Evaluate: ∫ cos⁻¹(sin x) dx

$\cos^{-1}(\sin x) = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - x\right)\right) = \frac{\pi}{2} - x$ for $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ $\int \left(\frac{\pi}{2} - x\right)dx = \frac{\pi x}{2} - \frac{x^2}{2} + C$ Hence the result is $\frac{\pi x}{2} - \frac{x^2}{2} + C$.