Evaluate $\int_0^\frac{\pi}{4} \cos^{-1}({\sin x}) \,dx$ I came across this integral in a math competition and couldn't solve it
$$\int_0^\frac{\pi}{4} \cos^{-1}({\sin x})\, dx$$
I tried a $u$-substitution, with $u=\sin x$ and ended up with the integral $$\int_0^\frac{\pi}{4} \cos^{-1}(\text{u}) \cdot \frac{1}{\sqrt{1-u^2}}\,du$$ which is not much simpler and I cannot figure out how to solve this. Any hints/solutions for this problems? 
I also tried drawing a triangle for the problem but it didn't really help with the solution.
 A: You can use the trigonometric formula :
$$\forall x \in\mathbb{R},\cos^{-1}(x)+\sin^{-1}(x)=\frac{\pi}{2}$$
Thus $\forall x\in\left[0,\frac{\pi}{4}\right]$ you have $\sin^{-1}(\sin(x))=x$ so :
$$\cos^{-1}(\sin(x))+\sin^{-1}(\sin(x))=\frac{\pi}{2}\Rightarrow \cos^{-1}({\sin x})=\frac{\pi}{2}-x$$
Finally :
$$\int_0^\frac{\pi}{4} \cos^{-1}({\sin x}) dx=\int_0^\frac{\pi}{4} \frac{\pi}{2}-xdx$$
Can you finish ?
A: HINT:
If $\cos^{-1}(\sin x)=y,0\le y\le\pi\ \ \ \ (1)$
and $\cos y=\sin x=\cos\left(\dfrac\pi2-x\right)$
$y=2m\pi\pm\left(\dfrac\pi2-x\right)$ where $m$ is an integer such that $(1)$ is satisfied 
For $0\le2m\pi+\dfrac\pi2-x\le\pi\iff 2m\pi-\dfrac\pi2\le x\le2m\pi+\dfrac\pi2$
Here $m=0$
A: Here's a barely different route to take, continuing from the direction you've taken.
First, note that substituting $u=\sin x$ would actually give
$$\int_{x=0}^{x=\pi/4}\cos^{-1}(\sin x)\,\mathrm dx=\int_{\color{red}{u=0}}^{\color{red}{u=1/\sqrt2}}\frac{\cos^{-1}u}{\sqrt{1-u^2}}\,\mathrm du$$
Now, recall that $\dfrac{\mathrm d}{\mathrm du}\cos^{-1}u=-\dfrac1{\sqrt{1-u^2}}$, which means you can make another intermediate substitution of, say, $v=\cos^{-1}u$. Then you have
$$-\int_{v=\cos^{-1}0=\pi/2}^{v=\cos^{-1}(1/\sqrt2)=\pi/4}v\,\mathrm dv=\int_{\pi/4}^{\pi/2}v\,\mathrm dv=\dfrac{3\pi^2}{32}$$
which agrees with the other answers above.
A: $$\int \cos^{-1}(\sin x) dx$$ $$= \int 1 \cdot \cos^{-1}(\sin x) dx = x \cdot \cos^{-1}(\sin x) - \int x \cdot -\frac{\cos x}{\sqrt{1-\sin^{2}x}} dx + C$$ (integration by parts)
$$ = x \cdot \cos^{-1}(\sin x) + \int x \cdot \frac{\cos x}{\sqrt{1-\sin^{2}x}} dx + C $$
$$ = x \cdot \cos^{-1}(\sin x) + \int x \cdot \frac{\cos x}{\sqrt{\cos^2 x}} dx + C $$
$$= x \cdot \cos^{-1}(\sin x) \pm \int x \cdot \frac{\cos x}{\cos x} dx + C$$
$$ = x \cdot \cos^{-1}(\sin x) \pm \frac{x^2}{2} + C$$
Now do it for the definite integral.
A: 1)
$(u^2)^{\prime}=2\times u\times u^{\prime}$
therefore $\dfrac{1}{2}u^2$ is an antiderivative of $u\times u^{\prime}$
2) derivative of $\text{cos}^{-1}(x)=-\dfrac{1}{\sqrt{1-x^2}}$
$\displaystyle J= \int_0^{\tfrac{\pi}{4}} \text{cos}^{-1}(\sin x)dx$
Perform the change of variable $y=\sin x$,
$\displaystyle J=\int_0^{\tfrac{\sqrt{2}}{2}} \dfrac{\text{cos}^{-1}(x)}{\sqrt{1-x^2}}dx$
Using $(1)$,
$\begin{align}J&=-\dfrac{1}{2}\Big[\text{cos}^{-1}
(x)^2\Big]_0^{\tfrac{\sqrt{2}}{2}}\\
&=-\dfrac{1}{2}\left[\dfrac{\pi^2}{16}-\dfrac{\pi^2}{4}\right]\\
&=\boxed{\dfrac{3}{32}\pi^2}
\end{align}$
