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Math Help - integration

  1. #1
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    integration

    ∫Sin2x/Sin^4x+cos^4x
    can u pls help me
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  2. #2
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    This is almost undreadable... is it \displaystyle \int{\frac{\sin{2x}}{\sin^4{x} + \cos^4{x}}\,dx} or \displaystyle \int{\frac{\sin{2x}}{\sin^4{x}} + \cos^4{x}\,dx}?
    Last edited by mr fantastic; March 8th 2011 at 04:33 AM.
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  3. #3
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    Undreadable? We shouldn't dread it then? I will assume that you mean
    \int \frac{sin(2x)}{sin^4(x)+ cos^4(x)}dx
    since the other would be almost trivial.

    Use trig identities:
    sin^2(x)= \frac{1}{2}(1- cos(2x)) and cos^2(x)= \frac{1}{2}(1+ cos(2x))

    sin^4(x)= \left(\frac{1}{2}(1- cos(2x))\right)^2= \frac{1}{4}(1- 2cos(2x)+ cos^2(2x)
    cos^4(x)= \left(\frac{1}{2}(1+ cos(2x))\right)^2= \frac{1}{4}(1+ 2cos(2x)+ cos^2(2x)
    sin^4(x)+ cos^4(x)= \frac{1}{2}(1+ cos^2(2x))
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Undreadable? We shouldn't dread it then? I will assume that you mean
    \int \frac{sin(2x)}{sin^4(x)+ cos^4(x)}dx
    since the other would be almost trivial.

    Use trig identities:
    sin^2(x)= \frac{1}{2}(1- cos(2x)) and cos^2(x)= \frac{1}{2}(1+ cos(2x))

    sin^4(x)= \left(\frac{1}{2}(1- cos(2x))\right)^2= \frac{1}{4}(1- 2cos(2x)+ cos^2(2x)
    cos^4(x)= \left(\frac{1}{2}(1+ cos(2x))\right)^2= \frac{1}{4}(1+ 2cos(2x)+ cos^2(2x)
    sin^4(x)+ cos^4(x)= \frac{1}{2}(1+ cos^2(2x))
    Alternatively:

    \displaystyle \int{\frac{\sin{2x}}{\sin^4{x} + \cos^4{x}}\,dx} = \int{\frac{2\sin{x}\cos{x}}{\sin^4{x} + (1 - \sin^2{x})^2}}

    \displaystyle = \int{\frac{2\sin{x}\cos{x}}{\sin^4{x} + 1 - 2\sin^2{x} + \sin^4{x}}\,dx}

    \displaystyle = \int{\frac{2\sin{x}\cos{x}}{2\sin^4{x} - 2\sin^2{x} + 1}\,dx}.

    Now make the substitution \displaystyle u = \sin^2{x} \implies du = 2\sin{x}\cos{x}\,dx and the integral becomes

    \displaystyle \int{\frac{1}{2u^2 - 2u + 1}\,du}

    \displaystyle = \frac{1}{2}\int{\frac{1}{u^2 - u + \frac{1}{2}}\,du}

    \displaystyle = \frac{1}{2}\int{\frac{1}{u^2 - u + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{1}{2}}\,du}

    \displaystyle = \frac{1}{2}\int{\frac{1}{\left(u - \frac{1}{2}\right)^2 + \frac{1}{4}}\,du}

    Now make the substitution \displaystyle u - \frac{1}{2} = \frac{1}{2}\tan{\theta} \implies du = \frac{1}{2}\sec^2{\theta}\,d\theta and the integral becomes

    \displaystyle \frac{1}{2}\int{\frac{\frac{1}{2}\sec^2{\theta}}{\  left(\frac{1}{2}\tan{\theta}\right)^2 + \frac{1}{4}}\,d\theta}

    \displaystyle = \frac{1}{4}\int{\frac{\sec^2{\theta}}{\frac{1}{4}\  tan^2{\theta} + \frac{1}{4}}\,d\theta}

    \displaystyle = \int{\frac{\sec^2{\theta}}{\tan^2{\theta} + 1}\,d\theta}

    \displaystyle = \int{\frac{\sec^2{\theta}}{\sec^2{\theta}}\,d\thet  a}

    \displaystyle = \int{1\,d\theta}

    \displaystyle = \theta + C

    \displaystyle = \arctan{(2u-1)} + C

    \displaystyle = \arctan{(2\sin^2{x} - 1)} + C.
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  5. #5
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    It was calculated in here (through many different methods).
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  6. #6
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    Quote Originally Posted by Ilsa View Post
    ∫Sin2x/Sin^4x+cos^4x
    can u pls help me
    thanku it was the first one by the way. thank u for ur help
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  7. #7
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    thanku mr fantastic it was the frst one by the way
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