# Integrate by Parts

• Feb 5th 2013, 06:46 PM
EliteAndoy
Integrate by Parts
Hi everyone! Well we are asked to do $\displaystyle \int_{}^{} sin(3x)cos(5x)dx$ exculsively by integration by parts. Doing it by trig identities is pretty easy, but once I do this one by integration by parts, including the integrand $\displaystyle \int_{}^{}vdu$, I always end up getting $\displaystyle 0=0$. Thanks you guys in advance! :D
• Feb 5th 2013, 07:08 PM
Prove It
Re: Integrate by Parts
\displaystyle \displaystyle \begin{align*} I &= \int{\sin{(3x)}\cos{(5x)}\,dx} \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} - \int{\frac{3}{5}\cos{(3x)}\sin{(5x)}\,dx} \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} - \frac{3}{5}\int{\cos{(3x)}\sin{(5x)}\,dx} \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} - \frac{3}{5} \left[ -\frac{1}{5}\cos{(3x)}\cos{(5x)} - \int{ \frac{3}{5}\sin{(3x)}\cos{(5x)} \,dx} \right] \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} + \frac{3}{25} \cos{(3x)}\cos{(5x)} + \frac{9}{25}\int{\sin{(3x)}\cos{(5x)}\,dx} \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} + \frac{3}{25}\cos{(3x)}\cos{(5x)} + \frac{9}{25}I \\ \frac{16}{25}I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} + \frac{3}{25}\cos{(3x)}\cos{(5x)} \\ I &= \frac{5}{16}\sin{(3x)}\sin{(5x)} + \frac{3}{16}\cos{(3x)}\cos{(5x)} \end{align*}

Therefore \displaystyle \displaystyle \begin{align*} \int{\sin{(3x)}\cos{(5x)}\,dx} = \frac{5}{16}\sin{(3x)}\sin{(5x)} + \frac{3}{16}\cos{(3x)}\cos{(5x)} + C \end{align*}
• Feb 5th 2013, 07:41 PM
EliteAndoy
Re: Integrate by Parts
Hmm... That's odd, I did the exact same thing and for some reason I am getting $\displaystyle 0=0$. Maybe just some arithmetic mistake. Anyways, thanks a lot, really helped me clarify that I'm going the right way. :D
• Feb 5th 2013, 08:57 PM
Prove It
Re: Integrate by Parts
Quote:

Originally Posted by EliteAndoy
Hmm... That's odd, I did the exact same thing and for some reason I am getting $\displaystyle 0=0$. Maybe just some arithmetic mistake. Anyways, thanks a lot, really helped me clarify that I'm going the right way. :D

You need to be consistent with the terms you choose as u and dv. Reversing them the second time you integrate by parts will simply result in getting back to where you started...