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

∫_( 0) ^(π/2) ((f(x))/(f(x)+f((π/2)−x))) dx =

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

Answered by tanmay last updated on 16/May/19

I=∫_0 ^(π/2) ((f(x))/(f(x)+f((π/2)−x)))dx t=(π/2)−x dt=−dx I=∫_(π/2) ^0 ((f((π/2)−t))/(f((π/2)−t)+f(t)))×−dt I=∫_0 ^(π/2) ((f((π/2)−t))/(f((π/2)−t)+f(t)))dt→∫_0 ^(π/2) ((f((π/2)−x))/(f((π/2)−x)+f(x)))dx now ∫_a ^b f(x)dx=∫_a ^b f(t)dt so 2I=∫_0 ^(π/2) ((f(x))/(f(x)+f((π/2)−x)))dx+∫_0 ^(π/2) ((f((π/2)−x))/(f((π/2)−x)+f(x)))dx 2I=∫_0 ^(π/2) dx 2I=(π/2)→I=(π/4) ■direct formula ∫_a ^b f(x)dx=∫_0 ^b f(a+b−x)dx I=∫_0 ^(π/2) ((f(x))/(f(x)+f((π/2)−x)))dx I=∫_0 ^(π/2) ((f((π/2)−x))/(f((π/2)−x)+f(x)))dx 2I=∫_0 ^(π/2) ((f((π/2)−x)+f(x))/(f(x)+f((π/2)−x)))dx 2I=∫_0 ^(π/2) dx I=(π/4)

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

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

I=∫_0 ^(π/2) ((f((π/2)−x))/(f((π/2)−x)+f(x)))dx ∴2I=∫_0 ^(π/2) ((f((π/2)−x)+f(x))/(f((π/2)−x)+f(x))) dx =∫_0 ^(π/2) dx =(π/2) ∴I =(π/4)

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

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