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

Mar 1, 2015

Hello,

Answer. $\frac{1}{4} \ln | \frac{\cos \left(2 x\right) + 1}{\cos \left(2 x\right) - 1} | + c$, where $c \in \mathbb{R}$.

Explanation.

• Because $\tan \left(2 x\right) = \sin \frac{2 x}{\cos} \left(2 x\right)$ and $\sec \left(2 x\right) = \frac{1}{\cos} \left(2 x\right)$, you have to calculate $\int \frac{\mathrm{dx}}{\sin} \left(2 x\right)$.

• Take $u = \cos \left(2 x\right)$. You have $\mathrm{du} = - 2 \sin \left(2 x\right) \mathrm{dx}$, so
$\int \frac{\mathrm{dx}}{\sin} \left(2 x\right) = - \frac{1}{2} \int \frac{\mathrm{du}}{\sin} ^ 2 \left(2 x\right) = - \frac{1}{2} \int \frac{\mathrm{du}}{1 - {\cos}^{2} \left(2 x\right)}$,
and so :
$\int \frac{\mathrm{dx}}{\sin} \left(2 x\right) = \frac{1}{2} \int \frac{\mathrm{du}}{{u}^{2} - 1}$.

• Decompose $\frac{1}{{u}^{2} - 1} = \frac{\frac{1}{2}}{u + 1} - \frac{\frac{1}{2}}{u - 1}$ and finally
int sec(2x)/tan(2x) dx = 1/4 (ln(|u+1| - ln(|u-1|) + c
$\int \sec \frac{2 x}{\tan} \left(2 x\right) \mathrm{dx} = \frac{1}{4} \ln | \frac{u + 1}{u - 1} | + c$ and you get the result because $u = \cos \left(2 x\right)$.