Question

How do you find the antiderivative of `Cosx/Sin^2x`?

Answer 1

`-cosecx+C`

Explanation:

`I=intcosx/sin^2xdx=int1/sinx*cosx/sinxdx`
`I=intcscx*cotxdx=-cscx+C`

Answer 2

`int\ cos(x)/sin^2(x)\ dx=-csc(x)+C`

Explanation:

`int\ cos(x)/sin^2(x)\ dx`

The trick to this integral is a u-substitution with `u=sin(x)`. We can see this is the right way to go because we've got the derivative of `u`, `cos(x)` in the denominator.

To integrate with respect to `u`, we need to divide by the derivative, `cos(x)`:

`int\ cos(x)/sin^2(x)\ dx=int\ cancel(cos(x))/(cancel(cos(x))u^2)\ du=int\ 1/u^2\ du=int\ u^-2\ du`

We can evaluate this integral using the reverse power rule:

`int\ x^n\ dx=x^(n+1)/(n+1)`

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

Now we resubstitute `u=sin(x)` to get the answer in terms of `x`:

`-1/u+C=-1/sin(x)+C=-csc(x)+C`