How do you integrate #int cosx/(1-cos^2x)dx#?

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Answer 1 · 2016-12-06T02:21:57

The integral is #-cscx + C#.

Rewrite #1 - cos^2x# as #sin^2x# using the pythagorean identity #sin^2x+ cos^2x= 1#.

#=>int((cosx)/(sin^2x))dx#

We use the identities #cotx= cosx/sinx# and #1/sinx = cscx#.

#=>int(cotxcscx)dx#

Use the famous integral that #int(-cotxcscx)dx = cscx + C#.

#=>-cscx + C#

Hopefully this helps!