Question

Integrate?

`csc^2x/cotxdx`

Answer 1

`intcsc^2x/cotxdx=-ln|cotx|+"C"`

Explanation:

We can make a u-substitution:

Let `u=cot(x)=>du=-csc^2xdx`

`intcsc^2x/cotxdx=>-int1/udu`

The integral of `1/u` is simply `ln|u|+"C"`

So `-int1/u=-ln|u|+"C"`

Undo the substitution:

`=-ln|cotx|+"C"`