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Math Help - [SOLVED] Evalute integral

  1. #1
    Senior Member Twig's Avatar
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    [SOLVED] Evalute integral

    Hi!

    Problem:  \int_{0}^{\frac{\pi}{4}} \frac{dx}{1+sin(x)cos(x)} \; dx

    Just point me in the right direction.

    Thx!
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  2. #2
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    \int_{0}^{\frac{\pi}{4}}\frac{1}{1+sin(x)cos(x)}dx

    Let x=tan^{-1}(u), \;\ dx=\frac{1}{u^{2}+1}du

    \int\frac{1}{u^{2}+u+1}du

    Now, let u=\frac{2w-1}{2}, \;\ du=dw

    Getting:

    4\int\frac{1}{3+4w^{2}}dw

    Now, it should be easier. Don't forget to change your limits of integration.
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  3. #3
    Senior Member Twig's Avatar
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    Hi!

    I dont follow from second to third line.
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  4. #4
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    Quote Originally Posted by Twig View Post
    Hi!

    I dont follow from second to third line.
    Because \sin(x)\cos(x)=\frac{\sin(2x)}{2}

    If you can't see it, first substitute t=2x
    And then apply Weierstrass substitution ( w=\tan\tfrac t2)
    It's equivalent.
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Twig View Post
    Hi!

    I dont follow from second to third line.
     \sin(x)\cos(x) = \frac{\tan x}{1+\tan^{2}x} = \frac {u}{1+u^{2}}

    so  \int \frac {dx}{1+ \cos x \sin x} = \int \frac {\frac{du}{1+u^{2}}}{1 + \frac {u}{1+u^{2}}}  = \int \frac {du}{1+u+u^{2}}
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  6. #6
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     \int \frac{dx}{ 1 + \sin{x} \cos{x}}

     = \int \frac{2dx}{2 + \sin2x }

    Sub  x = \frac{\pi}{4} - u

     - \int \frac{2du}{ \sin^2{u} + \cos^2{u} + 2\cos^2{u}}

     = -\int \frac{2 \sec^2{u} du}{3 + \tan^2{u}}

     =  \frac{-2 }{ \sqrt{3}} \tan^{-1}{\frac{tan(\pi/4 - x)}{\sqrt{3}}}
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