1. ## integration with sin

how do i integrate this 1 / [(sin x) - 1]

2. Originally Posted by crafty
how do i integrate this 1 / [(sin x) - 1]
$\displaystyle \frac{1}{{\sin \left( x \right) - 1}} = \frac{{\sin (x) + 1}} {{ - \cos ^2 (x)}} = \frac{{ - \sin (x)}} {{\cos ^2 (x)}} - \sec ^2 (x)$

3. how did you convert into those forms can you be a bit more explicit in your description???

4. Originally Posted by crafty
how did you convert into those forms can you be a bit more explicit in your description???
$\displaystyle \left( {\frac{1} {{\sin \left( x \right) - 1}}} \right)\left( {\frac{{\sin \left( x \right) + 1}} {{\sin \left( x \right) + 1}}} \right) = ??$

5. and how do i integrate (sin x)/1(cos x)^2??? do i change into tan form? and if so how do i proceed then??

6. $\displaystyle \frac{du}{u^2}$

7. I think you mean $\displaystyle 1+\cos^2x,$ so yes, put $\displaystyle u=\cos x$ and you'll get an arctangent.

8. Originally Posted by crafty
how do i integrate this 1 / [(sin x) - 1]
The same question (more or less) here: http://www.mathhelpforum.com/math-he...tegration.html