How do I integrate a multiplication equation?

**xEnOn** · October 25th 2010, 10:00 PM

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

$<br /> \int sin2x . e^{3x}<br />$

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

**Unknown008** · October 25th 2010, 10:21 PM

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!

**xEnOn** · October 26th 2010, 03:19 AM

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

**Unknown008** · October 26th 2010, 07:29 AM

You're welcome

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