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Thread: Integration

  1. #1
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    Integration

    I need help with the following : int. cosxsinxdx
    Thanks!
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  2. #2
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    $\displaystyle \int \cos^2(x) \sin^2(x) \ dx = \int \bigg{(} \cos(x) \sin(x) \bigg{)}^2 \ dx $

    now if we use the identity that $\displaystyle 2 \cos(x) \sin(x) = \sin(2x)$

    we have:

    $\displaystyle \int \bigg{(} \frac{1}{2} \sin(2x) \bigg{)}^2 \ dx$ which should be simpler to integrate
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  3. #3
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    Hello, nystudent2729!

    Another substitution . . .


    $\displaystyle \int \cos^2\!x \sin^2\!x\,dx$

    $\displaystyle \int\cos^2\!x\sin^2\!x\,dx \;=\;\int\left(\frac{1+\cos2x}{2}\right)\left(\fra c{1-\cos2x}{2}\right)\,dx$

    . . $\displaystyle =\;\frac{1}{4}\int(1-\cos^2\!2x)\,dx \;=\;\frac{1}{4}\int\sin^2\!2x\,dx$ .
    . . . which is exactly what $\displaystyle ^{|||||}$ got.


    Then: .$\displaystyle \frac{1}{4}\int\left(\frac{1-\cos4x}{2}\right)\,dx \;=\;\frac{1}{8}\int(1 - \cos4x)\,dx $ . . . . etc.

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