Evaluate the following integrals:

Let $I=\int \cos ^{-1}(\sin x) d x$

We know $\sin \theta=\cos \left(90^{\circ}-\theta\right)$

Therefore, we can write the integral as

$I=\int \cos ^{-1}\left[\cos \left(\frac{\pi}{2}-x\right)\right] d x$

$\Rightarrow I=\int\left(\frac{\pi}{2}-x\right) d x$

$\Rightarrow I=\int \frac{\pi}{2} d x-\int x d x$

$\Rightarrow I=\frac{\pi}{2} \int d x-\int x d x$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$ and $\int d x=x+c$

$\Rightarrow I=\frac{\pi}{2} \times x-\frac{x^{1+1}}{1+1}+c$

$\therefore I=\frac{\pi X}{2}-\frac{X^{2}}{2}+C$

Thus, $\int \cos ^{-1}(\sin x) d x=\frac{\pi x}{2}-\frac{x^{2}}{2}+c$