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I have this integral $\int_0^{2\pi}\frac{1}{1+k\cos{x}}{\rm d}x$,

which is definitely:

$\int_0^{2\pi}\frac{1}{1+k\cos{x}}{\rm d}x = 2\int_0^{\pi}\frac{1}{1+k\cos{x}}{\rm d}x$,

but if I do this:

$\int_0^{2\pi}\frac{1}{1+k\cos{x}}{\rm d}x = \int_0^{\pi}\frac{1}{1+k\cos{x}}{\rm d}x$ + $\int_\pi^{2\pi}\frac{1}{1+k\cos{x}}{\rm d}x$,

I get in the second integral after substitution $x = z + \pi$:

$\int_0^{\pi}\frac{1}{1+k\cos{(z+\pi)}}{\rm d}z=\int_0^{\pi}\frac{1}{1-k\cos{z}}{\rm d}z$.

What am I doing wrong?

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    I think the very first equality is incorrect. Cosine is $2\pi$-periodic, not $\pi$-periodic. – 2017-02-13
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    @JoshuaRuiter: If think it's correct because, $\cos$ beeing $2\pi-$periodic and also even, so is $f:x\mapsto\frac{1}{1+k\cos(x)}$ and we have $\int_0^{2\pi}f(x)\,dx=\int_{-\pi}^\pi f(x)\,dx=2\int_0^\pi f(x)\,dx$. – 2017-02-13
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    @JoshuaRuiter But note for example that$$\int_0^{2\pi} (x-\pi)^2\,dx=2\int_0^\pi(x-\pi)^2\,dx$$even though $(x-\pi)^2$ is not periodic at all. – 2017-02-13
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    @leosenko: You should mention the hypothesis $\vert k\vert<1$ which insures that the integral is well defined. – 2017-02-13
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    @Ieosenko Also, please stop using the word "trivial". When you haven't solved it up till yet then how can you call it trivial? It might be trivial for someone else, but not for you, unless you verify it yourself. Once you have done it, then only you can say it trivial. – 2017-02-13
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    As noted, your title was inappropriate. I have tried to make it express the question you seem to be asking at the end. But actually $\int_0^\pi\frac{1}{1-k\cos z}{\rm d}z = \int_0^\pi\frac{1}{1+k\cos z}{\rm d}z,$ and you did not make any mistake here. – 2017-02-13
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    For example, compare http://www.wolframalpha.com/input/?i=integral+of+1%2F(1%2B0.5+cos(x))+dx+for+x+%3D+0+to+pi and http://www.wolframalpha.com/input/?i=integral+of+1%2F(1-0.5+cos(x))+dx+for+x+%3D+0+to+pi – 2017-02-13

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What are you doing wrong? - nothing, both answers are correct.

But if you want to get the first answer, substitute $z=2\pi-x$ instead of $z=x-\pi$.