What is the Integral of a Trigonometric Function with an Indefinite Integral?

[planauts]

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.Thanks in advance!

 

[Dick]

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

 

[planauts]

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

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

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

Thanks for your help. I think that is the correct answer.

 

[gb7nash]

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

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

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

Thanks for your help. I think that is the correct answer.

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

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

 

[Dick]

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

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

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

Thanks for your help. I think that is the correct answer.

I'm really sure it's right.

 

[Char. Limit]

Don't forget the +C!

 

[planauts]

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!