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Question Number 100216 by Rio Michael last updated on 25/Jun/20

if I = ∫_0 ^(π/2) ((sin x)/(sin x + cos x))dx = ∫_0 ^(π/2) ((cos x)/(sin x +cos x))dx then I = ??

$$\mathrm{if}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}}{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}{dx}\: \\ $$$$\mathrm{then}\:{I}\:=\:?? \\ $$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

(π/4) (I think it should be)

$$\frac{\pi}{\mathrm{4}}\:\:\:\left(\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\right) \\ $$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

∫_0 ^(π/2) ((sinx)/(sinx+cosx))dx=∫_0 ^(π/2) ((sin((π/2)−x))/(sin((π/2)−x)+cos((π/2)−x)))dx=∫_0 ^(π/2) ((cosx)/(sinx+cosx))dx=I so, 2I=∫_0 ^(π/2) ((sinx+cosx)/(sinx+cosx))dx=(π/2) So, I=(π/4)

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sinx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)+\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)}\mathrm{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\mathrm{I} \\ $$$$\:\:\:\mathrm{so},\:\:\mathrm{2I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sinx}+\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{So},\:\:\:\:\:\mathrm{I}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 25/Jun/20

{ ((I=∫_0 ^(π/2) ((sinx)/(sinx+cosx))dx..........(1))),((I=∫_0 ^(π/2) ((cosx)/(sinx+cosx))dx...........(2))) :}⇒I+I=∫_0 ^(π/2) ((sinx+cosx)/(sinx+cosx))dx ⇒2I=∫_0 ^(π/2) dx=(π/2)⇒I=(π/4)

$$\begin{cases}{\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sinx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}..........\left(\mathrm{1}\right)}\\{\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}...........\left(\mathrm{2}\right)}\end{cases}\Rightarrow\mathcal{I}+\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sinx}+\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx} \\ $$$$\Rightarrow\mathrm{2}\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{dx}=\frac{\pi}{\mathrm{2}}\Rightarrow\mathcal{I}=\frac{\pi}{\mathrm{4}} \\ $$

Commented by Coronavirus last updated on 25/Jun/20

clean

$${clean} \\ $$

Commented by Ar Brandon last updated on 25/Jun/20

Ouais

Commented by Rio Michael last updated on 26/Jun/20

thank you′ll

$$\mathrm{thank}\:\mathrm{you}'\mathrm{ll} \\ $$

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