Integrating cos^{2m}(\theta) cos(2\theta) in Terms of A

[icystrike]

Homework Statement

Given \int_{0}^{k\pi} cos^{2m}(\theta) d\theta = A
Express \int_{0}^{k\pi} cos^{2m}(\theta) cos(2\theta) in terms of A.
Totally Stucked .. :XI've substituted cos^{2}\theta = \frac{1+cos(2\theta)}{2}

Then i get 1/2 A + another chunck of integral.

I've used integration by parts to tackle the chunck.

such that i integrate the cos(2\theta).

I've gotten pretty good simplification but i think I've made some mistakes here a there.

Homework Equations

The Attempt at a Solution

 

[Metaleer]

A method you could try is to remember that \cos (2 \theta) = 2 \cos ^2 (\theta) - 1. The second integral thus becomes

\int_0^{k \pi}\cos ^{2m} (\theta)(2 \cos ^2 (\theta) - 1) d \theta = 2 \int_0^{k \pi}\cos ^{2m+2} (\theta) d \theta - A,​

and you could try to use a http://en.wikipedia.org/wiki/Integration_by_reduction_formulae" on this last integral to try to get it as a function of A.

Hope this helps. :)

 

[icystrike]

But once agn we are stucked with the integral of cos^{2m+2}\theta

:)

 

[Metaleer]

...have you tried looking at the reduction formulas on that Wikipedia page I linked you to?

 

[icystrike]

...have you tried looking at the reduction formulas on that Wikipedia page I linked you to?

Are u referring to this? Seems to be great! Havent tried though...

 

[Metaleer]

Precisely the one! Remember that for us, our n is 2m + 2 and that since we have a definite integral, it just so happens that \sin(x) vanishes for 0 and k \pi, where k \in \mathbb{Z} ... :)

 

[icystrike]

Precisely the one! Remember that for us, our n is 2m + 2 and that since we have a definite integral, it just so happens that \sin(x) vanishes for 0 and k \pi, where k \in \mathbb{Z} ... :)

Fit perfectly well! Thanks a million!