# How do you find integral of ((secxtanx)/(secx-1))dx?

Apr 13, 2015

$\int \frac{\sec \left(x\right) . \tan \left(x\right)}{\sec \left(x\right) - 1} \mathrm{dx}$

We know that :

$\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$

So, we have :

$= \int \frac{1}{\cos} \left(x\right) . \sin \frac{x}{\cos} \left(x\right) \cdot \frac{1}{\frac{1}{\cos} \left(x\right) - 1} \mathrm{dx}$

$= \int \frac{\sin \left(x\right)}{\cos \left(x\right) - {\cos}^{2} \left(x\right)} \mathrm{dx}$

Let's $t = \cos \left(x\right)$

So $\mathrm{dt} = - \sin \left(x\right)$

$= \int \frac{1}{t - {t}^{2}} \left(- \mathrm{dt}\right)$

Factorize
$- \int \frac{1}{t \left(1 - t\right)} \mathrm{dx}$

With decomposition in simple elements, we have :

$\frac{\alpha}{t} + \frac{\beta}{1 - t}$

$= \frac{\alpha \left(1 - t\right)}{t \left(1 - t\right)} + \frac{\beta t}{t \left(1 - t\right)}$

So :

$\alpha - \alpha t + \beta t$
$\alpha + t \left(- \alpha + \beta\right)$

We have :

$\alpha = 1$
$- \alpha + \beta = 0$

$\alpha = \beta = 1$

We get :

$- \int \frac{1}{t} \mathrm{dt} + \int \frac{1}{1 - t} \mathrm{dt}$

$- \left[\log \left(t\right)\right] + \left[\log \left(1 - t\right)\right]$

$= - \left[\log \left(\cos \left(x\right)\right)\right] + \left[\log \left(1 - \cos \left(x\right)\right)\right]$

$= 2 \log \left(\sin \left(\frac{x}{2}\right)\right) - \log \left(\cos \left(x\right)\right) + C$

Apr 13, 2015

Another way of doing this is to consider that $d \left(\sec x\right) = \sec x \tan x \cdot \mathrm{dx}$

That is, the derivative of $\sec x$ is $\sec x \tan x$

$\implies \int \frac{\sec x \tan x}{\sec x - 1} \mathrm{dx} = \int d \frac{\sec x}{\sec x - 1}$

This is the same as letting $u = \sec x$

We then have,

$\int \frac{\mathrm{du}}{u - 1} = \ln \left(u - 1\right) + C = \ln \left(\sec x - 1\right) + C$