Need fast help on an integration problem.

**HclBr** · Nov 25th 2007, 10:14 PM

Hi, I need urgent help on this problem
int((e^(8x))*cos(9x))dx

I've tried integration by parts but you end up with a even more complicated integral. Tabular integration doesn't work since the e^8x never reduces to 0. I'm stumped , please help thanks!

**Opalg** · Nov 25th 2007, 11:35 PM

Originally Posted by HclBr

Hi, I need urgent help on this problem
int((e^(8x))*cos(9x))dx

I've tried integration by parts but you end up with a even more complicated integral. Tabular integration doesn't work since the e^8x never reduces to 0. I'm stumped , please help thanks!

Integrate by parts twice. It may look as though it's getting more complicated, but when you do it a second time it gets you back to a multiple of the integral that you started with. That gives you an equation for the integral. The answer should be of the form $e^{8x}(A\cos(9x)+B\sin(9x))$ (plus a constant of integration, of course).

**Krizalid** · Nov 26th 2007, 04:34 AM

I suggest (for this problems, 'cause they're well-known) take them more generally:

$\int e^{ax}\cos(bx)\,dx.$

As Opalg said, this requires a twice integration by parts.

**Krizalid** · Dec 1st 2007, 06:22 PM

Or we can use the following method:

$\int {e^{ax} \cos (bx)\,dx} = \text{Re} \int {e^{(a + bi)x} \,dx} = \text{Re} \,\frac{{e^{ax} \cdot (\cos (bx) + i\sin (bx))}}<br /> {{a + bi}}.$

After some simple calculations we have

$\int {e^{ax} \cos (bx)\,dx} = \frac{{e^{ax} (a\cos (bx) + b\sin (bx))}}<br /> {{a^2 + b^2 }} + k.$

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