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  1. #1
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    trig integral

    i need help with this integral....i seem to be getting no where with it


    (1/8)[integral]sin^2(2x)cos(2x)

    thanks
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  2. #2
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    Quote Originally Posted by johntuan View Post
    i need help with this integral....i seem to be getting no where with it


    (1/8)[integral]sin^2(2x)cos(2x)

    thanks
    substitute

    \sin 2x = t

    so that,

     <br />
2\cos 2x \;dx = dt<br />

     <br />
\cos 2x \;dx = \frac{dt}{2}<br />

    Spoiler:
    \frac{1}{8}\int \sin ^2 2x\; \cos 2x~dx=\frac{1}{16}\int t^2~dt
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  3. #3
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    Quote Originally Posted by johntuan View Post
    i need help with this integral....i seem to be getting no where with it


    (1/8)[integral]sin^2(2x)cos(2x)

    thanks
    sin^2(2x) = 1-cos^2(2x)

    \frac{1}{8} \int cos(2x) - cos^3(2x)

    Which should be easier to integrate
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    sin^2(2x) = 1-cos^2(2x)

    \frac{1}{8} \int cos(2x) - cos^3(2x)

    Which should be easier to integrate

    if i continued to integrate would it lead me to the answer of

    (1/8)[(1/3)(1/2)sin^3(2t)]......?

    thats the answer...I'm not sure...
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  5. #5
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    Quote Originally Posted by johntuan View Post
    if i continued to integrate would it lead me to the answer of

    (1/8)[(1/3)(1/2)sin^3(2t)]......?

    thats the answer...I'm not sure...
    Ignore my answer lol, the guy's above me is much simpler XD
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