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Math Help - Integration with Trigonometric Functions

  1. #1
    Junior Member xfriendsonfirex's Avatar
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    Integration with Trigonometric Functions

    Here's what I have so far, but I'm coming up with answers other than the book value. Mind pointing out my error?

    \int sin2xcos2xdx

    Letting u=sin2x, du=cos2xdx

    \int udu
    \frac{u^2}{2}+C
    \frac{sin^{2}2x}{2}+C
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  2. #2
    Super Member General's Avatar
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    Quote Originally Posted by xfriendsonfirex View Post
    Here's what I have so far, but I'm coming up with answers other than the book value. Mind pointing out my error?

    \int sin2xcos2xdx

    Letting u=sin2x, du=cos2xdx

    \int udu
    \frac{u^2}{2}+C
    \frac{sin^{2}2x}{2}+C
    If u=sin(2x), then du={\color{red}2}cos(2x)dx.
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  3. #3
    Junior Member xfriendsonfirex's Avatar
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    Could you walk me through why that is true?
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  4. #4
    Super Member General's Avatar
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    Surely, you learned the chain rule. Right?
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  5. #5
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    Quote Originally Posted by xfriendsonfirex View Post
    Could you walk me through why that is true?

    chain rule.

    sin(2x) is really the composition of f(x)=sin(x) and g(x)=2x, so that f(g(x)) = sin(2x). this then means that f(g(x))^{'} = f^{'}(g(x))g^{'}(x) by the chain rule. So f(g(x))^{'} = f^{'}(g(x))g^{'}(x) = cos(2x)dx*2
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  6. #6
    Junior Member xfriendsonfirex's Avatar
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    It has been a while. Haha.

    I just got the gist of it from Google, but a more problem specific view would surely help.
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  7. #7
    Junior Member xfriendsonfirex's Avatar
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    I think I got it down. Thanks guys.
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