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
