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Math Help - Prove

  1. #1
    Super Member dhiab's Avatar
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    Prove

    Prove :

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  2. #2
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    Try this.. tan(x)=sin(x)/cos(x) and cot(x) = cos(x)/sin(x) add them both you get c^2+s^2/c*s = 1/c*s

    then it would become integral between 0 and pi/2 --> ln (1/cos(x)*sin*(x)) dx ...

    now you know sin(2x)=2sin(x)cos(x)

    see what you get form here...
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  3. #3
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    I think of another method , old method is to transform it to this famous integral :  \int_0^{\pi/2} \ln[\sin(x)]~dx = - \frac{\pi}{2} \ln(2) .

    Let

     I(a) = \int_0^{\pi/2} \ln[ \tan(x) + a \cot(x) ] ~dx

    Then  I'(a) =  \int_0^{\pi/2} \frac{ \cot(x) }{  \tan(x) + a \cot(x) }~dx

     =  \int_0^{\pi/2} \frac{dx }{  \tan^2(x) + a  }

    Sub.  t = \tan(x) we have

     I'(a) = \int_0^{\infty} \frac{dt}{ (1+t^2)(t^2 + a ) }

     = \frac{1}{a-1 } \int_0^{\infty} \left[ \frac{1}{t^2 + 1 }  -  \frac{1}{t^2 + a} \right] ~dt

     = \frac{1}{a - 1 }  [ \frac{\pi}{2} - \frac{1}{\sqrt{a}} ~ \frac{\pi}{2} ]

     =  \frac{\pi}{2} \cdot \frac{\sqrt{a} - 1 }{\sqrt{a} (a-1) }

     = \frac{\pi}{2} \cdot \frac{1 }{\sqrt{a} (\sqrt{a}+1) }

    Therefore ,  I(1) - I(0) =  \frac{\pi}{2} ~ \int_0^1 \frac{da }{\sqrt{a} (\sqrt{a}+1) }


    Sub.  a = u^2

     = \frac{\pi}{2} \cdot 2\ln(2) = \pi \ln(2)

    Finally note that   \int_0^{\pi/2}  \ln[\tan(x)] ~dx   = 0 so


     \int_0^{\pi/2} \ln[ \tan(x) +  \cot(x) ] ~dx = \pi \ln(2)
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