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I'm trying to solve the following integral $$\int_{0}^{2\pi}\frac{1}{A\cos(2x+B)+C}\text{d}x$$ Any idea on how to approach this?

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"Solve" is not the right word here. "Evaluate" would fit. "Solve" and "equation" are words that get used as catch-alls by students who haven't learned definitions. – Michael Hardy Sep 26 '12 at 13:11
If $A$ and $C$ are such that $\{x \in [0,2\pi]:\ A\cos x+C=0\}\ne \emptyset$, then your integral diverges. – Mercy Sep 26 '12 at 13:20
Yes, you don't "solve" integrals. – Stefan Smith Sep 28 '12 at 0:10

1 Answer

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First notice that since $\cos$ is periodic, and you're integrating over (twice) its period, the $+B$ doesn't make a difference, and so $$\int_0^{2\pi} \dfrac{1}{A\cos(2x+B)+C}\, dx = \int_0^{2\pi} \dfrac{1}{A\cos 2x + C}\, dx$$ At this point, a $t$-substitution* should do the trick $-$ put $t=\tan x$, so that $\cos 2x = \dfrac{1-t^2}{1+t^2}$ and so forth.

(*Thanks Michael Hardy)

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Here is a better link for that substitution. – Michael Hardy Sep 26 '12 at 13:12
@MichaelHardy: Cheers, I thought I'd seen something better. – Clive Newstead Sep 26 '12 at 13:14
Thanks for the prompt answer! – Wox Sep 26 '12 at 13:26

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