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Math Help - integration by partial fractions

  1. #1
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    integration by partial fractions

    How do I integrate this integral using partial fractions?

    \int_ dx/( Asin[x] + Bcos[x])^2 where a =/ 0

    Thanks
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  2. #2
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    Quote Originally Posted by Befordxide View Post
    How do I integrate this integral using partial fractions?

    \int_ dx/( Asin[x] + Bcos[x])^2 where a =/ 0

    Thanks
    Say A,B\not = 0

    \int \frac{dx}{A\sin x+B\cos x}

    Write,

    \frac{1}{A\sin x + B\cos x} = \frac{1}{A\cdot \frac{2\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}} + B\cdot \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}} = \frac{\frac{1}{2}(1+\tan^2 \frac{x}{2})}{A\tan^2 \frac{x}{2} +\frac{1}{2}B(1-\tan^2 \frac{x}{2} ) }

    The substituion t = \tan \frac{x}{2} with t'= \frac{1}{2}(1+\tan^2 \frac{x}{2})

    Will convert the integral into,

    \int \frac{dt}{At^2+\frac{1}{2}(B-t^2)}

    Can you do it from there?
    Last edited by ThePerfectHacker; May 25th 2007 at 12:28 PM.
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  3. #3
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    Hello, Befordxide!

    I have a back-door approach . . .


    \int \frac{dx}{(A\!\cdot\!\sin x  + B\!\cdot\!\cos x )^2}

    In the denominator we have: . A\!\cdot\!\sin x + B\!\cdot\!\cos x

    Multiply top and bottom by \sqrt{A^2+B^2}: . \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}}\sin x + \frac{B}{\sqrt{A^2+B^2}}\cos x\right)


    Let \theta be an acute angle in a right triangle with: . adj = A,\:opp = B
    . . Then: . hyp = \sqrt{A^2+B^2}. . . Note that: . \tan\theta = \frac{B}{A}
    . . Hence: . \cos\theta \,= \,\frac{A}{\sqrt{A^2+B^2}},\;\;\sin\theta \,= \,\frac{B}{\sqrt{A^2+B^2}}


    Then we have: . \sqrt{A^2+B^2}\left(\cos\theta\!\cdot\!\sin x + \sin\theta\!\cdot\!\cos x\right) \;=\;\sqrt{A^2+B^2}\cdot\sin(x + \theta)


    The denominator becomes: . (A^2+B^2)\sin^2(x + \theta)

    The integral becomes: . \frac{1}{\sqrt{A^2+B^2}}\int\frac{dx}{\sin^2(x + \theta)} \;=\;\frac{1}{\sqrt{A^2+B^2}}\int\csc^2(x + \theta)\,dx


    Can you finish it now?

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