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Math Help - Integration

  1. #1
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    Integration

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

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

    we have:

    \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 . . .


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

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

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


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

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