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    Hi, I get zero for the answer of this integral but I think the answer is PiSqrt3/3.

    I = Integral (0 to Pi) dx/(2 + cosx)

    t = tan(x/2) => cosx = (1 - t^2)/(1 + t^2), dx = 2dt/(1 + t^2)
    x = Pi => t = Infinity, x = 0 => t = 0

    :. I = 2 . Integral (0 to Infinity) 1/(3 + t^2) dt
    Let t = Sqrt3.sinhu => dt = Sqrt3.coshu du
    t = Infinity => u = Infinity
    t = 0 => u = 0

    :. I = (2Sqrt3/3) Integral (0 to Infinity) sechu du
    = (2Sqrt3)[tanhu/coshu] (0 to Infinity)
    = 0

    I think my mistake is evaluating the limit of tanhu/coshu as u -> Infinity, seems like I need Pi/2 as the answer; I get 1/Infinity = 0 though!

    Help appreciated.
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    (Original post by Nima)
    :. I = (2Sqrt3/3) Integral (0 to Infinity) sechu du
    = (2Sqrt3)[tanhu/coshu] (0 to Infinity)
    Hmm?

    Also why did you use t=sqrt(3)sinh(u)? It seems to me that t=sqrt(3)tan(u) works better.
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    (Original post by dvs)
    Hmm?

    Also why did you use t=sqrt(3)sinh(u)? It seems to me that t=sqrt(3)tan(u) works better.
    True perhaps the tan sub would be better.

    But: I = 2 . Integral (0 to infinity) dt/(3 + t^2)
    = 2 . Integral (0 to infinity) 1/(3cosh^2u) . Sqrt3 cosh u du
    = 2Sqrt3/3 . Integral (0 to infinity) 1/coshu du
    = 2Sqrt3/3 . Integral (0 to infinity) sechu du
    = (2Sqrt3/3)[sechutanhu] (0 to infinity)
    = (2Sqrt3/3)[tanhu/coshu] (0 to infinity)

    and so on...
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    (Original post by Nima)
    = 2Sqrt3/3 . Integral (0 to infinity) sechu du
    = (2Sqrt3/3)[sechutanhu] (0 to infinity)
    You're not integrating sech(u) properly.
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    (Original post by dvs)
    You're not integrating sech(u) properly.
    Ooops I differentiated it instead Thanks for spotting that silly error...
    And yes tan sub works better.

    Thanks!
 
 
 
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