I'm supposed to integrate

pi

∫ (cos nx)(cosx)^2 dx

0

how do i integrate this?

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- Sep 23rd 2008, 09:37 AMpandaproblem in integration
I'm supposed to integrate

pi

∫ (cos nx)(cosx)^2 dx

0

how do i integrate this? - Sep 23rd 2008, 09:55 AMJameson
My first guess would be convert $\displaystyle \cos^2(x)$ to $\displaystyle 1-\sin^2(x)$ and then it becomes a trivial integral plus a u-sub one.

- Sep 23rd 2008, 09:55 AMChop Suey
First, use the half angle identity to get rid of the cos^2(x).

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

Multiplying this with \cos{nx} will get you

$\displaystyle \frac{1}{2} \int (\cos{nx} + \cos{nx}\cos{2x}) dx$

First one should be easy. For the second one, you can use either integration by parts or the product to sum identity:

$\displaystyle \cos{a}\cos{b} = \frac{1}{2}(\cos{(a-b)}+\cos{(a+b)})$ - Sep 23rd 2008, 09:59 AMJameson
I didn't see the (nx) term. Sorry. Obviously the above substitution is the one to go with now.