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Math Help - trig integrals

  1. #1
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    trig integrals

    \int sin^4(2x)dx

    I don't know what theorem to use or if to separate it...

    I just need a reference point for where to go.
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    Quote Originally Posted by saiyanmx89 View Post
    I don't know what theorem to use or if to separate it...

    I just need a reference point for where to go.
    Use \sin^2\theta=\frac{1-\cos2\theta}2 and \cos^2\theta=\frac{1+\cos2\theta}2\colon

    \int\sin^42x\,dx

    =\int\left(\sin^22x\right)^2\,dx

    =\int\left(\frac{1-\cos4x}2\right)^2\,dx

    =\frac14\int\left(1-2\cos4x+\cos^24x\right)\,dx

    =\frac14\left[\int dx-2\int\cos4x\,dx+\int\cos^24x\,dx\right]

    =\frac14\left[\int dx-2\int\cos4x\,dx+\frac12\int(1+\cos8x)\,dx\right]

    The rest should be easy.
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    I don't know how to integrate cos4xdx and (1+cos8x)
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    Quote Originally Posted by saiyanmx89 View Post
    I don't know how to integrate cos4xdx and (1+cos8x)
    Really? Let u=4x (and u=8x for the second). Substitute.

    And you should know that \int\cos x\,dx=\sin x+C (if not, brush up!)
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  5. #5
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    Is the integral of 1+cosu = cos^2u ?
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    I mean derivative...! not integral.
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    Quote Originally Posted by saiyanmx89 View Post
    I mean derivative...! not integral.
    Huh? \frac d{dx}\left[1+\cos u\right]=-(\sin u)\frac{du}{dx}\text.

    What does that have to do with your problem?
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