Question

How do you find the antiderivative of `int (cscx) dx`?

Answer 1

One way is by trickery.

Explanation:

`int cscx dx = int cscx/1*(cscx+cotx)/(cscx+cotx) dx`

`= int (csc^2x+cscxcotx)/(cscx+cotx) dx`

Let `u = cscx+cotx`, the `du = -(csc^2x+cscxcotx)`

So our integral becomes `-int (du)/u = -ln abs u +C`

So `int cscx dx = - ln abs(cscx+cotx)`