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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:

$\frac{1}{13} e^{2\theta} \left(2\sin{3\theta} - 3\cos{3\theta}\right)$

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    $\begingroup$ I know the intended purpose of this exercise is to practice integration by part, but there's a trick for this particular primitive. Write $\int e^{2 \theta} \sin(3 \theta) \, d \theta = \text{Im}\left(\int e^{(2+3i) \theta} \, d \theta \right)$. And compute $\int e^{(2+3i) \theta} \, d \theta = \frac{1}{2+3i} e^{(2+3i) \theta}$, whose imaginary part is $\frac{1}{13} e^{2 \theta} (2 \sin(3 \theta) - 3 \cos(3\theta))$. $\endgroup$ – Joel Cohen Mar 3 '12 at 18:22
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Ah ha! I got it!

$du = 3 \cos{3\theta} \ d\theta$ not $\frac{1}{3} \cos{3\theta} \ d\theta$

For some reason I must have been thinking of integrating the dv rather than differentiating.

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    $\begingroup$ You made the same mistake in the second integration by parts. Nice write up of the question and answer by the way. $\endgroup$ – David Mitra Mar 3 '12 at 18:19
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One could also use the right tool, which is the following fact:

For every nonzero complex number $z$, a primitive of $\theta\mapsto\mathrm e^{z\theta}$ is $\theta\mapsto z^{-1}\mathrm e^{z\theta}$.

Here, one is looking at the imaginary part of $\mathrm e^{z\theta}$ with $z=2+3\mathrm i$, hence the answer is the imaginary part of $$ \theta\mapsto z^{-1}\mathrm e^{z\theta}=|z|^{-2}\bar z\mathrm e^{z\theta}=|z|^{-2}(2-3\mathrm i)\mathrm e^{2\theta}(\cos(3\theta)+\mathrm i\sin(3\theta)). $$ Since $|z|^2=2^2+3^2=13$ and the imaginary part of $(2-3\mathrm i)(\cos(3\theta)+\mathrm i\sin(3\theta))$ is $2\sin(3\theta)-3\cos(3\theta)$, an answer is $\theta\mapsto \frac1{13}\mathrm e^{2\theta}(2\sin(3\theta)-3\cos(3\theta))$.

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  • $\begingroup$ An easy way is to use the method of undetermined coefficients, guess that the answer has the form $y=e^{2\theta}(A\sin(3\theta)+B\cos(3\theta))$, take the derivative of $y$, and figure out what $A$ and $B$ are. $\endgroup$ – Stefan Smith Jun 8 '12 at 23:04
  • $\begingroup$ I wouldn't say that using Euler's formula is the right tool. What the OP did above is a valid method, and it is good to have more than one method. $\endgroup$ – Baby Dragon Jun 19 '13 at 14:09
  • $\begingroup$ @BabyDragon In principle, sure, except that the question itself shows that the other method is more complicated and more error-prone (as well as, if I may add, less conceptual). $\endgroup$ – Did Jun 19 '13 at 15:32
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$$\int \mathrm{e}^{ax}\sin(bx)dx =\frac{\mathrm{e}^{ax}}{\mathrm{a^2+b^2}}[a\sin(bx)-b\cos(bx)]+c$$

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