# How do I evaluate the integral int(secx tanx) / (sec^2(x) - secx) dx?

Jan 26, 2015

First of all, let's rewrite the function to integrate in a more simple way: remembering that $\setminus \sec \left(x\right) = \frac{1}{\setminus} \cos \left(x\right)$, and of course that $\setminus \tan \left(x\right) = \setminus \sin \frac{x}{\setminus \cos \left(x\right)}$ we write the integrand as
$\setminus \frac{\setminus \frac{1}{\setminus \cos \left(x\right)} \setminus \frac{\setminus \sin \left(x\right)}{\setminus \cos \left(x\right)}}{\setminus \frac{1}{\setminus {\cos}^{2} \left(x\right)} - \setminus \frac{1}{\setminus \cos \left(x\right)}}$, which we can simplify into $\frac{\frac{\setminus \sin \left(x\right)}{\setminus {\cos}^{2} \left(x\right)}}{\frac{1 - \setminus \cos \left(x\right)}{\setminus {\cos}^{2} \left(x\right)}}$, and finally obtain $\setminus \frac{- \setminus \sin \left(x\right)}{\setminus \cos \left(x\right) - 1}$.

To integrate this function, we'll use a couple of substitutions: first of all, by choosing $t = \setminus \cos \left(x\right)$, one has $\mathrm{dt} = - \setminus \sin \left(x\right) \setminus \mathrm{dx}$, and so
$\setminus \int \setminus \frac{- \setminus \sin \left(x\right)}{\setminus \cos \left(x\right) - 1} \mathrm{dx}$ becomes
$\setminus \int \setminus \frac{\mathrm{dt}}{t - 1}$.
By choosing $y = t - 1$, one obtains $\mathrm{dy} = \mathrm{dt}$, and the integral becomes simply
$\setminus \int \setminus \frac{\mathrm{dy}}{y} = \setminus \log \left(y\right) + c$. Substituting back $y = t - 1$, one has $\setminus \log \left(t - 1\right) + c$, and again, plugging $t = \setminus \cos \left(x\right)$ into the equation, one has $\setminus \log \left(\setminus \cos \left(x\right) - 1\right) + c$.

By deriving, you can check that, infact, the following equation holds:

$\frac{d}{\mathrm{dx}} \setminus \log \left(\setminus \cos \left(x\right) - 1\right) + c = \setminus \frac{- \setminus \sin \left(x\right)}{\setminus \cos \left(x\right) - 1}$