Evaluate :$\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx$ How to evaluate
$$
\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx
$$
I know that $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ but after that I have no idea, so please help me. Thanks in advance.
I tried this way,
$$
\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\frac{\pi}{2}}dx
$$
then I put the value $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ so 
$$
\frac{2}{\pi}\int\left(\sin^{-1} \sqrt{x} -\left(\frac{\pi}{2}-\sin^{-1} \sqrt{x}\right)\right)dx
$$
Is this right?
after that I integrate by part and get,
$$ \int \frac{\sqrt{x}}{\sqrt{1-x}}$$ now,what can i do?
 A: Let $$ I_0=\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx $$$$\Rightarrow I_0=\int\frac{\frac{\pi}{2}-2\cos^{-1}\sqrt{x}}{\frac{\pi}{2}}dx$$$$\Rightarrow I_0=\int (1-\frac{4}{\pi}\cos^{-1}\sqrt{x})dx$$ $$\Rightarrow I_0=x-\frac{4}{\pi}\int\cos^{-1}\sqrt{x})dx$$ Now Consider $$I_1= \int\cos^{-1}\sqrt{x}dx$$ $$\Rightarrow I_1=\int 2z\cos^{-1} zdz$$ Where $$ x=z^2$$ Hence Integrating by parts we get  $$ I_1 = 2zcos^{-1}z+ \int \frac{z^2}{\sqrt{1-z^2}}dz$$ $$I_1 = 2zcos^{-1}z+ \int \frac{1}{\sqrt{1-z^2}}dz-\int\sqrt{1-z^2}dz$$ $$\int \frac{1}{\sqrt{1-z^2}}dz=-\cos^{-1}z$$ $$\int\sqrt{1-z^2}dz=\frac{z\sqrt{1-z^2}}{2}+\frac{1}{2}\sin^{-1}z$$
A: *

*Using the relation $\arcsin(\sqrt{x}) + \arccos(\sqrt{x}) = \frac{\pi}{2}$ (which is valid for $0 \leqslant x  \leqslant 1$, so this must an implicit assumption in your problem) solve for $\arccos(\sqrt{x})$ and substitute that into the integrand.

*After that make a $u$-substitution $u = \arcsin(\sqrt{x})$. This should lead to $\int \left( \frac{4 u}{\pi} - 1\right) \sin(2u) \mathrm{d} u$. This can be integrated by parts.

A: Given what you know, you should be able to get the answer if you can get $$\int\arcsin\sqrt x\,dx$$ and you can get that starting with the substitution $u=\arcsin\sqrt x$ ($\sin u=\sqrt x$, $x=\sin^2u$, $dx=2\sin u\cos u$, etc. )
