Thread: Need fast help on an integration problem.

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!

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 $\displaystyle e^{8x}(A\cos(9x)+B\sin(9x))$ (plus a constant of integration, of course).

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

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

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

Or we can use the following method:

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

After some simple calculations we have

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