Question

How do you find the antiderivative of `cosx/cscx`?

Answer 1

`-1/4cos2x+C.`

Explanation:

Since, `1/cscx=sinx, and, 2sinxcosx=sin2x`, we have,

`intcosx/cscxdx=intcosx*sinxdx=1/2int(2sinxcosx)dx`

`=1/2intsin2xdx=1/2((-cos2x)/2)=-1/4cos2x+C.`

Answer 2

Ratnaker M. has given the correct answer. Here are two other ways to get (and to write) the answer.

Explanation:

Once we realize that we need to evaluate

`int cosx sinx dx`

we have choices for how to evaluate this. One is shown in Ratnaker's answer.

OR let `u=cosx`, so that `du = -sinx dx` and the integral becomes

`- int u du = -u^2/2+C = -cos^2x +C`.

OR let `u=sinx`, so that `du = cosx dx` and the integral becomes

`int u du = u^2/2+C = sin^2x +C`.

Challenge show that all of these answers are "the same". (Hint: find the difference. -- it's in the `C`.)