# How do you use partial fractions to find the integral int (sinx)/(cosx(cosx-1))dx?

Feb 19, 2017

$\int \sin \frac{x}{\cos x \left(\cos x - 1\right)} \mathrm{dx} = \ln | \frac{\cos x}{\cos x - 1} | + C$

#### Explanation:

Let $u = \cos x$. Then $\mathrm{du} = - \sin x \mathrm{dx} \to \mathrm{dx} = \frac{\mathrm{du}}{- \sin x}$.

$\int \sin \frac{x}{u \left(u - 1\right)} \cdot \frac{\mathrm{du}}{- \sin x}$

$- \int \frac{1}{u \left(u - 1\right)} \mathrm{du}$

Now use partial fractions.

$\frac{A}{u} + \frac{B}{u - 1} = \frac{1}{u \left(u - 1\right)}$

$A \left(u - 1\right) + B \left(u\right) = 1$

$A u - A + B u = 1$

$\left(A + B\right) u - A = 1$

Write a system of equations.

$\left\{\begin{matrix}A + B = 0 \\ - A = 1\end{matrix}\right.$

Solving, we get $A = - 1$ and $B = 1$.

$- \int \frac{1}{u - 1} - \frac{1}{u} \mathrm{du}$

$\int \frac{1}{u} \mathrm{du} - \int \frac{1}{u - 1} \mathrm{du}$

$\ln | u | - \ln | u - 1 | + C$

$\ln | \cos x | - \ln | \cos x - 1 | + C$

$\ln | \frac{\cos x}{\cos x - 1} | + C$

Hopefully this helps!