# Another Indefinite Integral question

1. Jul 18, 2010

### neshepard

1. The problem statement, all variables and given/known data
∫cos(x)/sin^2(x)*dx

2. Relevant equations

3. The attempt at a solution
Based off my earlier question, where is my error please.
u=sinx du=cosx*dx
∫u^-1*du sin(x)^-1 1/sin(x) + C -or-csc(x) + C

Thanks

2. Jul 18, 2010

### hunt_mat

$$\int\frac{du}{u^{2}}=-\frac{1}{u}+C=-\frac{1}{\sin x}+C$$

3. Jul 18, 2010

### neshepard

But where does the -1/u com from? Since the dx of sin(x) is cos(x) and the sin(x) was in the denominator, I know to subtract 1 from the original sin^2(x) in the denominator instead of add. But I keep coming up with 1/sin(x)

Last edited: Jul 18, 2010
4. Jul 19, 2010

### hunt_mat

You should know that:
$$\int u^{n}du=\frac{1}{n+1}u^{n+1}+C\quad n\neq -1$$
Just take n=-2 for the answer you require.

5. Jul 19, 2010

### hunt_mat

I think I see your question (sorry)
The integral is:
$$\int\frac{\cos x}{\sin^{2}x}dx$$
let $$u=\sin x$$, then du/dx=cos x and the integral reduces to the one I posted earlier.