Question

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

Answer 1

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)`.