Solving $\int\frac{1}{\tan x}\frac{1}{\sin^2 x}dx$

Question

I calculated the value as $$-\frac{\cot^2 x}{2} + C$$ which differentiates back to the question.

When I checked for the answer on https://www.integral-calculator.com, I got the answer $$-\frac{1}{2\sin^2 x} + C$$ which also differentiates back to the question.

Are one of those answers wrong? or is it possible to have $2$ different integral values for the same function?