## Calculus: Early Transcendentals 8th Edition

$\frac{((-e^{-\theta}cos(2\theta))+(2e^{-\theta}sin(2\theta))}{5}$
$\int e^{-\theta}cos(2\theta)d\theta$ 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: $\frac{-cos(2\theta)e^{-\theta}+2sin(2\theta)e^{-\theta}}{5}= \int e^{-\theta}cos(2\theta)d\theta$