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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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    $\begingroup$ This is almost identical to a couple of questions asked before. See: math.stackexchange.com/questions/307995/… and also: math.stackexchange.com/questions/136595/… $\endgroup$ – Tom Oldfield Feb 27 '13 at 22:24
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    $\begingroup$ 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. $\endgroup$ – xavierm02 Feb 27 '13 at 22:38
  • $\begingroup$ I am a fan of this way and I do think it is a solid approach. $\endgroup$ – Eleven-Eleven Feb 22 '14 at 13:59
  • $\begingroup$ @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. $\endgroup$ – Jyrki Lahtonen Dec 20 '15 at 16:51
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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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HINT

Integrate by parts twice and rearrange.

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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

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

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