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Thread: integration with sin

  1. #1
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    integration with sin

    how do i integrate this 1 / [(sin x) - 1]
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  2. #2
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    Quote Originally Posted by crafty View Post
    how do i integrate this 1 / [(sin x) - 1]
    $\displaystyle \frac{1}{{\sin \left( x \right) - 1}} = \frac{{\sin (x) + 1}}
    {{ - \cos ^2 (x)}} = \frac{{ - \sin (x)}}
    {{\cos ^2 (x)}} - \sec ^2 (x)$
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    how did you convert into those forms can you be a bit more explicit in your description???
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  4. #4
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    Quote Originally Posted by crafty View Post
    how did you convert into those forms can you be a bit more explicit in your description???
    $\displaystyle \left( {\frac{1}
    {{\sin \left( x \right) - 1}}} \right)\left( {\frac{{\sin \left( x \right) + 1}}
    {{\sin \left( x \right) + 1}}} \right) = ??$
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  5. #5
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    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??
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  6. #6
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    $\displaystyle \frac{du}{u^2}$
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  7. #7
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    I think you mean $\displaystyle 1+\cos^2x,$ so yes, put $\displaystyle u=\cos x$ and you'll get an arctangent.
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  8. #8
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    Quote Originally Posted by crafty View Post
    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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