Show that # int \ csc x \ dx = ln|tan(x/2)| + C #?
1 Answer
We have:
# int \ csc x \ dx = int \ cscx * (cscx-cotx)/(cscx-cotx) \ dx #
# " "= int \ (csc^2x-cscxcotx)/(cscx-cotx) \ dx #
We now use the following substitution:
#u=cscx-cotx => (du)/dx=-cscxcotx+csc^2x# ,
And so our integral becomes:
# int \ csc x \ dx = int \ 1/u \ du #
# " "= ln|u| + C #
# " "= ln|cscx-cotx| + C #
This is the calculus part of the question complete, It now remains to show that this solution is equivalent to the given solution;
We have;
# cscx-cotx = 1/sinx-cosx/sinx #
# " "= (1-cos(2*1/2x))/sin(2*1/2x) #
# " "= (1-(1 - 2sin^2(x/2)))/(2sin(x/2)cos(x/2)) #
# " "= (2sin^2(x/2))/(2sin(x/2)cos(x/2)) #
# " "= (sin(x/2))/(cos(x/2)) #
# " "= tan(x/2) #
And so;
# int \ csc x \ dx = ln|cscx-cotx| + C #
# " "= ln|tan(x/2)| + C " "# , QED
