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Math Help - Trigonometric substitution problem

  1. #1
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    Trigonometric substitution problem

    I'm supposed to integrate 8*cos^3(2x)*sin(2x).

    I'm whittling it down. I factor out a cosine and use the property cos^2(2x)=1-sin^2(2x).

    This gives me the integral of 8*((1-sin^2(2x))*cos(2x)*sin(2x)).

    Here I substitute u=sin(2x) and du=2cos(2x)

    So I have 1/2 the integral of (1-u^2)*u*du

    I suspect that by now I've messed it up. The final answer I SHOULD get is -(cos(2x)^4) What's going on?


    Thanks in advance!
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  2. #2
    MHF Contributor harish21's Avatar
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    let u=\cos(2x) and proceed
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  3. #3
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    Make \displaystyle u=\cos 2x
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  4. #4
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    How hard should I hit myself for missing that?
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  5. #5
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    Quote Originally Posted by Wolvenmoon View Post
    I'm supposed to integrate 8*cos^3(2x)*sin(2x).

    I'm whittling it down. I factor out a cosine and use the property cos^2(2x)=1-sin^2(2x).

    This gives me the integral of 8*((1-sin^2(2x))*cos(2x)*sin(2x)).

    Here I substitute u=sin(2x) and du=2cos(2x)

    So I have 1/2 the integral of (1-u^2)*u*du

    I suspect that by now I've messed it up. The final answer I SHOULD get is -(cos(2x)^4) What's going on?


    Thanks in advance!
    Your method, though more involved, was still correct.

    Continuing....

    \displaystyle\ 4\int{\left(u-u^3\right)du=4\left(\frac{u^2}{2}-\frac{u^4}{4}\right)+C

    =2u^2-u^4+C=2sin^2(2x)-sin^4(2x)+C=2\left[1-cos^2(2x)\right]-\left[1-cos^2(2x)\right]^2+C

    =\displaystyle\ 2-2cos^2(2x)-\left[1-2cos^2(2x)+cos^4(2x)\right]+C

    =2-2cos^2(2x)-1+2cos^2(2x)-cos^4(2x)+C

    =1-cos^4(2x)+C=-cos^4(2x)+C_1
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  6. #6
    Master Of Puppets
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    Quote Originally Posted by Wolvenmoon View Post
    How hard should I hit myself for missing that?
    Hahaha, no don't hit yourself yet, you are only at the beginning of your journey into integration, there will be plenty of oppurtunities, I promise...
    Last edited by pickslides; February 15th 2011 at 02:34 PM. Reason: sp
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