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Math Help - Trigonometric Integral

  1. #1
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    Trigonometric Integral

    Sorry, I'm new and I don't know how to show mathematical symbols on here.

    The integral from 0 to pi/2 of sin^4x dx. The exponent is just 4, not 4x.

    Thanks a lot!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by element View Post
    Sorry, I'm new and I don't know how to show mathematical symbols on here.

    The integral from 0 to pi/2 of sin^4x dx. The exponent is just 4, not 4x.

    Thanks a lot!
    \int_0^{\frac{\pi}{2}} \sin^4 x\, dx = \int_0^{\frac{\pi}{2}} \sin^2 x(1 - \cos^2x)\, dx

    you can use the following identities:

    \cos 2x = \cos^2 x - \sin^2x = 2\cos^2x - 1 = 1 - 2\sin^2x
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kalagota View Post
    \int_0^{\frac{\pi}{2}} \sin^4 x\, dx = \int_0^{\frac{\pi}{2}} \sin^2 x(1 - \cos^2x)\, dx

    you can use the following identities:

    \cos 2x = \cos^2 x - \sin^2x = 2\cos^2x - 1 = 1 - 2\sin^2x
    Uhm...why rewrite it as that?

    \int\sin^4(x)dx=\int\left(\frac{1-\cos(2x)}{2}\right)^2dx

    Expand that and see that

    \cos^2(nx)=\frac{1+\cos(2nx)}{2}
    Last edited by Mathstud28; June 23rd 2008 at 06:14 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Uhm...why rewrite it as that?

    \int\sin^4(x)dx=\int\left(\frac{1-\cos(2x)}{2}\right)dx

    Expand that and see that

    \cos^2(nx)=\frac{1+\cos(2nx)}{2}
    How is
    sin^4(x) = \frac{1 - cos(2x)}{2}?

    -Dan
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