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Math Help - Difficult Trigonometric Integral

  1. #1
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    Difficult Trigonometric Integral

    I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.

    \int {sin^2  x \over \sqrt {1+cosx}}

    Thanks for the help!
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  2. #2
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    Quote Originally Posted by Tylerzzz View Post
    I was working through my homework and stumbled across this problem. I have tried many methods, but still have no luck. Rationalizing the denominator didn't seem to get me anywhere. WolframAlpha wasn't much either help in showing me the steps.

    \int {sin^2  x \over \sqrt {1+cosx}}
    Substitute u = \cos x. Then du = -\sin x\,dx and the integral becomes

    -\int \frac{\sin  x}{\sqrt {1+cosx}}(-\sin x)dx = -\int \frac{\sqrt{1-u^2}}{\sqrt{1+u}}du = -\int\sqrt{1-u}\,du = \frac23(1-u)^{3/2} =  \frac23(1-\cos x)^{3/2} (plus a constant).
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  3. #3
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    Wow! I never really thought about solving it that way. Thanks a lot!
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    EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]
    Last edited by TheCoffeeMachine; October 11th 2010 at 07:47 AM.
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  5. #5
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    Quote Originally Posted by TheCoffeeMachine View Post
    EDIT: Apologies to Opalg, I mistook it for [LaTeX ERROR: Convert failed]
    You had me worried for a moment.
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  6. #6
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    Hello, Tylerzzz!

    Another approach . . .


    \displaystyle \int \frac{\sin^2\!x}{\sqrt{1+\cos x}}\,dx

    We have: . \dfrac{\sin^2\!x}{\sqrt{1+\cos x}}


    Multiply by \frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}\!:\;\;\dfrac{\sin^2\!x}{\sqrt{1+\cos x}} \cdot\dfrac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{1-\cos^2\!x}}


    . . . \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sqrt{\sin^2\!x}} \;=\;\dfrac{\sin^2\!x\sqrt{1-\cos x}}{\sin x} \;=\;\sin x\sqrt{1-\cos x}


    The integral becomes: . \displaystyle \int(1-\cos x)^{\frac{1}{2}}(\sin x\,dx)


    Let: . u \:=\:1-\cos x \quad\Rightarrow\quad du \:=\:\sin x\,dx


    Substitute: . \displaystyle \int u^{\frac{1}{2}}\,du \;=\;\tfrac{2}{3}u^{\frac{3}{2}} + C


    Back-substitute: . \frac{2}{3}(1-\cos x)^{\frac{3}{2}} + C
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  7. #7
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    Quote Originally Posted by Opalg View Post
    You had me worried for a moment.
    Haha! In my defence, I blame the UCAS for the plight of my eyesight today.
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