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I've tried using:

x = sin(u) which leads me to the integral being arcsin(x) + c

x = cos(u) which leads me to the integral being -arccos(x) + c

however I have drawn arcsin(x) and -arccos(x) and they are not the same, is this because of the constant C? I.e the constant is not necessarily the same in both of the integrals? (The graphs are on *close*.

Also, I've tried using the substitution x = tanh(u), this lead me to [itex] \displaystyle\int \dfrac{sech^2u}{1-sech^2u}\ du [/itex], where can I go from here?