Answer
$\frac{((-e^{-\theta}cos(2\theta))+(2e^{-\theta}sin(2\theta))}{5}$
Work Step by Step
$\int e^{-\theta}cos(2\theta)d\theta$
Apply integration by parts:
$\int f(x)g'(x)=f(x)g(x)-\int f'(x)g(x)dx$
$f(x)=cos(2\theta)$
$f'(x)= -2sin(2\theta)$
$g(x)=-e^{-\theta}$
$g'(x)= e^{-\theta}$
$-cos(2\theta)e^{-\theta}-\int (-2sin(2\theta))(-e^{-\theta})d\theta$
Clean up the integral by taking out constants and negatives:
$-cos(2\theta)e^{-\theta}-2\int sin(2\theta)e^{-\theta}d\theta$
Apply integration by parts again:
$\int sin(2\theta)e^{-\theta}d\theta$
$f(x)=sin(2\theta)$
$f'(x)= 2cos(2\theta)$
$g(x)=-e^{-\theta}$
$g'(x)= e^{-\theta}$
*Note that now you have the same integral you started with*
$-sin(2\theta)e^{-\theta}+2\int cos(2\theta)e^{-\theta}d\theta$
Put Everything Together and make it equal to what you started with:
$-cos(2\theta)e^{-\theta}-2[-sin(2\theta)e^{-\theta}+2\int
cos(2\theta)e^{-\theta}d\theta]= \int e^{-\theta}cos(2\theta)d\theta$
Distribute:
$-cos(2\theta)e^{-\theta}+2sin(2\theta)e^{-\theta}-4\int e^{-\theta}cos(2\theta)d\theta= \int e^{-\theta}cos(2\theta)d\theta$
"Combine like terms":
$-cos(2\theta)e^{-\theta}+2sin(2\theta)e^{-\theta}=5 \int e^{-\theta}cos(2\theta)d\theta$
Answer:
$\frac{-cos(2\theta)e^{-\theta}+2sin(2\theta)e^{-\theta}}{5}= \int e^{-\theta}cos(2\theta)d\theta$
