Question

1÷(tanx+cotx)(tanx+cotx)dx

Answer 1

I assume you want the integral, but it's not clear just what you are working with.

(tanx+cotx)(tanx+cotx)
= tan^2x + 2 + cot^2x
= tan^2x+1 + cot^2x+1
= sec^2x + csc^2x
= 1/(sin^2x cos^2x)

so 1 over all that is just

1/4 sin^2(2x) = 1/8 (1-cos(4x))

and that's easy to integrate, right?