Thread: How do I integrate a multiplication equation?

I always have problems integrating a multiplication equation. How do I integrate an equation like this one:

$\displaystyle
\int sin2x . e^{3x}
$

I tried doing it by parts but still canot solve it.

You could not use by parts?

$\displaystyle \int e^{3x}\sin2x\ dx = \sin 2x. \dfrac{e^{3x}}{3} - \int \dfrac{e^{3x}}{3} .2\cos 2x\ dx$

Clean up;

$\displaystyle \int e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x - \dfrac23 \int e^{3x}\cos 2x\ dx$

By parts for the second part;

$\displaystyle \int e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x - \dfrac23\left[\cos2x. \dfrac{e^{3x}}{3} - \int \dfrac{e^{3x}}{3}.-2\sin2x\ dx\right]$

Does the last part of the integral seem familiar?

Clean up:

$\displaystyle \int e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x - \dfrac23\left[\cos2x. \dfrac{e^{3x}}{3} + \dfrac23\int e^{3x}\sin2x\ dx\right]$

Expand the brackets;

$\displaystyle \int e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x - \dfrac{2 e^{3x}}{9}\cos2x - \dfrac49\int e^{3x}\sin2x\ dx$

The trick now is to send the last part to the left.

$\displaystyle \int e^{3x}\sin2x\ dx + \dfrac49\int e^{3x}\sin2x\ dx= \dfrac{e^{3x}}{3}\sin 2x - \dfrac{2 e^{3x}}{9}\cos2x $

Simplify;

$\displaystyle \dfrac{13}{9}\int e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x - \dfrac{2 e^{3x}}{9}\cos2x $

Multiply by 9/13;

$\displaystyle \int e^{3x}\sin2x\ dx = \dfrac{9}{13}\left[\dfrac{e^{3x}}{3}\sin 2x - \dfrac{2 e^{3x}}{9}\cos2x\right] $

Expand and there you are!

wow! That's a nice trick! Thanks man!

You're welcome