Answer
$$ - \frac{1}{{20}}{e^{ - 2\theta }}sin6\theta - \frac{3}{{20}}{e^{ - 2\theta }}\cos 6\theta + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{ - 2\theta }}sin6\theta } d\theta \cr
& {\text{substitute }}u = sin6\theta ,{\text{ }}du = 6\cos 6\theta d\theta \cr
& dv = {e^{ - 2\theta }}d\theta ,{\text{ }}v = - \frac{1}{2}{e^{ - 2\theta }} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int {{e^{ - 2\theta }}sin6\theta } d\theta = - \frac{1}{2}{e^{ - 2\theta }}sin6\theta - \int {\left( { - \frac{1}{2}{e^{ - 2\theta }}} \right)\left( {6\cos 6\theta d\theta } \right)} \cr
& \int {{e^{ - 2\theta }}sin6\theta } d\theta = - \frac{1}{2}{e^{ - 2\theta }}sin6\theta + 3\int {{e^{ - 2\theta }}\cos 6\theta d\theta } \cr
& {\text{substitute }}u = \cos 6\theta ,{\text{ }}du = - 6\cos 6\theta d\theta \cr
& dv = {e^{ - 2\theta }}d\theta ,{\text{ }}v = - \frac{1}{2}{e^{ - 2\theta }} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int {{e^{ - 2\theta }}sin6\theta } d\theta = - \frac{1}{2}{e^{ - 2\theta }}sin6\theta + 3\left( { - \frac{1}{2}{e^{ - 2\theta }}\cos 6\theta - \int {\left( { - \frac{1}{2}{e^{ - 2\theta }}} \right)\left( { - 6\cos 6\theta d\theta } \right)} } \right) \cr
& \int {{e^{ - 2\theta }}sin6\theta } d\theta = - \frac{1}{2}{e^{ - 2\theta }}sin6\theta + 3\left( { - \frac{1}{2}{e^{ - 2\theta }}\cos 6\theta - 3\int {{e^{ - 2\theta }}\cos 6\theta d\theta } } \right) \cr
& \int {{e^{ - 2\theta }}sin6\theta } d\theta = - \frac{1}{2}{e^{ - 2\theta }}sin6\theta - \frac{3}{2}{e^{ - 2\theta }}\cos 6\theta - 9\int {{e^{ - 2\theta }}\cos 6\theta d\theta } \cr
& {\text{solving for }}\int {{e^{ - 2\theta }}sin6\theta } d\theta \cr
& \int {{e^{ - 2\theta }}sin6\theta } d\theta + 9\int {{e^{ - 2\theta }}\cos 6\theta d\theta } = - \frac{1}{2}{e^{ - 2\theta }}sin6\theta - \frac{3}{2}{e^{ - 2\theta }}\cos 6\theta \cr
& 10\int {{e^{ - 2\theta }}\cos 6\theta d\theta } = - \frac{1}{2}{e^{ - 2\theta }}sin6\theta - \frac{3}{2}{e^{ - 2\theta }}\cos 6\theta + C \cr
& \int {{e^{ - 2\theta }}\cos 6\theta d\theta } = - \frac{1}{{20}}{e^{ - 2\theta }}sin6\theta - \frac{3}{{20}}{e^{ - 2\theta }}\cos 6\theta + C \cr} $$
