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The two integrals differ by a constant,
sage: log(-1)
I*pi
and integration is defined only up to a constant. So there is no problem, seen from this point of view. From the point of view of a Mathematical Olympiad -say - there is some problem, since at that level there is no log(−1), and such a calculus my get negative points, if things are not explained in full detail. As a rule of thumb for such particular cases, one can replace all terms log( f(x) ) by the corresponding log| f(x) | so tht the computations "remain in the school". In our case, "adding the modulus" changes the sign inside the log of cosx−1 to get log(1−cosx). In our case we can then ask sage for
sage: bool( (cot(x)+csc(x))^2 == (1+cos(x)) / (1-cos(x)) )
True
or even better check it with bare hands by the substitution t=tan(x/2) s in the post.