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Thread: integration

  1. #1
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    integration

    $\displaystyle

    \int x sin2x cos2x dx

    $

    i started of by making u=2x, du=x, which will cancel out the x and give me

    $\displaystyle
    \int sinucosu du

    $

    how would i go from here
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  2. #2
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    Quote Originally Posted by simsima_1 View Post
    $\displaystyle

    \int x sin2x cos2x dx

    $

    i started of by making u=2x, du=x, which will cancel out the x and give me

    $\displaystyle
    \int sinucosu du

    $

    how would i go from here
    if u=2x then du=2dx not x sorry

    try this trig fomula

    $\displaystyle 2\sin(\theta)\cos(\theta)=\sin(2\theta)$

    so you would get

    $\displaystyle \sin(2x)\cos(2x)=\frac{1}{2}\sin(4x)$

    now we have the integral

    $\displaystyle \frac{1}{2}\int x \sin(4x)dx$

    by parts we get...

    $\displaystyle \frac{1}{2}\int x \sin(4x)dx=\frac{1}{2}[-\frac{x}{4}\cos(4x)+\frac{1}{16}\sin(4x)]=-\frac{x}{8}\cos(4x)+\frac{1}{32}\sin(4x)+C$
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