Question

How do I find the antiderivative of `y=csc(x)cot(x)`?

Answer 1

You can write:
`intcsc(x)cot(x)dx=` as:
`int1/sin(x)cos(x)/sin(x)dx=intcos(x)/sin^2(x)dx=`

But: `d[sin(x)]=cos(x)dx` so your integral becomes:

`intcos(x)/sin^2(x)dx=intsin^(-2)(x)d[sin(x)]=-1/sin(x)+c`

Where you integrate `sin^-2(x)` as if it was `x^2` in a normal integral where you have `dx`.