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$\int e^{2\theta}\sin(3\theta)d\theta$ seems to be leading me in circles. The integral I get when I use integration by parts, $\int e^{2\theta}\cos(3\theta)d\theta$ just leads me back to $\int e^{2\theta}\sin(3\theta)d\theta$. I am not sure how to solve it.

My Steps:

$\int e^{2\theta}\sin(3\theta)d\theta$

Let $u = \sin(3\theta)$ and $dv=e^{2\theta}d\theta$

Then $du = 3\cos(3\theta)d\theta$ and $v = \frac{1}{2}e^{2\theta}$ \begin{align*} \int e^{2\theta} \sin(3 \theta)d\theta &= \frac{1}{2} e^{2\theta}\sin(3\theta) - \int\frac{1}{2}e^{2\theta}3\cos(3\theta)d\theta\\ &=e^{2\theta}\sin(3\theta) - \frac{3}{2}\int e^{2\theta}\cos(3\theta)d\theta\\ \end{align*}

$\int e^{2\theta}\cos(3\theta)d\theta$

Let $u = \cos(3\theta)$ and $dv = e^{2\theta}d\theta$

Then $du = -3\sin(3\theta)d\theta$ and $v=\frac{1}{2}e^{2\theta}$

\begin{align*} \int e^{2\theta}\cos(3\theta) &= \frac{1}{2}e^{2\theta}\cos(3\theta)-\int (\frac{1}{2}e^{2\theta}\cdot-3\sin(3\theta))d\theta\\ &=\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} \int e^{2\theta}\sin(3\theta)d\theta \end{align*}

So you can see I just keep going in circles. How can I break out of this loop?

LuminousNutria
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    You have done it correctly. The very last term in your last step is actually (3/2)I where I=original integral. Substitute that and you will get the answer. – Siddhant Sep 10 '19 at 18:03
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    The method is actually explained quite well in this question. – Peter Foreman Sep 10 '19 at 18:05
  • @PeterForeman Thank you both. I have read both comments and looked in the link provided. I am not sure I understand completely, but basically the question says to "rearrange and solve" $\int e^x \sin x, dx=g(x)-n\int e^x \sin x ,dx$. I am not sure what they mean by this. I could put the right-most integral to the left of the equals sign, but this doesn't help me solve the problem, so I'm not sure what you mean. – LuminousNutria Sep 10 '19 at 18:14
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    All the calculus is over. You now have to do ALGEBRA!!! $I=blabla+\frac{3}{2}I$ and $I$ is the unknown you want to solve for. – MasB Sep 10 '19 at 18:18
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    You could evaluate it faster with complex exponentials – IMOPUTFIE Sep 10 '19 at 18:18
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    Just replace your entire integral by some letter, say $I$, then you have$$I=e^{2\theta}\sin{(3\theta)}-\frac32\left(\frac12e^{2\theta}\cos{(3\theta)}+\frac32I\right)$$ – Peter Foreman Sep 10 '19 at 18:18
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    You started with $\int e^{2\theta}\sin(3\theta)d\theta$ and you got and expression involving $\frac32\int e^{2\theta}\sin(3\theta)d\theta$ So if you subtract the latter from both sides, you'll get an equation for $\frac12\int e^{2\theta}\sin(3\theta)d\theta$ – saulspatz Sep 10 '19 at 18:18
  • @IMOPUTFIE I was going to put an answer along those lines when I saw your comment. – Mark Bennet Sep 10 '19 at 18:45

4 Answers4

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Take your examples together, \begin{align*} \int e^{2\theta}\sin(3\theta)d\theta &=\frac{1}{2}e^{2\theta}\sin(3\theta)- \frac{3}{2} \left(\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} \int e^{2\theta}\sin(3\theta)d\theta\right) \end{align*} Substituting the integral for a variable, say $X$, gives you: $$X=\frac{1}{2}e^{2\theta}\sin(3\theta)- \frac{3}{2} \left(\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} X\right)$$ simplifying gives you: $$X=\frac{1}{2}e^{2\theta}\sin(3\theta)- \frac{3}{4}e^{2\theta}\cos(3\theta)- \frac{9}{4} X$$ so your answer is $$X=\int e^{2\theta}\sin(3\theta)=\frac{4}{13}\left(\frac{1}{2}e^{2\theta}\sin(3\theta)- \frac{3}{4}e^{2\theta}\cos(3\theta)\right)=\frac{e^{2\theta}\left(2\sin(3\theta)-3\cos(3\theta)\right)}{13}$$ and a simple derivative check shows this to be true. Note your first example last line, you are missing a $\frac{1}{2}$ on the right hand side.

Gabe
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As lab bhattacharjee answered, in case integration by parts is not mandatory, you can make life easier considering that what you need is the imaginary part of $$I=\int e^{2\theta} e^{3i \theta}\,d\theta=\int e^{(2+3i)\theta}\,d\theta=\frac {e^{(2+3i)\theta}}{(2+3i)}=\frac{2-3i}{13}e^{(2+3i)\theta}$$ $$I=\frac{3}{13} e^{2 \theta } \sin (3 \theta )+\frac{2}{13} e^{2 \theta } \cos (3 \theta )+i \left(\frac{2}{13} e^{2 \theta } \sin (3 \theta )-\frac{3}{13} e^{2 \theta } \cos (3 \theta )\right)$$

Claude Leibovici
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Another method is to predict the answer: $$\int e^{2x}\sin(3x)dx=Ae^{2x}\sin (3x)+Be^{2x}\cos (3x)+C \Rightarrow \\ e^{2x}\sin (3x)=2Ae^{2x}\sin (3x)+3Ae^{2x}\cos (3x)+2Be^{2x}\cos (3x)-3Be^{2x}\sin (3x) \Rightarrow \\ \begin{cases} 2A-3B=1\\ 3A+2B=0\end{cases}\Rightarrow A=\frac2{13};B=-\frac3{13}$$ Hence, the final answer is: $$\int e^{2x}\sin(3x)dx=\frac2{13}e^{2x}\sin (3x)-\frac3{13}e^{2x}\cos (3x)+C.$$

farruhota
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Hint

In case integration by parts is not mandatory,

find $$\dfrac{e^{2x}(a\cos3x+b\sin3x)}{dx}$$ and compare with $e^{2x}\sin3x$ to find the values of $a,b$

lab bhattacharjee
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