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

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Answer 1 · 2015-03-01T15:44:55

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 $u=cos(2x)$ .

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