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Math Help - calculate

  1. #1
    Super Member dhiab's Avatar
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  2. #2
    MHF Contributor Amer's Avatar
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    \int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+sinx}dx=

    \int_{0}^{\frac{\pi}{2}}\frac{sinx}{1+sinx}\left(\  frac{1-sinx}{1-sinx}\right)dx

    \int_{0}^{\frac{\pi}{2}}\frac{sinx}{cos^2x}dx-\int_{0}^{\frac{\pi}{2}}\frac{sin^2x}{cos^2x}dx

    the first one substitute u=cosx the other tan^2x+1=sec^x .... tan^2x=sec^2-1 and now the integrate for sec^2x =tanx and integrate of -1 =-x

    thats all
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  3. #3
    Super Member dhiab's Avatar
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    they cannot multiply and divide on 1-sinx; because 1-sinx is can equal to zero
    thankyou
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  4. #4
    Super Member dhiab's Avatar
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    help

    help: let us pose t= Tan(x/2)
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  5. #5
    MHF Contributor Amer's Avatar
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    you know the answer right ??

    but if you derive my answer you will get the integral I try it I think there is no problem if I multiply with 1-sinx

    or if the integral have limits then we should look at the interval [0,pi/2]
    I cant multiply with 1-sinx or 1-cosx or 1-tanx .....
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  6. #6
    Super Member dhiab's Avatar
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    Quote Originally Posted by Amer View Post
    you know the answer right ??

    but if you derive my answer you will get the integral I try it I think there is no problem if I multiply with 1-sinx

    or if the integral have limits then we should look at the interval [0,pi/2]
    I cant multiply with 1-sinx or 1-cosx or 1-tanx .....
    remark : 1-sinx=0
    sinx=1
    x=pi/2?????
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  7. #7
    MHF Contributor Amer's Avatar
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    I know I ask if the integral have not limits then I can multiply with 1-sinx
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