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Thread: Integration Problem

  1. #1
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    Integration Problem

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

    So: $\displaystyle
    2 \cdot I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx} + \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\left( {\tfrac{{x \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}} + \tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}} \right)dx}
    $

    We get $\displaystyle
    2 \cdot I = \pi \cdot \int_0^\pi {\left( {\tfrac{{\sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}} \right)dx}
    $ 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 $\displaystyle \pi - x$?
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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 $\displaystyle \pi - x$?
    The substitution $\displaystyle u = \pi - x$ is made and then the dummy variable is switched back to x:


    $\displaystyle I = \int_0^\pi {\frac{{x \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}} \, dx}$


    Substitute $\displaystyle u = \pi - x$:


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


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


    Now switch the dummy variable back to x:


    $\displaystyle = \int_0^\pi {\frac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}}
    {{1 + \sin \left( {\pi - x} \right)}} \, dx}$

    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 $\displaystyle (\pi - u)$ into $\displaystyle (\pi - x)$

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




    $\displaystyle
    I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}}
    {{1 + \sin \left( {\pi - x} \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx}
    $

    $\displaystyle
    2 \cdot I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx} + \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx}

    $

    tell me hw u got this conversion ..............
    Last edited by varunnayudu; Dec 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,
    $\displaystyle
    I=\int^{a}_{b}{f(x)}$

    $\displaystyle I=\int^{a}_{b}{f(a+b-x)}
    $

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

    $\displaystyle 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
    $\displaystyle
    \int^{a}_{b}{f(x)dx}=-\int^{b}_{a}{f(a+b-t)dt}
    $$\displaystyle
    =\int^{a}_{b}{f(a+b-t)dt}
    $
    As we know
    $\displaystyle \int^{x}_{y}{g(\alpha)d{\alpha }}
    =\int^{x}_{y}{g(\beta)d{\beta }}
    $
    So$\displaystyle

    =\int^{a}_{b}{f(a+b-x)dx}=\int^{a}_{b}{f(x)dx}
    $
    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 $\displaystyle (\pi - u)$ into $\displaystyle (\pi - x)$

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




    $\displaystyle
    I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( {\pi - x} \right)}}
    {{1 + \sin \left( {\pi - x} \right)}}dx} = \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx}
    $

    $\displaystyle
    2 \cdot I = \int_0^\pi {\tfrac{{x \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx} + \int_0^\pi {\tfrac{{\left( {\pi - x} \right) \cdot \sin \left( x \right)}}
    {{1 + \sin \left( x \right)}}dx}

    $

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

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

    using the fact that

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

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

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

    or, upon expanding,

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

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

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

    or

    $\displaystyle 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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