Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 44

Find the definite integral $\displaystyle \int^{\frac{\pi}{2}}_0 \cos x \sin (\sin x) dx$

Let $u = \sin x$, then $du = \cos x dx$. When $x = 0, u = 0$ and when $\displaystyle x = \frac{\pi}{2}, u = 1$. Thus,



$\begin{equation} \begin{aligned} \int^{\frac{\pi}{2}}_0 \cos x \sin (\sin x) dx =& \int^{\frac{\pi}{2}}_0 \sin(\sin x) \cos x dx \\ \\ \int^{\frac{\pi}{2}}_0 \cos x \sin (\sin x) dx =& \int^{\frac{\pi}{2}}_0 \sin u du \\ \\ \int^{\frac{\pi}{2}}_0 \cos x \sin (\sin x) dx =& \left. - \cos u \right|^{\frac{\pi}{2}}_0 \\ \\ \int^{\frac{\pi}{2}}_0 \cos x \sin (\sin x) dx =& - \cos (1) - (- \cos 0) \\ \\ \int^{\frac{\pi}{2}}_0 \cos x \sin (\sin x) dx =& - \cos(1) + 1 \\ \\ \int^{\frac{\pi}{2}}_0 \cos x \sin (\sin x) dx =& 0.4597 \end{aligned} \end{equation}$