# How do I integrate a multiplication equation?

• October 25th 2010, 09:00 PM
xEnOn
How do I integrate a multiplication equation?
I always have problems integrating a multiplication equation. How do I integrate an equation like this one:

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

I tried doing it by parts but still canot solve it.
• October 25th 2010, 09:21 PM
Unknown008
You could not use by parts?

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

Clean up;

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

By parts for the second part;

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

Does the last part of the integral seem familiar?

Clean up:

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

Expand the brackets;

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

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

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

Simplify;

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

Multiply by 9/13;

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

Expand and there you are! (Smile)
• October 26th 2010, 02:19 AM
xEnOn
wow! That's a nice trick! Thanks man! :D
• October 26th 2010, 06:29 AM
Unknown008
You're welcome (Happy)