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Thread: Another interesting integral

  1. #1
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    Another interesting integral

    evaluate the integral $\displaystyle \int_{0}^{\frac{\pi}{2}} x \cot(x) ~dx $

    In fact , it is not a challenge problem for an experienced integrator
    but i believe that this may help remind us of some techniques
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  2. #2
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    Quote Originally Posted by simplependulum View Post
    evaluate the integral $\displaystyle \int_{0}^{\frac{\pi}{2}} x \cot(x) ~dx $

    In fact , it is not a challenge problem for an experienced integrator
    but i believe that this may help remind us of some techniques
    let $\displaystyle x=u$ and $\displaystyle \cot x \ dx = dv.$ then integration by parts gives us: $\displaystyle I=\int_0^{\frac{\pi}{2}} x \cot x \ dx = -\int_0^{\frac{\pi}{2}} \ln(\sin x) \ dx.$ thus:
    $\displaystyle -I=\frac{\pi}{2} \ln 2 + \int_0^{\frac{\pi}{2}} \ln(\sin(x/2)) \ dx + \int_0^{\frac{\pi}{2}} \ln(\cos(x/2)) \ dx$

    $\displaystyle =\frac{\pi}{2} \ln 2 + \int_0^{\frac{\pi}{2}} \ln(\sin(x/2)) \ dx + \int_{\frac{\pi}{2}}^{\pi} \ln(\sin(x/2)) \ dx$

    $\displaystyle =\frac{\pi}{2} \ln 2 + \int_0^{\pi} \ln(\sin(x/2)) \ dx=\frac{\pi}{2} \ln 2 + 2\int_0^{\frac{\pi}{2}} \ln(\sin x) \ dx$

    $\displaystyle =\frac{\pi}{2} \ln 2 - 2I.$ therefore: $\displaystyle I=\frac{\pi}{2}\ln 2.$
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  3. #3
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    terrific !
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