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

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

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

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

$\Rightarrow I=a\int \theta ·2\tan \theta {\sec }^2\theta d\theta$

Using by parts considering $\theta$ as the first function, we get,

$I=a\left(\theta {\tan }^2\theta -\tan \theta +\theta \right)+c$, where $c$ is the constant of integration.

$\Rightarrow I=\left(x+a\right){\tan }^{-1}\sqrt{\frac{x}{a}}-\sqrt{ax}+c$