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 $u=\arcsin t, t=\sin u, dt =\cos u du\implies \dfrac{du}{dt}=\dfrac1{\cos u}=\dfrac1{\sqrt{1-t^2}}$
Similarly, $\dfrac{d\arccos t}{dt}=-\dfrac1{\sqrt{1-t^2}}$
So, $$\dfrac{d(\arccos t+\arcsin t)}{dt}=0$$
$$\implies \arccos t+\arcsin t=C$$ where $C$ is an arbitrary constant of indefinite integral.
Now, $C=\arccos 0+\arcsin 0=\dfrac\pi2+0$
