Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
1  
This is almost identical to a couple of questions asked before. See: math.stackexchange.com/questions/307995/… and also: math.stackexchange.com/questions/136595/… – Tom Oldfield Feb 27 '13 at 22:24
2  
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
    
@H.R. Did your edit really substantially improve this question from 3 years ago? Why did you waste frontpage real estate by bumping this question, and knocking a new question our of view? Edits to old questions should IMO not be minor. – Jyrki Lahtonen Dec 20 '15 at 16:51
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:

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

share|cite|improve this answer

HINT

Integrate by parts twice and rearrange.

share|cite|improve this answer

I just generalize your question a little and put $a$ instead of $2$ and $b$ instead of $3$ in the integrand. You should do the integration by parts twice

$$\eqalign{ & I = \int {{e^{a\theta }}} \;\sin b\theta \;d\theta \cr & \,\,\,\, = {1 \over a}{e^{a\theta }}\sin b\theta - {b \over a}\int {{e^{a\theta }}\cos b\theta \,d\theta } \cr & \,\,\,\, = {1 \over a}{e^{a\theta }}\sin b\theta - {b \over a}\left[ {{1 \over a}{e^{a\theta }}\cos b\theta + {b \over a}\int {{e^{a\theta }}\sin b\theta d\theta} } \right] \cr & \,\,\,\, = {1 \over a}{e^{a\theta }}\sin b\theta - {b \over a}\left[ {{1 \over a}{e^{a\theta }}\cos b\theta + {b \over a}I} \right] \cr & \,\,\,\, = {1 \over a}{e^{a\theta }}\sin b\theta - {b \over {{a^2}}}{e^{a\theta }}\cos b\theta - {{{b^2}} \over {{a^2}}}I \cr} $$

and then solve for $I$. This will result in

$$\eqalign{ & {{{a^2} + {b^2}} \over {{a^2}}}I = {1 \over a}{e^{a\theta }}\sin b\theta - {b \over {{a^2}}}{e^{a\theta }}\cos b\theta \cr & I = {a \over {{a^2} + {b^2}}}{e^{a\theta }}\sin b\theta - {b \over {{a^2} + {b^2}}}{e^{a\theta }}\cos b\theta \cr} $$

Or equivalently

$$\boxed{I = {1 \over {{a^2} + {b^2}}}{e^{a\theta }}\left[ {a\sin b\theta - b\cos b\theta } \right]}$$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.