1. ## help

intergrate

$dx/1-cos2x$

2. Rewrite using half-angle identity.

$\int \frac {1}{1 - cos2x}dx = \int \frac {1}{2sin^2x}dx = 1/2 \int csc^2xdx = \frac{-cotx}{2} + c$

3. Originally Posted by manalive04
intergrate

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

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

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

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

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