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Thread: A Trig Integral

  1. #1
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    A Trig Integral

    Hi everyone. I did a trig integral problem. Can someone please check my work for me? Thanks.

    $\displaystyle \int sin(3x)^3cos(3x) $


    $\displaystyle

    u=sin(3x) du=cos3x dx

    $

    $\displaystyle \int u^3 du $


    $\displaystyle u^4/4+c$

    $\displaystyle sin(4x)^4/4+c$


    Thanks a lot, everyone!
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  2. #2
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    Quote Originally Posted by zhangxupage2 View Post
    [FONT=Verdana]Hi everyone. I did a trig integral problem. Can someone please check my work for me? Thanks.

    $\displaystyle \int sin(3x)^3cos(3x) $


    $\displaystyle

    u=sin(3x) du=cos3x dx

    \int u^3 du

    u^4/4+c

    sin(4x)^4/4+c
    $

    Thanks a lot, everyone!
    $\displaystyle du=3\cos{(3x)}dx$
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  3. #3
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    Thanks a lot. I thought I made a mistake.
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  4. #4
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    Here's my revised answer

    $\displaystyle u=sin(3x) du=3cos(3x)dx$

    $\displaystyle
    \int u^3 du/3
    $

    $\displaystyle u^4/4*1/3$

    $\displaystyle sin(3x)^4/12+c$

    Is this correct?
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  5. #5
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    Quote Originally Posted by zhangxupage2 View Post
    Here's my revised answer

    $\displaystyle u=sin(3x) du=3cos(3x)dx$

    $\displaystyle
    \int u^3 du/3
    $

    $\displaystyle u^4/4*1/3$

    $\displaystyle sin(3x)^4/12+c$

    Is this correct?
    Correct
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  6. #6
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    Thanks a lot.
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