Relation between $\tan^{-1}(x)$ and $\cot^{-1} (x)$

Question

Suppose we've got $$I_1=\int_{-1}^{1} \tan^{-1}(x) + \tan^{-1} \left(\frac{1}{x}\right)$$ and $$ I_2=\int_{-1}^{1} \cot^{-1}(x) + \cot^{-1}\left(\frac{1}{x}\right)$$ So how can we relate $I_1$ and $I_2$? I know that $$ cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1}(x)$$ and $$ \tan^{-1}(x)+ \cot^{-1}(x)=π/2$$ but is that always true?? And $I_1 = I_2$??

Answer 1

Draw the graphs of $\arctan x$ and $\mathrm{arccot} \;x$. That should help answer about their relation.

edit: I had said

I seem to recall that there are two sensible (but different) conventions for $\mathrm{arccot} \;x$, with each convention used in some calculus texts.

I looked it up. This different convention thing is for $\mathrm{arcsec}\;x$, however.