Question Number 109709 by nimnim last updated on 25/Aug/20
$$\:\:\:\:\:\int_{\mathrm{0}\:} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tanx}+\mathrm{1}\right)\mathrm{dx} \\ $$
Answered by mnjuly1970 last updated on 25/Aug/20
$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tanx}\right){dx}\overset{\int_{{a}} ^{\:{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{\:{b}} {f}\left({a}+{b}−{x}\right){dx}} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right){dx} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}−{tan}\left({x}\right)}{\mathrm{1}+{tan}\left({x}\right)}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{\mathrm{2}}{\mathrm{1}+{tan}\left({x}\right)}\right){dx} \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{2}\right){dx}\:−\mathrm{I}\Rightarrow\:\mathrm{2I}=\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{I}:=\:\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)\:\:\:…..\:\:\: \\ $$$$\:\:\:{solved}\:{by}\:\mathscr{M}.\mathscr{N}.{july}\:\mathrm{1970}# \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 25/Aug/20
$${grateful}.. \\ $$
Commented by nimnim last updated on 25/Aug/20
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$
Answered by mathmax by abdo last updated on 25/Aug/20
$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{tant}\right)\mathrm{dt}\:\:\Rightarrow\mathrm{I}\:=_{\mathrm{t}=\frac{\pi}{\mathrm{4}}−\mathrm{x}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\mathrm{x}\right)\right)\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}−\mathrm{tanx}}{\mathrm{1}+\mathrm{tanx}}\right)\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tanx}}\right)\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)−\mathrm{I}\:\Rightarrow\mathrm{2I}\:=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{I}\:=\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$