I would like to show that
$\int_{0}^{\pi} \cos\left(\sqrt{1-\sin^2(\theta)}\right) d\theta = \int_{-1}^{1} \frac{\cos\left(\sqrt{1-u^2}\right)}{\sqrt{1-u^2}} du$ by substituting $u=\sin(\theta)$.
Substituting $u=\sin(\theta)$ into $\int_{0}^{\pi} \cos\left(\sqrt{1-\sin^2(\theta)}\right) d\theta$ causes the limits of integration to be $\sin(0)$ to $\sin(\pi)$ which are both $0$. So $\int_{0}^{\pi} \cos\left(\sqrt{1-\sin^2(\theta)}\right) d\theta = \int_{0}^{0} \frac{\cos\left(\sqrt{1-u^2}\right)}{\sqrt{1-u^2}} du =0$?
What am I missing here?
