Question

What's the integral of `int (cscx)^2 dx`?

Answer 1

`int (cscx)^2 dx = -cotx+C`

Explanation:

The derivative of `cotx` is `csc^2x`,

so the integral of `csc^2x` is `cotx + C`

If you really want the integral of the integral, then use:

`int[int (cscx)^2 dx] dx = int [-cotx+C] dx` where `C` is an arbitrary constant

`= int (-cosx/sinx + C) dx` where `C` is an arbitrary constant

`= -lnabssinx + Cx +D` where `C,D` are arbitrary constants

`= lnabscscx + Cx +D` where `C,D` are arbitrary constants