Alternative way to $\int \sin^{-1}\sqrt{ \frac{x}{a+x}} dx$ . 
$$\int \sin^{-1}\sqrt{ \frac{x}{a+x}} dx$$

We can substitute it as $x=a\tan^2 (\theta)$ . Then:
$$2a\int \theta \tan (\theta)\sec^2 (\theta) d\theta$$
Using integration by parts will be enough here.  But I wanted to know if this particular problem can be solved by any other method. Because the above method is quite lengthy. 
 A: I don't see anything lengthy here.
$$\int \theta\frac{\sin\theta}{\cos^3\theta}d\theta=-\frac\theta{2\cos^2\theta}+\frac12\int\frac{d\theta}{\cos^2\theta}$$ and the last integral is immediate.
A: With Integration by parts we get
$$x\arcsin(\sqrt{\frac{x}{a+x}})-\int\frac{1}{2}\sqrt{\frac{ax}{(a+x)^2}}dx$$
now we get by substituting
$$u=\sqrt{x},du=\frac{1}{2\sqrt{x}}dx$$ we get
$$x\arcsin(\sqrt{\frac{x}{a+x}})-\int\frac{\sqrt{a}u^2}{a+u^2}du$$
Write the last Integrand in the form
$$\sqrt{a}\int 1-\frac{a}{u^2+a}du$$
Can you proceed?
A: Method 1: 
$$I=2a\int \theta\cdot \tan \theta\cdot  \sec^2 \theta d\theta$$
write it as 
$$I=2a\int \theta\cdot (\sec \theta\cdot  \sec \theta \tan \theta) d\theta$$
now apply integration by parts 
Method 2:
Directly apply integration by parts
thus $$I=x \sin^{-1}\sqrt{ \frac{x}{a+x}} -\int \frac{1}{2} \sqrt{\frac{ax}{(a+x)^2}} dx$$
put $t=\sqrt x$ thus $dt=\frac{dx}{2 \sqrt x}$
$$I=x \sin^{-1}\sqrt{ \frac{x}{a+x}} -\sqrt{a}\int \frac{t^2}{a+t^2}dt$$
$$I=x \sin^{-1}\sqrt{ \frac{x}{a+x}} -\sqrt{a}t+a tan^{-1}(\frac{t}{\sqrt a})+C$$
$$I=x \sin^{-1}\sqrt{ \frac{x}{a+x}} -\sqrt{ax}+a tan^{-1}(\frac{\sqrt{x}}{\sqrt a})+C$$
A: Hint:
Let $\sin^{-1}\sqrt{\dfrac x{x+a}}=y\ge0\implies\tan^2y=\dfrac xa\implies y=\tan^{-1}\sqrt{\dfrac xa}$
$$\int\sin^{-1}\sqrt{\dfrac x{x+a}}dx=\int\tan^{-1}\sqrt{\dfrac xa}dx$$
$$=\tan^{-1}\sqrt{\dfrac xa}\int dx-\int\left(\dfrac{d \tan^{-1}\sqrt{\dfrac xa}}{dx}\int dx\right)dx$$
$$=x\tan^{-1}\sqrt{\dfrac xa}-\int\dfrac{x}{2\sqrt a\sqrt x\left(1+\dfrac xa\right)}dx$$
For the last part, choose $\sqrt x=u$
