How do you integrate #(sec2x) / (tan2x) dx# using substitution?
1 Answer
Mar 1, 2015
Hello,
Answer.
Explanation.
-
Because
#tan(2x)=sin(2x)/cos(2x)# and#sec(2x) = 1/cos(2x)# , you have to calculate#int (dx)/sin(2x)# . -
Take
#u = cos(2x)# . You have#du = -2 sin(2x) dx# , so
#int dx/sin(2x) = -1/2 int (du)/sin^2(2x) = -1/2 int (du)/(1-cos^2(2x))# ,
and so :
#int dx/sin(2x) = 1/2 int (du)/(u^2-1)# . -
Decompose
#1/(u^2-1) = (1/2)/(u+1) - (1/2)/(u-1)# and finally
#int sec(2x)/tan(2x) dx = 1/4 (ln(|u+1| - ln(|u-1|) + c#
#int sec(2x)/tan(2x) dx = 1/4 ln |(u+1)/(u-1)| + c# and you get the result because#u=cos(2x)# .
