How do you integrate #cscx #?
1 Answer
Feb 18, 2017
# int \ csc x \ dx = - ln|csc(x) + cot(x)| +C #
Explanation:
There are many ways to prove this result. The quickest method that I am aware of is as follows:
# int \ csc x \ dx = int \ cscx \ (cscx + cotx)/(cscx + cotx) \ dx #
# " "= int \ (csc^2x + cscxcotx)/(cscx + cotx) \ dx #
Then we perform simple substitution, Let
#u = cscx + cotx => (du)/dx = -cscxcotx - csc^2x #
# " "= -(cscxcotx + csc^2x) #
And so:
# int \ csc x \ dx = int \ (-1/u) \ du #
# " "= - int \ 1/u \ du #
# " "= - ln|u| +C #
# " "= - ln|cscx + cotx| +C #
