I am working on an integration by parts problem that, compared to the student solutions manual, my answer is pretty close. Could someone please point out where I went wrong?
Find $\int e^{2\theta} \cdot \sin{3\theta} \ d\theta$
$u_1 = \sin{3\theta}$
$du_1 = \frac{1}{3}\cos{3\theta} \ d\theta$
$v_1 = \frac{1}{2} e^{2\theta}$
$dv_1 = e^{2\theta} \ d\theta$
$\underbrace{\sin{3\theta}}_{u_1} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_1} – \int\underbrace{\frac{1}{2} e^{2\theta}}_{v_1} \cdot \underbrace{\frac{1}{3} \cos{3\theta} \ d\theta}_{du_1}$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} – \frac{1}{6} \int \cos{3\theta} \cdot e^{2\theta} \ d\theta$
Doing integration by parts again…
$u_2 = \cos{3\theta}$
$du_2 = -\frac{1}{3} \sin{3\theta} \ d\theta$
$v_2 = \frac{1}{2} e^{2\theta}$
$dv_2 = e^{2\theta} \ d\theta$
$\underbrace{\sin{3\theta}}_{u_1} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_1} – \frac{1}{6}\left(\underbrace{\cos{3\theta}}_{u_2} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_2} – \int \underbrace{\frac{1}{2} e^{2\theta}}_{v_2} \cdot \underbrace{-\frac{1}{3} \sin{3\theta} \ d\theta}_{du_2}\right)$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} – \frac{1}{6}\left(\frac{1}{2}\cos{3\theta} \cdot e^{2\theta} + \frac{1}{6} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta\right)$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} – \frac{1}{12}\cos{3\theta} \cdot e^{2\theta} – \frac{1}{36} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta$
$\frac{37}{36} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta = \frac{1}{2}\sin{3\theta} \cdot e^{2\theta} – \frac{1}{12}\cos{3\theta} \cdot e^{2\theta}$
$\int \sin{3\theta} \cdot e^{2\theta} \ d\theta = \begin{equation} \boxed{\frac{18}{37}\sin{3\theta} \cdot e^{2\theta} – \frac{1}{37}\cos{3\theta} \cdot e^{2\theta}} \end{equation}$
However, the boxed answer is incorrect. The answer should read:
$\frac{1}{13} e^{2\theta} \left(2\sin{3\theta} – 3\cos{3\theta}\right)$
Best Answer
Ah ha! I got it!
For some reason I must have been thinking of integrating the
dvrather than differentiating.