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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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    Yes, you don't "solve" integrals.2012-09-28

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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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    Thanks for the prompt answer!2012-09-26