# Homework Help: Partial Fractions (with trig functions)

1. Oct 24, 2009

### zooboodoo

1. The problem statement, all variables and given/known data
Integral(sinx(x)dx/(cos^2(x)+cos(x)-2)

2. Relevant equations

3. The attempt at a solution
What I tried to do first was factor the denominator, so i got (cos(x)-1)(cos(x)+2)
from there, I set up my partial fractions equation trying to solve B(cos(x)-1) + A(cos(x)+2) = sin(x)

at this point i tried plugging in 0 for x -> A(0) + B(3) = 0 ? i didn't feel good about this, after working through the remainder using this i solved for A=0 and B=(-1), which left me with
integral( -dx/(cos(x)+2), at this point I thought there may have been a 1/2 / double angle formula that I could use to try to simplify the equation; cosx^2 = 1+cos2x / 2, however I havn't felt good about the solving for A=0 and losing a term? any insight would be great, thanks

2. Oct 24, 2009

### Bohrok

As far as I know, partial fractions only work with polynomials in the numerator and denominator.
Try a u-substitution with u=cosx and then partial fractions.

3. Oct 25, 2009

### lurflurf

Partial fractions work perfectly here. A substitution is not needed to apply partial fractions, here after partial fractions is a good time to change variable because different changes can be used for each term.
A(0) + B(3) = 0 this is good, but there is a mistake elsewhere as A=0 B=-1 does not work you need another equation.
I would have started with the identity
$$\frac{\sin(x)}{\cos^2(x)+\cos(x)-2}=\frac{-\cos\left(\frac{x}{2}\right)}{3\sin\left(\frac{x}{2}\right)-2\sin^3\left(\frac{x}{2}\right)}$$

Last edited: Oct 25, 2009