Evaluating: $I_1 = \int\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx$

Question

$$I_1 =\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx= ?$$

I tried substitution: $\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) = \Xi$, but then I'm not able to do anything after the resulting integral.

Could someone help? There must be a simple way to solve this...

Answer 1

Hint. Assume $a>0,\,x+a>0$. Integrating by parts gives $$\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) -\int x\left( \frac1{2x} \frac{\sqrt{ax}}{x+a}\right)dx \tag1 $$ and the last integral is easy to evaluate $$\int \frac{\sqrt{x}}{x+a}\:dx=2\int \frac{u^2}{u^2+a}\:du\quad (\sqrt{x}=u). \tag2$$ Bringing $(1)$ and $(2)$ together leads to

$$\int \sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) dx=x\sin^{-1} \left(\sqrt{\frac{x}{x+a}}\;\right) -\sqrt{ax}+ a\sqrt{x}\arctan \left( \sqrt{\frac{x}{a}}\right)+C.$$

Answer 2

A classic substitution in cases where $\sqrt\frac{x}{x+a}$ is present is $x = a \tan^2(u)$. Then $dx = 2a \tan(u) \sec^2 (u) du$,

$$\int u (2a \tan(u) \sec^2(u)) du$$

which can be evaluated easily using integration by parts.