How do you integrate #(sec2x) / (tan2x) dx# using substitution?

Question

Isabella Ackerman · Accepted Answer

Hello, Answer. #1/4 ln |(cos(2x)  + 1)/(cos(2x) - 1)| + c#, where #c in RR#. 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)#.

Paisley Johnson · Answer

undefined To integrate $ \frac{{\sec^2 x}}{{\tan^2 x}} $ using substitution, let $ u = \tan x $. Then $ du = \sec^2 x \, dx $. Substituting $ u = \tan x $ and $ du = \sec^2 x \, dx $ into the integral yields:

$$ \int \frac{{\sec^2 x}}{{\tan^2 x}} \, dx = \int \frac{1}{u^2} \, du = -\frac{1}{u} + C = -\frac{1}{\tan x} + C = -\cot x + C $$