Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 36

Evaluate $\displaystyle \int^\pi_0 e^{\cos t} \sin 2t dt$ by making a substitution first, then by using Integration by parts.
Recall that $\sin 2t = 2 \sin t \cos t$ so,
$\displaystyle \int^\pi_0 e^{\cos t} \sin 2 t dt = \int^\pi_0 e^{\cos t} (2 \sin t \cos t) dt$

if we let $z = \cos t$, then $dz = - \sin t dt$
Make sure that the upper and lower limits are also in terms of $z$, so...
$\displaystyle \int^\pi_0 e^{\cos t} (2 \sin t \cos t) dt = -2 \int^{\cos \pi}_{\cos 0} z e^zdz = -2\int^{-1}_1 ze^z dz$

By using integration by parts,
If we let $u = z$ and $dv = e^z dz$. Then,
$du = dz$ and $\displaystyle v = \int e^z dz = e^z$

So,

$
\begin{equation}
\begin{aligned}
-2 \int^{-1}_1 ze^z dz = uv - \int v du &= -2 \left[ ze^z - \int e^z dz \right]\\
\\
&= -2 \left[ ze^z - e^z\right]\\
\\
&= -2e^z [z-1]
\end{aligned}
\end{equation}
$


Evaluating from 1 to -1,
$\displaystyle = \frac{-4}{e}$