TITLE: Integrate $\int_0^1 \sin^{-1}{\frac{x^2}{1+x^2}}dx$
QUESTION [6 upvotes]: Integrate $\int_0^1 \sin^{-1}{\frac{x^2}{1+x^2}}dx$

I tried to put $x=\tan\theta$, which gives
$\int_0^{\frac{\pi}{4}} {\sin^{-1}({\sin^2\theta}})\sec^2\theta d\theta$, but I don't know how to proceed after this.
Is there something I am missing here?

REPLY [6 votes]: Hint: 
You might know the property that $$\int_a^b f(x) dx+\int_{f(a)}^{f(b)} f^{-1} (x) dx=bf(b)-af(a)$$
Using this we get $$\int_0^1 \arcsin \left( \frac {x^2}{1+x^2}\right) dx+\int_0^{\frac {\pi}{6}} \sqrt {\frac {\sin x}{1-\sin x}} dx=\frac {\pi}{6}$$
Therefore $$\int_0^1 \arcsin \left( \frac {x^2}{1+x^2}\right) dx=\frac {\pi}{6}- \int_0^{\frac {\pi}{6}} \sqrt {\frac {\sin x}{1-\sin x}} dx$$ 
Now for the integral on right hand side use the  substitution $u=\sin x$ which will lead you to a very simple integral $$\int_0^{\frac 12} \sqrt {\frac {u}{1-u}} \frac {du}{\sqrt {1-u^2}}$$ Which can be solved as stated by Jack D'Aurizio