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    Hi

    I have this integral \displaystyle\int^\pi_0 \frac{dx}{1 + acos(x)}\ \mathrm{, where\ } a < 1 , the answer to which is
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     \frac{\pi}{(1-a^2)^\frac{1}{2}}
    I seem to be useless with the normal tangent half-angle substitution because I can't get my head around the fact that one of my limits becomes undefined. Is there a 'usual' way to deal with this? I have a method to do it by considering the function from 0 to 2\pi and halving the result, but is there a better way?

    (I also suspect the paper meant  |a| < 1 , any thoughts?

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    (Original post by klegend02)
    Hi

    I have this integral \displaystyle\int^\pi_0 \frac{dx}{1 + acos(x)}\ \mathrm{, where\ } a < 1 , the answer to which is
    Spoiler:
    Show
     \frac{\pi}{(1-a^2)^\frac{1}{2}}
    The t = tan(x/2) substitution should work fine - it transforms the integral into a standard inverse trigonometric one.

    I seem to be useless with the normal tangent half-angle substitution because I can't get my head around the fact that one of my limits becomes undefined.
    One of the limits becomes +infinity but that is not a problem ultimately, as its arc-tangent is defined.

    Is there a 'usual' way to deal with this?
    At a more elevated (university) level this is a standard integral to approach with residue theory/complex analysis.

    (I also suspect the paper meant  |a| < 1 , any thoughts?
    Yes, it should have stated that.
 
 
 

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