Proving an integral relation for the following Question, i had to prove this : 
that for every $$-1\le y \le 1 \\
\arcsin(y) + \arccos(y) = \frac{\pi}{2}$$
NOTE: this I've shown this using basic trigonometric id's
and (probably) somehow use this to prove the following : 
$$\int_0^{\sin^2(x)}\arcsin(\sqrt {t})dt + \int_0^{\cos^2(x)}\arccos(\sqrt {t})dt = \frac{\pi}{4} $$ 
I've been working quite some time on this one, and will appreciate your help on proving this, thank you.

Let
$$f(x)=\int_0^{\sin^2(x)}\arcsin(\sqrt {t})dt + \int_0^{\cos^2(x)}\arccos(\sqrt {t})dt  $$ 
then we can see that
$$f'(x)=2\sin(x)\cos(x)\arcsin(\sin x)-2\sin(x)\cos(x)\arccos(\cos x)=0$$
so 
$$f(x)=f\left(\frac{\pi}{4}\right)=\int_0^{1/2}\arccos(\sqrt{t})dt+\int_0^{1/2}\arcsin(\sqrt{t})dt=\frac{1}{2}\times\frac{\pi}{2}=\frac{\pi}{4}$$
