Integral cos^2(mx)

**MatteNoob** · Sep 11th 2008, 09:15 PM

I'd love to see various ways to solve this integral, please.

$\int_0^{2\pi} \cos^2(mx)\rm{d}x \qquad m\in \mathbb N$

I have thought about using that

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

as I know that $m\in \mathbb N$ , but I'm not sure if it is OK. However, I hope someone feels like showing me various ways of doing it.

In advance, thank you for your time and help.

**o_O** · Sep 11th 2008, 09:26 PM

$u = mx \ \Rightarrow \ du = m dx$

So: $\int_{0}^{2\pi} \cos^2 (mx) \ dx = \frac{1}{m} \int_{u(0)}^{u(2\pi)} \cos^2 u \ du$

Now use the identity: $\cos^2 x = \frac{1 + \cos 2x}{2}$

**Shyam** · Sep 11th 2008, 09:33 PM

Originally Posted by MatteNoob

I'd love to see various ways to solve this integral, please.

$\int_0^{2\pi} \cos^2(mx)\rm{d}x \qquad m\in \mathbb N$

I have thought about using that

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

as I know that $m\in \mathbb N$ , but I'm not sure if it is OK. However, I hope someone feels like showing me various ways of doing it.

In advance, thank you for your time and help.

$\int_0^{2\pi} \cos^2(mx)\rm{d}x \qquad m\in \mathbb N$

$=\int_0^{2\pi} \frac{1+\cos(2mx)}{2}~dx$

$=\frac{1}{2}\int_0^{2\pi} \left[ 1+\cos(2mx)\right]~dx$

**Showcase_22** · Sep 12th 2008, 12:56 AM

I have this way:

I'd like to point out that I don't usually use such a long and windy method, it's just that you wanted various ways of solving it.

I just noticed there's a typo on my image. For the second substitution it should have a -x/2m. See my last post.

**Showcase_22** · Sep 12th 2008, 12:58 AM

This is another long and windy method:

Just use the cos^2(x)+sin^2(x)=1 and plug in some numbers to finish it off.

**Showcase_22** · Sep 12th 2008, 01:03 AM

(sorry about 3 posts in a row, it's just that I can't get the image thing to pop up when I use "edit").

This is less of an integrating method and more of just rewriting the question:

One of the above methods would need to be used to solve it though.

**Showcase_22** · Sep 12th 2008, 01:07 AM

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