# Tri Integration

## 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 .. :X

I'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.

## The Attempt at a Solution

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
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" [Broken] on this last integral to try to get it as a function of A.

Hope this helps. :)

Last edited by a moderator:
But once agn we are stucked with the integral of $$cos^{2m+2}\theta$$

:)

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 $$k \in \mathbb{Z}$$ ... :)
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 $$k \in \mathbb{Z}$$ ... :)