Another Indefinite Integral question

[neshepard]

Homework Statement

∫cos(x)/sin^2(x)*dx

Homework Equations

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

 

[hunt_mat]

your integral now becomes:
[tex]
\int\frac{du}{u^{2}}=-\frac{1}{u}+C=-\frac{1}{\sin x}+C
[/tex]

 

[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)

 

[hunt_mat]

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

 

[hunt_mat]

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