Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 18

Evaluate $\displaystyle \int e^{-\theta} \cos 2 \theta d \theta$
If we let $u = e^{-\theta}$ and $dv = \cos 2 \theta d \theta$ then
$du = e^{-\theta} (-1) d \theta$ and $\displaystyle v = \int \cos 2 \theta d \theta = \frac{1}{2} \sin 2 \theta$

So,

$
\begin{equation}
\begin{aligned}
\int e^{-\theta} \cos 2 \theta d \theta = uv - \int v du &= \frac{e^{-\theta}}{2} \sin 2 \theta - \int \frac{1}{2} \sin 2 \theta \left( -e^{-\theta} \right)\\
\\
&= \frac{e^{-\theta}}{2} \sin 2 \theta + \frac{1}{2} \int e^{-\theta} \sin 2 \theta d \theta
\end{aligned}
\end{equation}
$



To evaluate $\displaystyle \int e^{-\theta} \sin 2 \theta d \theta$, we must apply integration by parts once more... so,
If we let $u_1 = e^{-\theta}$ and $dv_1 = \sin 2 \theta d \theta$, then
$du_1 = e^{-\theta} (-1) d \theta$ and $ \displaystyle v_1 = \int \sin 2 \theta d \theta = -\frac{1}{2} \cos 2 \theta$

So,
$\displaystyle \int e^{- \theta} \sin 2 \theta d \theta = u_1 v_1 - \int v_1 d u_1 = \frac{e^{-\theta}\cos 2 \theta}{2}-\int \frac{e^{-\theta}\cos 2 \theta d \theta}{2}$

Going back from the equation,

$
\begin{equation}
\begin{aligned}
\int e^{- \theta} \cos 2 \theta d \theta &= \frac{e^{-\theta}}{2} \sin 2 \theta + \frac{1}{2} \left[ -\frac{e^{- \theta} \cos 2 \theta }{2} - \int \frac{e^{-\theta}\cos 2\theta}{2} d \theta \right]\\
\\
\int e^{- \theta} \cos 2 \theta d \theta &= \frac{e^{-\theta}}{2} \sin 2 \theta - \frac{e^{-\theta}\cos 2 \theta}{4} - \frac{1}{4} \int e^{-\theta}\cos 2\theta d \theta
\end{aligned}
\end{equation}
$


Let's continue the like terms.

$
\begin{equation}
\begin{aligned}
\int e^{- \theta} \cos 2 \theta d \theta + \frac{1}{4} \int e^{-\theta} \cos 2\theta d \theta &= \frac{e^{-\theta}}{2} \sin 2 \theta - \frac{e^{-\theta}\cos 2 \theta}{4}\\
\frac{5}{4} \int e^{-\theta} \cos 2 \theta d \theta &= \frac{e^{-\theta}}{2} \sin 2 \theta - \frac{e^{-\theta}\cos 2\theta}{4}\\
\\
\int e^{-\theta} \cos 2 \theta d \theta &= \left[ \frac{e^{-\theta}}{2} \sin 2 \theta - \frac{e^{-\theta}}{4} \cos 2 \theta \right] \frac{4}{5}\\
\\
\int e^{-\theta} \cos 2 \theta d \theta &= \frac{2e^{-\theta}\sin 2 \theta}{5} - \frac{e^{-\theta}\cos 2 \theta}{5} + c
\end{aligned}
\end{equation}
$