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Math Help - Integral trig

  1. #1
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    Integral trig

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

    How do I evaluate this?
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  2. #2
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    I=\int x\sin^2(x)\cos(x)dx

    Use product rule: \int uv' = uv -\int u'v

    Let \ u=x \ \ so: \ \ u'=1

    Let \ v'=sin^2xcosx \ \ so: \ \ v=\frac{1}{3}sin^3x

    I=uv - \int u'v = \frac{1}{3}xsin^3x - J
    where
    J=\frac{1}{3}\int sin^3xdx

    =\frac{1}{3}\int sinx(1-cos^2x)dx

    =\frac{1}{3}\int sinx-cos^2xsinxdx

    =\frac{1}{3}((-cosx)-(-\frac{1}{3}cos^3x)) + C

    I=\frac{1}{3}(xsin^3x+cosx-\frac{1}{3}cos^3x) + C
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  3. #3
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    Quote Originally Posted by usagi_killer View Post
    \int x\sin^2(x)\cos(x)dx

    How do I evaluate this?
    Let u=x

    dv=Sin^2xCosxdx=w^2dw

    w=Sinx,\  dw=Cosxdx

    v=\frac{w^3}{3}

    \int{xSin^2xCosx}dx=x\ \frac{Sin^3x}{3}-\frac{1}{3}\int{w^3}dx=x\ \frac{Sin^3x}{3}-\frac{1}{3}\int{Sin^3x}dx

    Evaluating the resulting integral...

    \int{Sin^3x}dx=\int{Sin^2xSinx}dx=\int{(1-Cos^2x)Sinx}dx

    let u=Cosx,\ -du=Sinxdx

    gives -\int{(1-u^2)}du=\int{(u^2-1)}du=\frac{1}{3}u^2-u+C=\frac{1}{3}Cos^3x-Cosx+C

    \int{xSin^2xCosx}dx=x\ \frac{Sin^3x}{3}-\frac{1}{9}Cos^3x+\frac{1}{3}Cosx+C
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