integration with sin

**crafty** · Feb 5th 2009, 12:24 PM

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

**Plato** · Feb 5th 2009, 12:30 PM

Originally Posted by crafty

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

$\frac{1}{{\sin \left( x \right) - 1}} = \frac{{\sin (x) + 1}}<br /> {{ - \cos ^2 (x)}} = \frac{{ - \sin (x)}}<br /> {{\cos ^2 (x)}} - \sec ^2 (x)$

**crafty** · Feb 5th 2009, 01:02 PM

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

**Plato** · Feb 5th 2009, 01:06 PM

Originally Posted by crafty

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

$\left( {\frac{1}<br /> {{\sin \left( x \right) - 1}}} \right)\left( {\frac{{\sin \left( x \right) + 1}}<br /> {{\sin \left( x \right) + 1}}} \right) = ??$

**crafty** · Feb 5th 2009, 01:16 PM

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??

**Plato** · Feb 5th 2009, 01:22 PM

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

**Krizalid** · Feb 5th 2009, 02:35 PM

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

**mr fantastic** · Feb 5th 2009, 04:44 PM

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

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