Asked by Courtney
How do I find the indefinite integral of cosx/(1-cos^2 x)dx?
Answers
Answered by
Reiny
cosx/(1-cos^2 x)
= cosx/(sin^2 x)
= (cosx/sinx)(1/sinx)
= cotx cscx
I just happen to know, as should you, that the derivative of cscx = -cotx cscx
so ∫cosx/(1-cos^2 x) dx = - cscx
= cosx/(sin^2 x)
= (cosx/sinx)(1/sinx)
= cotx cscx
I just happen to know, as should you, that the derivative of cscx = -cotx cscx
so ∫cosx/(1-cos^2 x) dx = - cscx
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