Thread: integration with sin

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

Originally Posted by crafty

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

$\displaystyle \frac{1}{{\sin \left( x \right) - 1}} = \frac{{\sin (x) + 1}}
{{ - \cos ^2 (x)}} = \frac{{ - \sin (x)}}
{{\cos ^2 (x)}} - \sec ^2 (x)$

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

Originally Posted by crafty

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

$\displaystyle \left( {\frac{1}
{{\sin \left( x \right) - 1}}} \right)\left( {\frac{{\sin \left( x \right) + 1}}
{{\sin \left( x \right) + 1}}} \right) = ??$

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??

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

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

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