What is the integral of #int cot^2(x)secxdx#?

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Answer 1 · 2016-03-17T03:14:37

#-cscx+C#

Using the definitions of #cotx# and #secx#, we see that

#intcot^2xsecxdx=int(cos^2x/sin^2x)(1/cosx)dx#

#=intcosx/sin^2xdx=intcosx/sinx(1/sinx)dx#

#=intcotxcscxdx#

This is a common integral that equals

#=-cscx+C#

Another method we could have used was to use substitution at #intcosx/sin^2xdx#:

If #u=sinx#, then #du=cosxdx#, so we have

#intcosx/sin^2xdx=int1/u^2du=intu^-2du#

Then, through the rule

#intu^ndu=u^(n+1)/(n+1)+C#

We obtain

#intu^-2du=u^-1/(-1)+C=-1/u+C#

#=-1/sinx+C=-cscx+C#