$\displaystyle

\int xcos^2(2x)~dx

$

I can tell that this needs some sort of trig identity in order to solve, but I may be wrong.

How would I set this problem up with regards to u and dv?

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- Sep 2nd 2008, 12:08 PMredman223[SOLVED] Integration by parts and trig identity help
$\displaystyle

\int xcos^2(2x)~dx

$

I can tell that this needs some sort of trig identity in order to solve, but I may be wrong.

How would I set this problem up with regards to u and dv? - Sep 2nd 2008, 12:36 PMChop Suey
$\displaystyle \cos^2{2x} = \frac{1 + \cos{4x}}{2}$

Try it now. :D Choose u and dv in a way so that the next integral would only have Sine. - Sep 2nd 2008, 07:19 PMredman223
So how would that problem look if it was multiplied out? Or should I not do that?

$\displaystyle

\int x*\frac {1+cos4x}{2}~dx = \int \frac {x+cos4x^2}{2}~dx

$

Because that looks confusing and I don't know what to do next. - Sep 2nd 2008, 07:22 PMChop Suey
You cannot multiply x with the inside terms of cosine!

$\displaystyle \int x\cos^2(2x)~dx$

$\displaystyle = \int x(\frac{1+\cos{4x}}{2})$

$\displaystyle = \int \frac{x+x\cos{4x}}{2}$

Now split.

$\displaystyle = \int \frac{x}{2} + \int \frac{x\cos{4x}}{2}$

I presume you know what to do next.