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answered 7 years ago

dan_fulea gravatar image

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 cosx1 to get log(1cosx). 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.