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Math Help - Some trouble with trig integration.

  1. #1
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    Some trouble with trig integration.

    ∫sin^3(x)cos^2(x)dx.

    I know you have to use the sin rule where you can change the sin into a 1-cos.

    I'm not sure how to utilize that.

    I've done u substitution where u=cos(x) and du=sin(x) but that's as far as I got before I was stuck.

    Thanks!
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  2. #2
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    Quote Originally Posted by jelloish View Post
    ∫sin^3(x)cos^2(x)dx.

    I know you have to use the sin rule where you can change the sin into a 1-cos.

    I'm not sure how to utilize that.

    I've done u substitution where u=cos(x) and du=sin(x) but that's as far as I got before I was stuck.

    Thanks!
     \int \sin^3(x)cos^2(x)dx

     \int \sin(x)\sin^2(x)\cos^2(x)dx

     \int \sin(x)(1-\cos^2(x))\cos^2(x)dx

     \int \sin(x)(\cos^2(x)-\cos^4(x))dx

     u = \cos(x)

     -du = \sin(x)

     -\int (u^2-u^4)du

     \int (u^4-u^2)du
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