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Math Help - Integral Problem ?

  1. #1
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    Integral Problem ?

    I have difficulty to solve this problem...plz help me
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  2. #2
    MHF Contributor harish21's Avatar
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    use \sin (x) = 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})

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

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

    let\;u=\frac{x}{2} \rightarrow dx = 2du

    \displaystyle \int_0^{\frac{\pi}{2}} \ln(\sin(x))  =  \dfrac{\pi\ln(2)}{2}+ 2 \int_0^{\frac{\pi}{4}} \ln(\sin(u))\;du+2\int_0^{\frac{\pi}{4}}\ln(\cos(u  ))\;du........(I)

    In the last integral Let u=\frac{\pi}{2}-t

    \displaystyle \int_0^{\frac{\pi}{4}}\ln(\cos(u))\;du= -\int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \ln(\sin(t)\;dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln(\sin(t)\;dt

    now substitute this back in (I) to finish..
    Last edited by harish21; November 30th 2010 at 04:04 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by harish21 View Post
    use \sin (x) = 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})

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

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

    let\;u=\frac{x}{2} \rightarrow dx = 2du

    \displaystyle \int_0^{\frac{\pi}{2}} \ln(\sin(x))  =  \dfrac{\pi\ln(2)}{2}+ 2 \int_0^{\frac{\pi}{4}} \ln(\sin(u))\;du+2\int_0^{\frac{\pi}{4}}\ln(\cos(u  ))\;du........(I)

    In the last integral Let u=\frac{\pi}{2}-t

    \displaystyle \int_0^{\frac{\pi}{4}}\ln(\cos(u))\;du= -\int_{\frac{-\pi}{2}}^{\frac{\pi}{4}} \ln(\sin(t)\;dt = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln(\sin(t)\;dt

    now substitute this back in (I) to finish..

    Very nice, indeed! As I knew it took at least twice more calculations. Just a little observation: in the last line, first integral from the right,

    the lower limit must be \frac{\pi}{2} , not minust this.

    Tonio
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  4. #4
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    great work
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