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Math Help - Trigonometric Integrals

  1. #1
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    Trigonometric Integrals

    Can anyone help me or give me a good idea of how to approach the definite integral ([sinx]^2)*([cosx]^2) with the limits of integration from 0 to pi/2? I'm pretty sure I should somehow be able to simplify it using half-angle identities but I'm stuck. Thanks
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\sin ^2 x\cos ^2 x = \frac{1}<br />
{4}\sin ^2 2x = \frac{1}<br />
{4}\left( {\frac{{1 - \cos 4x}}<br />
{2}} \right) = \frac{{1 - \cos 4x}}<br />
{8}

    <br /> <br />
^* \sin 2x = 2\sin x\cos x<br /> <br /> <br />
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  3. #3
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    Use the identities sin^{2}(x)=\frac{1-cos(2x)}{2} and cos^{2}(x)=\frac{(1+cos(2x)}{2}

    Then, \frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos(2x))(1+cos(2x))dx

    =\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos^{2}(2x))dx =\frac{1}{4}\int_{0}^{\frac{\pi}{2}}sin^{2}(2x)dx =\frac{1}{8}\int_{0}^{\frac{\pi}{2}}(1-cos(4x))dx

    Now continue.
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  4. #4
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    Is that an identity? That [1-cos(2x)^2] = sin(2x)^2 ? I'm not questioning you, I'm just making sure I understand what you did. Thanks
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  5. #5
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    Yes, that is an identity. Try deriving it.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thejabronisayz View Post
    Is that an identity? That [1-cos(2x)^2] = sin(2x)^2 ? I'm not questioning you, I'm just making sure I understand what you did. Thanks
    You know that 1 - cos^2(y) = sin^2(y) right? Well, let y = 2x ...

    -Dan
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