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Integration by parts - unless I'm not thinking straight - doesn't seem to help here.. If I pick either of the functions $e^{2x}$ or $\sin(3x)$ to be $u$ or $dv$, they don't change to anything easier... $e^{2x}$ stays in the form $e^x$ and $\sin(3x)$ flip-flops between $\sin(3x)$ and $\cos(3x)$, neglecting the constant that gets introduced.

So how can I approach solving this? u-substitution doesn't seem to be something I can use either.

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This is almost identical to a couple of questions asked before. See:… and also:… – Tom Oldfield Feb 27 '13 at 22:24
I guess you could also use $\int e^{2x}\sin(3x)\ dx = \int e^{2x}\Im(\cos(3x)+i\sin(3x))\ dx =\int e^{2x}\Im(e^{3ix})\ dx=\int \Im(e^{2x}e^{3ix})\ dx =\Im(\int e^{(2+3i)x})\ dx$. Quite sure it's not the most efficient way though. – xavierm02 Feb 27 '13 at 22:38
I am a fan of this way and I do think it is a solid approach. – Eleven-Eleven Feb 22 '14 at 13:59

3 Answers 3

up vote 2 down vote accepted

The key here is to set $v' = e^{2x}$ and integrate by parts twice. This will make $\sin$ become $\cos$ and then $\sin$ again. The same integral will show up on the right side but with a different factor (so it won't cancel out).

Call the integral $I$ and integrate by parts twice to get: \begin{align} I &= \int e^{2x} \sin(3x) \,dx = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \int e^{2x} \cos(3x) \,dx \\ &= \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \left(\frac{1}{2} e^{2x}\cos(3x) + \frac{3}{2} \int e^{2x} \sin(3x) \,dx\right) \end{align}

Thus: $$ I = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \left(\frac{1}{2} e^{2x}\cos(3x) + \frac{3}{2} I \right) $$

Solve for $I$ to get:

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

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Integrate by parts twice and rearrange.

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let S be our integrate...........that is, S e2x, let u=sin(3x), dv=e2x ..which implies...du=3cos3x dx and v=1/2e"2x..........The integration by parts formula gives. S e"2x sin3x dx=uv-S vdu. that is......... 1/2sin(3x)e"2x -S 3cos(3x)1/2e"2x dx. which gives us. S cos(3x)e"2x dx.. we have to apply another integration for S cos(3x)e"2x dx... let u=cos3x and dv=e"2x dx.. which implies, du=-3sin3x and v=1/2e"2x. Hence, S cos(3x)e"2 dx=1/2cos(3x)e"2x-S-3sin(3x)1/2e"2x. combining boths formulas we get. S sin(3x)e"2x dx=1/2sin(3x)e"2x-3/2(1/2cos(3x)1/2e"2x-S-3sin(3x)1/2e"2x dx). Easy calculation give. S sin(3x)e"2x dx=1/2sin(3x)e"2x-3/4cos(3x)1/2e"2x+9/4 S sin(3x)e"2x dx. finally easy algebraic manipulation gives. S sin(3x)e"2x dx = 2/13sin(3x)-3/13cos(3x)1/2e"2x + C.. our answer give us..... 1/13e"2x(2sin(3x)-3cos(3x)). I hope this will help. By O'john

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You should really consider putting this into LaTeX form – Jean-Sébastien Aug 15 '13 at 15:46

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