Question

How do you integrate `int (sinx) / ((cos^2x + cosx -2)) dx` using partial fractions?

Answer 1

`1/3 ln|(cos x + 2)/(cos x - 1)|`

Explanation:

When dealing with trigonometric functions during partial fraction expansions, it can be very helpful to make some substitutions to simplify the problem. We will let `u = cosx`. Then we have

`int sin x /(cos^2 x + cos x - 2)dx = int sin x / (u^2 + u - 2) dx`

We may actually take this one step further and eliminate the `sin` from the numerator. In this case we will note that `(du)/(dx) = -sin x`.

One can think of this as being equivalent to `-du = sin x dx`.

After replacing `sin x dx` with `-du` we have

`int sin x / (u^2 + u - 2) dx = int (-du) / (u^2 + u - 2)`

Now we can simply expand our new, trig-less expression and integrate normally.

We will factor the denominator, and set the expression equal to a general expression involving these factors to begin the partial fraction expansion:

`-1/((u + 2)(u - 1)) = A/(u + 2) + B/(u - 1)`

To continue our expansion we will multiply through by `(u + 2)(u - 1)` as follows,

`-1 = A(u - 1) + B(u + 2) = Au - A + Bu + 2B`

We may now find `A` and `B` by equating coefficients. To show more clearly what we're working with, I'll rearrange everything a bit:

`-1 = (A + B)u + (2B - A)`

On the left-hand side, we have no multiples of `u`. Thus, `A + B` must equal zero. `2B - A` then must equal `-1`.

`A + B = 0`
`2B - A = -1`

This is a simple linear system which I will not bother with in too much detail, so you'll just have to trust me when I say the solution is
`A = 1/3, B=-1/3`

The expanded integral, then, is

`int (-du) / (u^2 + u - 2) = int 1/(3(u+2))du - int 1/(3(u-1))du`

From here the answer should be fairly clear:

`int (du)/(3(u+2)) - int (du)/(3(u-1)) = 1/3ln|u + 2| - 1/3ln|u - 1| + C`

However, we must substitute `u` back to get our answer in terms of `x`. Our final answer then is

`1/3 ln |cos x + 2| - 1/3 ln|cos x - 1|`

After a small tweak for the sake of aesthetics:

`1/3 ln|(cos x + 2)/(cos x - 1)|`