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I have been staring at this integral for a while and I don't quite know how to take the first step. Is there some trick that I have to use to manipulate the function first?

Integral in question:

$$\int^{2\pi}_0 \frac{dx}{a + \cos x}$$

Any help/hints/insights is deeply appreciated.

kennytm
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some1fromhell
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  • @kennytm, I think from the tags the OP is interested in a complex analysis approach. – Zaid Alyafeai Mar 21 '17 at 03:33
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    Note the on the unit circle we have $z+z^{-1} =2 \cos(\theta)$ – Zaid Alyafeai Mar 21 '17 at 03:34
  • @ZaidAlyafeai OK maybe http://math.stackexchange.com/questions/146761/compute-integral-int-02-pi-frac1z-cos-phi-d-phi and http://math.stackexchange.com/questions/2030472/trigonometrical-integral-of-1-b-cos-theta and http://math.stackexchange.com/questions/1760520/int2-pi-0-frac1-sqrt5-cos-tdt-int2-pi-0-frac-cos2t-5 and http://math.stackexchange.com/questions/1027486/integral-int-02-pi-fracdx2-cosx and http://math.stackexchange.com/questions/833442/contour-integral-of-int-02-pi-frac1a-cos-theta-d-theta – kennytm Mar 21 '17 at 03:40

2 Answers2

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Hint this can be rewritten as

$$\int_{|z|=1}\frac{\frac{dz}{iz}}{a+\left(z+z^{-1} \right)/2} = -2i\int_{|z|=1}\frac{dz}{z^2+2az+1} $$

Now you need to look at the roots of $z^2+2az+1 =0$ and some conditions on $a$ to see if the poles lie inside the circle or outside.

Zaid Alyafeai
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A possible way to do this: Let $t= tan(x/2)$ and the integral will be transformed to $2 \int \frac{dt}{((a-1)t^2+(a+1))}$. Then use the Newton-Leibniz formula with a careful discussion at $x= \pi$

Johnny
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