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}
$