Answer
$$\displaystyle\int e^{2\theta}\sin{3\theta}d\theta= \frac{2}{13}e^{2\theta}\sin{3\theta}-\frac{3}{13}e^{2\theta}\cos{3\theta}+C$$
Work Step by Step
We use integration by parts to obtain:
$u= \sin{3\theta}\quad du=3cos{3\theta}d\theta$
$dv = e^{2\theta}d\theta\quad v=\frac{1}{2}e^{2\theta}$
$\int u\thinspace dv = uv - \int v\thinspace du$ $$\int e^{2\theta}\sin{3\theta}=\sin{3\theta}\frac{1}{2}e^{2\theta}-\int \frac{1}{2} e^{2\theta}3\cos{3\theta}\thinspace d\theta$$ $$\int e^{2\theta}\sin{3\theta}=\sin{3\theta}\frac{1}{2}\thinspace e^{2\theta}-\frac{3}{2}\int e^{2\theta}\cos{3\theta}\thinspace d\theta$$
$u=\cos{3\theta}\quad du = -3\sin{3\theta}$
$dv=e^{2}{d\theta}\quad v=\frac{1}{2}e^{2\theta}$
$$\int e^{2\theta}\sin{3\theta}\thinspace \frac{1}{2}e^{2\theta}-\frac{3}{2}(uv-\int v\thinspace du)$$ $$= \sin{3\theta}\thinspace\frac{1}{2}e^{2\theta}-\frac{3}{2}\bigg(\cos({3\theta})\thinspace\frac{1}{2}e^{2\theta}-\int\frac{1}{2}e^{2\theta}\times -3\sin{3\theta}\thinspace d\theta\bigg)$$ $$=\sin{3\theta}\times\frac{1}{2}e^{2\theta}-\frac{3}{2}\bigg(\frac{1}{2}e^{2\theta}\cos{3\theta}+\frac{3}{2}\int e^{2\theta}\sin{3\theta}\thinspace d\theta\bigg)$$ $$=\frac{1}{2}e^{2\theta}\sin{3\theta}-\frac{3}{4}e^{2\theta}\cos{3\theta}-\frac{9}{4}\int e^{2\theta}\sin{3\theta}\thinspace d\theta$$ $$\frac{13}{4}\int e^{2\theta}\sin{3\theta}\thinspace d\theta= \frac{1}{2}e^{2\theta}\sin{3\theta}-\frac{3}{4}e^{2\theta}\cos{3\theta}$$ $$= \frac{2}{13}e^{2\theta}\sin{3\theta}-\frac{3}{13}e^{2\theta}\cos{3\theta}+C$$
