prove that $\int_{0}^{\cos^2{x}}\arccos\sqrt{t}\ dt + \int_{0}^{\sin^2{x}}\arcsin\sqrt{t}\ dt =\frac{\pi}{4}$

Question

I have to prove the following problem:

$$\int_{0}^{\cos^2{x}}\arccos\sqrt{t}\ dt + \int_{0}^{\sin^2{x}}\arcsin\sqrt{t}\ dt =\frac{\pi}{4}$$

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I know that $\arccos(x)+\arcsin(x)=\frac{\pi}{2}$ but now, I don't know what more to do...

Answer 1

Let $ \arccos (\sqrt t) = z \Rightarrow t = \cos^2(z) $. Now the limits become, $z : \frac\pi2 \to x$

Similarly do with the second one by using $\arcsin$.

So, $$\color{blue}{I = \int_\frac\pi2^x z (-2\cos z\sin z)dz+\int_0^x z (2\cos z\sin z)dz }$$

Using the facts that,

$\int_a^b-f(x)dx = \int_b^af(x)dx$
and
$\int_a^bf(x)dx+\int_b^cf(x)dx = \int_a^cf(x)dx$

$$\color{blue}{I = \int_0^{\frac\pi2} (2z\cos z\sin z)dz = \int_0^{\frac\pi2} z\sin(2z)dz = \frac\pi4} $$