# How do you find the antiderivative of dx/(cos(x) - 1)?

Jun 18, 2016

Do some conjugate multiplication, apply some trig, and finish to get a result of $\int \frac{1}{\cos x - 1} \mathrm{dx} = \csc x + \cot x + C$

#### Explanation:

As with most problems of this type, we'll solve it using a conjugate multiplication trick. Whenever you have something divided by something plus/minus something (as in $\frac{1}{\cos x - 1}$), it's always helpful to try conjugate multiplication, especially with trig functions.

We will begin by multiplying $\frac{1}{\cos x - 1}$ by the conjugate of $\cos x - 1$, which is $\cos x + 1$:
$\frac{1}{\cos x - 1} \cdot \frac{\cos x + 1}{\cos x + 1}$

You may wonder why we do this. It's so we can apply the difference of squares property, $\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$, in the denominator, to simplify it a little. Back to the problem:
$\frac{1}{\cos x - 1} \cdot \frac{\cos x + 1}{\cos x + 1} = \frac{\cos x + 1}{\left(\cos x - 1\right) \left(\cos x + 1\right)}$
(underbrace(cosx)-underbrace(1))(underbrace(cosx)+underbrace1))
$\textcolor{w h i t e}{I I I} a \textcolor{w h i t e}{X X X} b \textcolor{w h i t e}{X X X} a \textcolor{w h i t e}{X X X} b$
Notice how this is essentially $\left(a - b\right) \left(a + b\right)$.
$= \frac{\cos x + 1}{{\cos}^{2} x - 1}$

Now, what about ${\cos}^{2} x - 1$? Well, we know ${\sin}^{2} x = 1 - {\cos}^{2} x$. Let's multiply that by $- 1$ and see what we get:
$- 1 \left({\sin}^{2} x = 1 - {\cos}^{2} x\right) \to - {\sin}^{2} x = - 1 + {\cos}^{2} x$
$= {\cos}^{2} - 1$

It turns out that $- {\sin}^{2} x = {\cos}^{2} x - 1$, so let's replace ${\cos}^{2} x - 1$:
(cosx+1)/(-sin^2x

This is equivalent to $\cos \frac{x}{-} {\sin}^{2} x + \frac{1}{-} {\sin}^{2} x$, which, using some trig, boils down to $- \cot x \csc x - {\csc}^{2} x$.

At this point, we've simplified to integral $\int \frac{1}{\cos x - 1} \mathrm{dx}$ to $\int - \cot x \csc x - {\csc}^{2} x \mathrm{dx}$. Using the sum rule, this becomes:
$\int - \cot x \csc x \mathrm{dx} + \int - {\csc}^{2} x \mathrm{dx}$

The first of these is $\csc x$ (because the derivative of $\csc x$ is $- \cot x \csc x$) and the second is $\cot x$ (because the derivative of $\cot x$ is $- {\csc}^{2} x$). Add on the constant of integration $C$ and you have your solution:
$\int \frac{1}{\cos x - 1} \mathrm{dx} = \csc x + \cot x + C$