm{ Let's put }}\,\,\,\,\,\,\,\,\,\,\tan \theta = t}\\{ \Rightarrow \sec \theta \cdot \tan \theta \cdot d\theta = dt}\\{ \Rightarrow \int {\tan } \theta {{\sec }^2}\theta d\theta = \int t dt}\end{array}} \right]$

$= 2a\left[ {\theta \cdot \frac{{{{\tan }^2}\theta }}{2} - \int {\frac{{{{\tan }^2}\theta }}{2}} d\theta } \right]$

$= a\theta {\tan ^2}\theta - a\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta$

$= a\theta \cdot {\tan ^2}\theta - a\tan \theta + a\theta + C$

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