What is the integral of #int (csc2x)dx #?
1 Answer
Both answers are equivalent. Differentiate them if you have any doubt.
Explanation:
First, use the
#int csc(2x) "d"x = int 1/sin(2x) "d"x#
#= int 1/(2sin(x)cos(x)) "d"x#
#= int cos(x)/(2sin(x)cos^2(x)) "d"x#
#= 1/2 int cos(x)/(sin(x)(1-sin^2(x))) "d"x#
Next, substitute
#frac{"d"u}{"d"x} = cos(x)#
#frac{"d"u}{"d"x} * "d"x = cos(x)*"d"x#
#1/2 int cos(x)/(sin(x)(1-sin^2(x))) "d"x = 1/2 int 1/(u(1-u^2)) "d"u#
Next write the partial fractions.
#1/2 int 1/(u(1-u^2)) "d"u = 1/2 int 1/u "d"u #
#- 1/4 int 1/(1+u) "d"u - 1/4 int 1/(u-1) "d"u#
#= 1/4 ln|u^2/(1-u^2)| + "constant"#
Now revert
#1/4 ln|u^2/(1-u^2)| = 1/4 ln|sin^2(x)/(1-sin^2(x))| #
#= 1/4 ln|sin^2(x)/cos^2(x)| #
#= 1/4 ln|tan^2(x)| #
#= 1/2 ln|tan(x)| #