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Math Help - Integral

  1. #1
    Senior Member slevvio's Avatar
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    Integral

    Hello, some of you may remember me from the previous question.

    <br />
\int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx<br />

    Hence evaluate

    <br />
\int^{\frac{\pi}{2}}_{0} xsin^2xcos^2xdx<br />

    After messing about with it I can never seem to get it into a nicer form... thanks any help would be appreciated
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  2. #2
    Eater of Worlds
    galactus's Avatar
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    You could try rewriting your integral.

    xsin^{2}xcos^{2}x=\frac{x^{2}}{8}-\frac{x^{2}cos(4x)}{8}
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  3. #3
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by slevvio View Post
    Hello, some of you may remember me from the previous question.

    <br />
\int^{a}_{0} f(x)dx = \int^{a}_{0}f(a-x)dx<br />

    Hence evaluate

    <br />
\int^{\frac{\pi}{2}}_{0} xsin^2xcos^2xdx<br />

    After messing about with it I can never seem to get it into a nicer form... thanks any help would be appreciated
    Do you just want to evaluate the integral?

    Try substituting 1-cos^2(x)=sin^2(x) to get:

    <br />
\int^{\frac{\pi}{2}}_{0} x(1-cos^4x)dx<br />

    I then suggest using tabular integration to solve that.
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by slevvio View Post
    evaluate

    <br />
\int^{\frac{\pi}{2}}_{0} xsin^2xcos^2xdx<br />
    Flip it around and substitute u = \frac{\pi }<br />
{2} - x, \varphi becomes (say the integral):

    \varphi  = \int_0^{\pi /2} {\bigg( {\frac{\pi }<br />
{2} - x} \bigg)\sin ^2 x\cos ^2 x\,dx} .

    Hence \varphi  = \frac{\pi }<br />
{2}\int_0^{\pi /2} {\sin ^2 x\cos ^2 x\,dx}  - \int_0^{\pi /2} {x\sin ^2 x\cos ^2 x\,dx} .

    The rest follows easily, the remaining integral can be computed without problems.
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