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Thread: Help (trig intergration)

  1. #1
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    Help (trig intergration)

    $\displaystyle \int$ (o to pie/2) $\displaystyle sin^2xcos^2x dx$

    please if your gonna help dont jump steps like crazy. cause i probably wont understand..

    this problem i believe u have to use $\displaystyle sin^2x = \frac{1-cos2x}{2} $ or $\displaystyle cos^2x= \frac{1+cos2x}{2} $
    Last edited by Legendsn3verdie; May 2nd 2008 at 01:20 PM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    Using what you've written :

    $\displaystyle
    I=\int_0^{\frac{\pi}{2}}\sin^2x\cos^2x\,\mathrm{d} x=
    \int_0^{\frac{\pi}{2}}\frac{1-\cos (2x)}{2}\cdot \frac{1+\cos (2x)}{2}\,\mathrm{d}x = \int_0^{\frac{\pi}{2}}\frac{1-\cos^2(2x)}{4}\,\mathrm{d}x$ because $\displaystyle (a-b)(a+b)=a^2-b^2$
    $\displaystyle I=\frac{1}{4}\int_0^{\frac{\pi}{2}}1-\cos^2(2x)\,\mathrm{d}x$

    I let you take it from here : you have to find a way of getting rid off the remaining $\displaystyle \cos^2$.

    Hope that helps.
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  3. #3
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    Hello, Legendsn3verdie!

    $\displaystyle \int^{\frac{\pi}{2}}_0\sin^2\!x\cos^2\!x\,dx$
    I think it's easier if we use: . $\displaystyle 2\sin\theta \cos\theta \;=\;\sin2\theta\quad\text{ and }\quad\sin^2\!\theta \:=\:\frac{1-\cos2\theta}{2}
    $

    We have: .$\displaystyle \sin^2\!x\cos^2\!x \;=\;\frac{1}{4}(4\sin^2\!x\cos^2\!x) \;=\;\frac{1}{4}(2\sin x\cos x)^2 \;=\;\frac{1}{4}(\sin2x)^2 $

    . . $\displaystyle = \;\frac{1}{4}\sin^2\!2x \;=\;\frac{1}{4}\left(\frac{1-\cos4x}{2}\right) \;=\;\frac{1}{8}(1 - \cos4x) $


    Then we have: .$\displaystyle \frac{1}{8}\int^{\frac{\pi}{2}}_0(1 - \cos4x)\,dx \quad\hdots\quad \text{okay?}$

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