Question Number 120064 by bemath last updated on 29/Oct/20
$${I}\:=\:\int\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\:}}\left(\frac{\mathrm{1}}{\mathrm{ln}\:\left(\mathrm{tan}\:{r}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{tan}\:{r}}\:\right)\:{dr} \\ $$
Answered by mnjuly1970 last updated on 29/Oct/20
$$\mathrm{I}\overset{\int_{{a}} ^{\:{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{\:{b}} {f}\left({a}+{b}−{x}\right){dx}} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left\{\frac{\mathrm{1}}{−{ln}\left({tan}\left({r}\right)\right)}+\frac{{tan}\left({r}\right)}{{tan}\left({r}\right)−\mathrm{1}}\right\}{dr} \\ $$$$=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{−\mathrm{1}}{{ln}\left({tan}\left({r}\right)\right)}\:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{tan}\left({r}\right)}{dr} \\ $$$$=−\mathrm{I}+\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{2I}=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\mathrm{I}=\frac{\pi}{\mathrm{4}}\:\checkmark \\ $$$$\:\:\:{m}.{n}.\mathrm{1970}… \\ $$$$\:\:\:\:\: \\ $$