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Math Help - integral

  1. #1
    Member akhayoon's Avatar
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    integral

    \int sin^{4}x

    so what I did was trying to cut it up

    \int sin^{2}x(1-cos^{2}x)

    eventually I realized this method was wrong because I got a big fat zero


    so any ideas???
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by akhayoon View Post
    \int sin^{4}x

    so what I did was trying to cut it up

    \int sin^{2}x(1-cos^{2}x)

    eventually I realized this method was wrong because I got a big fat zero


    so any ideas???
    \int{sin^4(x)dx}=\int\bigg(\frac{1-cos(2x)}{2}\bigg)^2dx=\frac{1}{4}\int[1-2cos(2x)+cos^2(2x)]dx =\frac{1}{4}\int\bigg[1-2cos(2x)+\frac{1+cos(4x)}{2}\bigg]dx...I think you can go from there
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by akhayoon View Post
    \int sin^{4}x

    so what I did was trying to cut it up

    \int sin^{2}x(1-cos^{2}x)

    eventually I realized this method was wrong because I got a big fat zero


    so any ideas???
    try the identity

    \sin^2(x)=\frac{1}{2}(1-\cos(2x))

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

    \frac{1}{4}\int 1-2\cos(2x)+\cos^2(2x)dx

    Now use the cos version of the above

    \cos^{2}(2x)=\frac{1}{2}(1+\cos(4x))

    Then just integrate.

    Good luck.
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