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  1. #1
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    help

    intergrate

    $\displaystyle dx/1-cos2x$
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  2. #2
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    Rewrite using half-angle identity.

    $\displaystyle \int \frac {1}{1 - cos2x}dx = \int \frac {1}{2sin^2x}dx = 1/2 \int csc^2xdx = \frac{-cotx}{2} + c$
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  3. #3
    Air
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    Quote Originally Posted by manalive04 View Post
    intergrate

    $\displaystyle dx/1-cos2x$
    $\displaystyle \int \frac{1}{1-\cos \left(2x\right)}\ \mathrm{d}x$

    Let $\displaystyle u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}$

    $\displaystyle \frac12 \int \frac{1}{1-\cos \left(u\right)}\ \mathrm{d}u$

    You can use T-Substitution and solve: $\displaystyle t=\tan \left(\frac{u}{2}\right)$

    Note: $\displaystyle \cos u = \frac{1-t^2}{1+t^2}$ and $\displaystyle \sin u = \frac{2t}{1+t^2}$.
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