What is #int cot^2 (3-x)dx#?
1 Answer
Jan 29, 2017
Explanation:
#I=intcot^2(3-x)dx#
Use the identity
Then:
#I=intcsc^2(3-x)dx-intdx#
#I=intcsc^2(3-x)dx-x#
For the remaining integral, let
#I=-intcsc^2(u)du-x#
This is a standard integral, since
#I=cot(u)-x#
#I=cot(3-x)-x+C#