# Indefinite Trig Integral

• planauts
In summary, the conversation discusses the steps taken to solve the integral of cosx divided by the denominator sin^2(x) - sin(x) - 6. The individual first attempted to factor the denominator and then substituted u = sin(x) to simplify the problem. After using partial fractions, the resulting answer was ln(sin(x)-3)/5 - ln(sin(x)+2)/5. The conversation concludes with a reminder to include the +C term and a variation of the final answer.

## Homework Statement

$$\int \frac{cos x}{sin^2(x) - sin(x)-6}$$

2. The attempt at a solution
I first tried factoring the denominator.
$$\int \frac{cos x}{(sin(x) -3)(sin(x)+2)}$$

The first thing that came to my mind was Partial Factoring but I don't think it would work in this case.

I would start by substituting u=sin(x) and then worry about the partial fractions in the variable u.

Last edited:
Let:
u = sin(x)
du = cos(x) dx

$$\int \frac{du}{(u-3)(u+2)}$$

After doing Partial Fractions, I ended up with:
$$\frac{1}{5} ln(sinx-3)-\frac{1}{5} ln(sinx+2)$$

planauts said:
Let:
u = sin(x)
du = cos(x) dx

$$\int \frac{du}{(u-3)(u+2)}$$

After doing Partial Fractions, I ended up with:
$$\frac{1}{5} ln(sinx-3)-\frac{1}{5} ln(sinx+2)$$

Also note you can factor out a 1/5 and you get:

$$\frac{1}{5} ln(\frac{sinx-3}{sinx+2})$$

planauts said:
Let:
u = sin(x)
du = cos(x) dx

$$\int \frac{du}{(u-3)(u+2)}$$

After doing Partial Fractions, I ended up with:
$$\frac{1}{5} ln(sinx-3)-\frac{1}{5} ln(sinx+2)$$

I'm really sure it's right.

Don't forget the +C!

Char. Limit said:
Don't forget the +C!

Yes you are right, I would have lost 1/2 a mark on a test.

$$\frac{1}{5} ln(sinx-3)-\frac{1}{5} ln(sinx+2) + C$$

OR as gb7nash pointed out
$$\frac{1}{5} ln(\frac{sinx-3}{sinx+2} + C)$$

Thanks everyone!