What is #int cot^2 (3-x)dx#?

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Answer 1 · 2017-01-29T23:05:05

#cot(3-x)-x+C#

#I=intcot^2(3-x)dx#

Use the identity #1+cot^2(theta)=csc^2(theta)# to say that #cot^2(theta)=csc^2(theta)-1#.

Then:

#I=intcsc^2(3-x)dx-intdx#

#I=intcsc^2(3-x)dx-x#

For the remaining integral, let #u=3-x#. This implies that #du=-dx#. Then:

#I=-intcsc^2(u)du-x#

This is a standard integral, since #d/(du)cot(u)=-csc^2(u)#:

#I=cot(u)-x#

#I=cot(3-x)-x+C#