Integration by Parts: Struggling with e^xcos(x)dx

[ranger1716]

I'm kind of lost on where to go next with this integration by parts problem.

I have to integrate e^xcos(x)dx.

I've gotten as far as one step of integration by parts, but I can't understand how this will help. It seems I'll just be going in circles. I have:

e^xsin(x) - int(e^xsin(x))dx. If I do a second integration by parts, will I not just get back to where I started?

 

[math-chick_41]

do parts twice, combine like terms and you can solve in terms of the original integral in question.

 

[Hurkyl]

In other words, you do know how to solve z = a - z for z.

(P.S. doesn't your textbook do this as an example?)

 

[VietDao29]

I'm kind of lost on where to go next with this integration by parts problem.

I have to integrate e^xcos(x)dx.

I've gotten as far as one step of integration by parts, but I can't understand how this will help. It seems I'll just be going in circles. I have:

e^xsin(x) - int(e^xsin(x))dx. If I do a second integration by parts, will I not just get back to where I started?

One thing that can prevents you from getting an obvious equation like 0 = 0 is that: If you previously assigned u = e^x, and dv = cos(x) => v = sin(x), and get to:
\int e ^ x \cos x dx = e ^ x \sin x - \int e ^ x \sin x dx
Then what you should do next is to let u = e^x, and dv = sin(x).
After that, just do some little rearrangement, and you'll arrive at the answer.
Do NOT do the reverse (i.e, let u = sin(x), and dv = e^x). If you want to see why, then just try it. Don't be surprise if you get an equation 0 = 0, or \int 0 dx = C.
Can you go from here? :)

 

[fourier jr]

If I do a second integration by parts, will I not just get back to where I started?

that's the trick! get back where you started, and combine like terms, as in ac + bc = (a+b)c. that's the only trick to this problem.