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Thread: Trouble with Trig integral

  1. #1
    Junior Member Shadow236's Avatar
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    Trouble with Trig integral

    For some reason I am having trouble with the integral: sin2xcos2x dx
    I have tried a few identities with no luck.

    1st try
    --------
    (1/2) | (2sinxcosx)2 dx

    2nd try
    --------
    (1/4) | (1 - cos(2x))(1 + cos(2x)) dx

    The website I am looking at keeps saying to use integration by parts, which makes no sense to me.

    Thanks for your help.
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  2. #2
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    Re: Trouble with Trig integral

    Your approach is absolutely correct.
    first step : Integral sin^2x cos^2x dx = 1/4 Integral sin^2 ( 2x ) dx
    Now use the identity cos 2x = 1 - sin^2 x
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  3. #3
    Newbie Naranja's Avatar
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    Re: Trouble with Trig integral

    Hello,

    $\displaystyle \sin^2x\cos^2x = (\sin x \cos x)^2$

    You can use the double angle formula:

    $\displaystyle (\sin x \cos x) = \frac{\sin (2x)}{2}$

    Thus

    $\displaystyle (\sin x \cos x)^2 = \frac{\sin^2(2x)}{4}$

    Now write $\displaystyle \sin^2(2x) = 1 - \cos^2(2x)$ and subsequently $\displaystyle \cos^2(2x) = \frac{1}{2} + \frac{\cos (4x)}{2}$. So, $\displaystyle \sin^2(2x) = 1 - (\frac{1}{2} + \frac{\cos (4x)}{2}) = \frac{1}{2} - \frac{\cos (4x)}{2}$.

    So, the integral is now: $\displaystyle \frac{1}{4}(\frac{1}{2} - \frac{\cos (4x)}{2}) = \frac{1}{8} - \frac{\cos (4x)}{8}$.

    Integrate that and you get $\displaystyle \frac{x}{8} - \frac{\sin (4x)}{32}$. The only identities you need are $\displaystyle \sin^2x + \cos^2x = 1$ and the double angle formulae $\displaystyle \sin (2x) = 2\sin x\cos x$, and $\displaystyle \cos (2x) = \cos^2x - \sin^2x$. No integration by parts is necessary.
    Last edited by Naranja; May 6th 2013 at 07:13 AM.
    Thanks from Shadow236
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  4. #4
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    Re: Trouble with Trig integral

    I think you're along the right lines for one possible solution with your first try.

    Using the identity:

    $\displaystyle \sin2x = 2\sin x\cos x$

    we can say:

    $\displaystyle \int \sin^2x\cos^2x \ dx = \int (\sin x\cos x)^2 \ dx = \frac{1}{4}\int (2\sin x\cos x)^2 \ dx = \frac{1}{4}\int \sin^2 2x \ dx$

    now, in my understanding we can't integrate $\displaystyle \sin^2x$, so:

    $\displaystyle \cos 4x = \cos^2 2x - \sin^2 2x$

    $\displaystyle \cos 4x = 1 - 2\sin^2 2x$

    $\displaystyle \sin^2 2x = \frac{1-\cos 4x}{2}$

    therefore the integral now becomes:

    $\displaystyle \frac{1}{8} \int (1 - \cos 4x) \ dx = \frac{1}{8}[x - \frac{1}{4} \sin 4x] = \frac{1}{8}x - \frac{1}{32}\sin 4x + c$
    Last edited by Qwob; May 6th 2013 at 07:22 AM.
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  5. #5
    Newbie Naranja's Avatar
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    Re: Trouble with Trig integral

    Qwob, you missed that the angle is doubled twice so you end up with $\displaystyle \sin(4x)$, and of course the coefficient then halves.
    Thanks from topsquark
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  6. #6
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    Re: Trouble with Trig integral

    Thanks Naranja, fixed it now.
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  7. #7
    Junior Member Shadow236's Avatar
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    Re: Trouble with Trig integral

    It looks like it was a little more in depth than I thought, thanks guys! =)
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  8. #8
    Forum Admin topsquark's Avatar
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    Re: Trouble with Trig integral

    This link gives some quick advice about how to proceed with combined sine/cosine integrals. It doesn't actually explain why these rules of thumb work, but if you do several examples I think you'll figure out the why's by yourself. (Or we can help, too!)

    -Dan
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