# Trying to calculate $\displaystyle\int e^{2x}\sin(3x)\, \mathrm{d}x$

How can I calculate this integral?

$$\int {\exp(2x)}{\sin(3x)}\, \mathrm{d}x$$

I tried using integration by parts, but it doesn't lead me any improvement. So I made ​​an attempt through the replacement $$\cos(3x) = t$$ and it becomes $$\frac{1}{-3}\int \exp\left(2\left(\dfrac{\arccos(t)}{3}\right)\right)\, \mathrm{d}t$$ but I still can not calculate the new integral. Any ideas?

SOLUTION:

$$\int {\exp(2x)}{\sin(3x)}\, \mathrm{d}x = \int {\sin(3x)}\, \mathrm{d(\frac{\exp(2x)}{2}))}=$$

$$\frac{1}{2}{\sin(3x)}{\exp(2x)}-\frac{1}{2}\int {\exp(2x)}\, \mathrm{d}(\sin(3x))=$$

$$\frac{1}{2}{\sin(3x)}{\exp(2x)}-\frac{3}{2}\int {\exp(2x)}{\cos(3x)}\mathrm{d}x=$$

$$\frac{1}{2}{\sin(3x)}{\exp(2x)}-\frac{3}{2}\int {\cos(3x)}\mathrm{d(\frac{\exp(2x)}{2}))}=$$

$$\frac{1}{2}{\sin(3x)}{\exp(2x)}-\frac{3}{4}{\cos(3x)}{\exp(2x)}+\frac{3}{4}\int {\exp(2x)}\mathrm{d({\cos(3x)})}=$$

$$\frac{1}{2}{\sin(3x)}{\exp(2x)}-\frac{3}{4}{\cos(3x)}{\exp(2x)}-\frac{9}{4}\int {\sin(3x)}{\exp(2x)}\mathrm{d}x$$

$$=>(1+\frac{9}{4})\int {\exp(2x)}{\sin(3x)}\, \mathrm{d}x= \frac{1}{2}{\sin(3x)}{\exp(2x)}-\frac{3}{4}{\cos(3x)}{\exp(2x)}+c$$

$$=\frac{1}{13}\exp(2x)(2\sin(3x)-3\cos(3x))+c$$

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By $sen$, do you mean $\sin$? – Ahaan S. Rungta Dec 23 '13 at 13:48
Note that $e^{2x}\sin (3x)$ is the imaginary part of $e^{?}$. – 1015 Dec 23 '13 at 13:49
@AhaanRungta Yep, sorry. I just edited. – Gabriele Salvatori Dec 23 '13 at 13:49
Let $I$ be your integral. Integrate by parts twice. You'll wind up with an equation of the form $I=$ stuff $+cI$, where $c$ is a constant. Solve for $I$... – David Mitra Dec 23 '13 at 14:03

You need to use integration by parts or use the identity

$$\sin y = \frac{e^{iy} - e^{-iy} }{2i}$$

Which makes the integral easier.

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An alternative, as julien noticed: write $e^{(2+3i)x}=e^{2x}\cos(3x)+ie^{2x}\sin(3x)$, then integrate the LHS and you get two primitives in the RHS (taking real and imaginary parts). The imaginary part is the answer. – Jean-Claude Arbaut Dec 23 '13 at 14:07

You need to integrate by parts twice, either differentiating the exponential each time or differentiating the trig function each time. You will be able to solve for the integral.

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Hint:

$$\displaystyle\int e^{2x}\sin(3x)\, \mathrm{d}x=Ae^{2x}\sin(3x)+Be^{2x}\cos(3x)$$

where $A$ and $B$ are constant.

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