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Math Help - Integration Problem

  1. #1
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    Integration Problem

    Evaluate
    \int\limits_{0}^{\pi} \frac{xsinx}{1+sinx}dx.
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  2. #2
    Super Member PaulRS's Avatar
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    Set u=\pi-x then <br />
I = \int_0^\pi  {\tfrac{{x \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx}  = \int_0^\pi  {\tfrac{{\left( {\pi  - x} \right) \cdot \sin \left( {\pi  - x} \right)}}<br />
{{1 + \sin \left( {\pi  - x} \right)}}dx}  = \int_0^\pi  {\tfrac{{\left( {\pi  - x} \right) \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx} <br />

    So: <br />
2 \cdot I = \int_0^\pi  {\tfrac{{x \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx}  + \int_0^\pi  {\tfrac{{\left( {\pi  - x} \right) \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx}  = \int_0^\pi  {\left( {\tfrac{{x \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}} + \tfrac{{\left( {\pi  - x} \right) \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}} \right)dx} <br />

    We get <br />
2 \cdot I = \pi  \cdot \int_0^\pi  {\left( {\tfrac{{\sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}} \right)dx} <br />
and this can be solved by standard methods
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  3. #3
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    unable to understand ..

    I am unable to understand hw u carried out the steps.................
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  4. #4
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    Quote Originally Posted by varunnayudu View Post
    I am unable to understand hw u carried out the steps.................
    Read the reply again - carefully. Now try and be more specific - what part or parts don't you understand?
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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    And kindly show your try
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  6. #6
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    I'm also having trouble following.

    Where did u come into it? i.e. how did x become \pi - x?
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  7. #7
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    Quote Originally Posted by Prove It View Post
    I'm also having trouble following.

    Where did u come into it? i.e. how did x become \pi - x?
    The substitution u = \pi - x is made and then the dummy variable is switched back to x:


    I = \int_0^\pi {\frac{{x \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}} \, dx}


    Substitute u = \pi - x:


    = \int_\pi^0 {\frac{{\left( {\pi - u} \right) \cdot \sin \left( {\pi - u} \right)}}{{1 + \sin \left( {\pi - u} \right)}} \, (-du) }


     = \int_0^\pi {\frac{{\left( {\pi - u} \right) \cdot \sin \left( {\pi - u} \right)}}{{1 + \sin \left( {\pi - u} \right)}} \, du }


    Now switch the dummy variable back to x:


     = \int_0^\pi {\frac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}}<br />
{{1 + \sin \left( {\pi - x} \right)}} \, dx}

    etc.

    This is a standard technique.
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  8. #8
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    Still hve a problem




    Now switch the dummy variable back to x:




    ---------------------------------------------

    Hw can one substitute (\pi - u) into (\pi - x)

    and secondly even after doing the sub the sum is not evaluated it is still the same ...........only instead of x u have (\pi - x)




    <br />
I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}}<br />
{{1 + \sin \left( {\pi - x} \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx}<br />

    <br />
2 \cdot I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx} + \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx} <br /> <br />

    tell me hw u got this conversion ..............
    Last edited by varunnayudu; December 28th 2008 at 04:59 AM.
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  9. #9
    Like a stone-audioslave ADARSH's Avatar
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    Hi Varun,
     <br />
I=\int^{a}_{b}{f(x)}

     I=\int^{a}_{b}{f(a+b-x)}<br />

    I hope you know the above property
    Here  <br />
f(x)=\frac{xsinx}{1-sin(x)}<br />
    Hence
    the two equations are
    added

    I+I= 2I
    which is given




    ----------------------------------------------------------------
    to prove that property
    Put t=a+b-x
    dt=-dx
    When x=a ,t=b
    and when x=b,t=a
    Now
     <br />
\int^{a}_{b}{f(x)dx}=-\int^{b}_{a}{f(a+b-t)dt}<br />
 <br />
=\int^{a}_{b}{f(a+b-t)dt}<br />
    As we know
     \int^{x}_{y}{g(\alpha)d{\alpha }} <br />
=\int^{x}_{y}{g(\beta)d{\beta }} <br />
    So  <br /> <br />
=\int^{a}_{b}{f(a+b-x)dx}=\int^{a}_{b}{f(x)dx}<br />
    And that's the end of that show..
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  10. #10
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    Quote Originally Posted by varunnayudu View Post



    Now switch the dummy variable back to x:




    ---------------------------------------------

    Hw can one substitute (\pi - u) into (\pi - x)

    and secondly even after doing the sub the sum is not evaluated it is still the same ...........only instead of x u have (\pi - x)




    <br />
I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}}<br />
{{1 + \sin \left( {\pi - x} \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx}<br />

    <br />
2 \cdot I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx} + \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}<br />
{{1 + \sin \left( x \right)}}dx} <br /> <br />

    tell me hw u got this conversion ..............
    Maybe I can help. After you make the substitution  x = \pi - u your integral, as Mr. F says is

     I = \int_{\pi}^0 \frac{(\pi-u) \sin (\pi-u)}{1 + \sin(\pi-u)}(-du)

    using the fact that

     \sin(\pi-u) = \sin u

    then (I used the negative to switch the limits of integration)

     I = \int_0^{\pi} \frac{(\pi-u) \sin (u)}{1 + \sin(u)}\,du

    or, upon expanding,

     I = \int_0^{\pi} \frac{\pi \sin (u)}{1 + \sin(u)}\,du - \int_0^{\pi} \frac{u \sin (u)}{1 + \sin(u)}\,du

    If you look at your original integral and the second one of the right these are the same except for the variable u. As this is only a dummy variable (it is used and never appears in the answer) then we can replace u with x giving

     I = \int_0^{\pi} \frac{\pi \sin (u)}{1 + \sin(u)}\,du - I

    or

     2I = \int_0^{\pi} \frac{\pi \sin (u)}{1 + \sin(u)}\,du

    as others have said. Others can correct me if I'm wrong but I believe this question appears in Stewart's Calculus book.
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