UNKNOWN Questions

Question Number 59976 by soufiane last updated on 16/May/19

$$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}\:{dx}\:= \\$$

Answered by tanmay last updated on 16/May/19

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\$$$${t}=\frac{\pi}{\mathrm{2}}−{x} \\$$$${dt}=−{dx} \\$$$${I}=\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \:\frac{{f}\left(\frac{\pi}{\mathrm{2}}−{t}\right)}{{f}\left(\frac{\pi}{\mathrm{2}}−{t}\right)+{f}\left({t}\right)}×−{dt} \\$$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left(\frac{\pi}{\mathrm{2}}−{t}\right)}{{f}\left(\frac{\pi}{\mathrm{2}}−{t}\right)+{f}\left({t}\right)}{dt}\rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)+{f}\left({x}\right)}{dx} \\$$$${now}\:\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{{a}} ^{{b}} {f}\left({t}\right){dt} \\$$$${so} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)+{f}\left({x}\right)}{dx} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\$$$$\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\rightarrow{I}=\frac{\pi}{\mathrm{4}} \\$$$$\blacksquare{direct}\:{formula} \\$$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{b}} {f}\left({a}+{b}−{x}\right){dx} \\$$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\$$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)+{f}\left({x}\right)}{dx} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)+{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\$$$${I}=\frac{\pi}{\mathrm{4}} \\$$

Answered by Prithwish sen last updated on 16/May/19

$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{f}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}{\mathrm{f}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)}\mathrm{dx} \\$$$$\therefore\mathrm{2I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{f}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{f}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)}\:\mathrm{dx} \\$$$$\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{dx}\:=\frac{\pi}{\mathrm{2}} \\$$$$\therefore\mathrm{I}\:=\frac{\pi}{\mathrm{4}} \\$$