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.
 A: 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 
$$f(x)=f\left(\frac{\pi}{4}\right)=\int_0^{1/2}\arccos(\sqrt{t})dt+\int_0^{1/2}\arcsin(\sqrt{t})dt=\frac{1}{2}\times\frac{\pi}{2}=\frac{\pi}{4}$$
A: 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, $C=\arccos 0+\arcsin 0=\dfrac\pi2+0$
A: Here is a proof
$$  \arcsin(y) + \arccos(y) = \frac{\pi}{2} \implies \sin( \arcsin(y) + \arccos(y) ) = 1 $$
$$ \implies y.y+\sqrt{1-y^2}. \sqrt{1-y^2} = 1 $$
$$ \implies 1=1. $$
Note: We used the identities 
1) 
$$ \sin(a+b)= \sin(a)\cos(b)+\sin(b)\cos(a), $$
$$ \sin(\arcsin(x))=x,\quad  \sin(\arccos(x))=\sqrt{1-x^2}= \cos(\arcsin(x)).$$
