Question

How do you integrate `int csc^2(x/2) dx`?

Answer 1

`" "intcsc^2(x/2)dx=-2cot(x/2)+C`

Explanation:

`d/(dx)(cotx)=-csc^2x`

so`" "d/(dx)(cot(x/2))=-1/2csc^2x`

inspecting `" "intcsc^2(x/2)dx`

we have to adjust for the `" "-1/2`

`" "intcsc^2(x/2)dx=-2cot(x/2)+C`