Integrate by Parts

**EliteAndoy** · February 5th 2013, 06:46 PM

Hi everyone! Well we are asked to do $\int_{}^{} sin(3x)cos(5x)dx$ 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 $\int_{}^{}vdu$ , I always end up getting $0=0$ . Thanks you guys in advance!

**Prove It** · February 5th 2013, 07:08 PM

$\displaystyle \begin{align*} I &= \int{\sin{(3x)}\cos{(5x)}\,dx} \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} - \int{\frac{3}{5}\cos{(3x)}\sin{(5x)}\,dx} \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} - \frac{3}{5}\int{\cos{(3x)}\sin{(5x)}\,dx} \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} - \frac{3}{5} \left[ -\frac{1}{5}\cos{(3x)}\cos{(5x)} - \int{ \frac{3}{5}\sin{(3x)}\cos{(5x)} \,dx} \right] \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} + \frac{3}{25} \cos{(3x)}\cos{(5x)} + \frac{9}{25}\int{\sin{(3x)}\cos{(5x)}\,dx} \\ I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} + \frac{3}{25}\cos{(3x)}\cos{(5x)} + \frac{9}{25}I \\ \frac{16}{25}I &= \frac{1}{5}\sin{(3x)}\sin{(5x)} + \frac{3}{25}\cos{(3x)}\cos{(5x)} \\ I &= \frac{5}{16}\sin{(3x)}\sin{(5x)} + \frac{3}{16}\cos{(3x)}\cos{(5x)} \end{align*}$

Therefore $\displaystyle \begin{align*} \int{\sin{(3x)}\cos{(5x)}\,dx} = \frac{5}{16}\sin{(3x)}\sin{(5x)} + \frac{3}{16}\cos{(3x)}\cos{(5x)} + C \end{align*}$

**EliteAndoy** · February 5th 2013, 07:41 PM

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

**Prove It** · February 5th 2013, 08:57 PM

Originally Posted by EliteAndoy

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

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...

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