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Thread: 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?

    $\displaystyle \int sin2xcos2xdx$

    Letting $\displaystyle u=sin2x, du=cos2xdx$

    $\displaystyle \int udu$
    $\displaystyle \frac{u^2}{2}+C$
    $\displaystyle \frac{sin^{2}2x}{2}+C$
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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?

    $\displaystyle \int sin2xcos2xdx$

    Letting $\displaystyle u=sin2x, du=cos2xdx$

    $\displaystyle \int udu$
    $\displaystyle \frac{u^2}{2}+C$
    $\displaystyle \frac{sin^{2}2x}{2}+C$
    If $\displaystyle u=sin(2x)$, then $\displaystyle du={\color{red}2}cos(2x)dx$.
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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
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    Surely, you learned the chain rule. Right?
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    Quote Originally Posted by xfriendsonfirex View Post
    Could you walk me through why that is true?

    chain rule.

    $\displaystyle sin(2x)$ is really the composition of $\displaystyle f(x)=sin(x)$ and $\displaystyle g(x)=2x$, so that $\displaystyle f(g(x)) = sin(2x)$. this then means that $\displaystyle f(g(x))^{'} = f^{'}(g(x))g^{'}(x)$ by the chain rule. So $\displaystyle f(g(x))^{'} = f^{'}(g(x))g^{'}(x) = cos(2x)dx*2$
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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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