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

$$\mathrm{if}I={\int }_0^{\frac{\pi }{2}}\frac{\sin x}{\sin x+\cos x}dx={\int }_0^{\frac{\pi }{2}}\frac{\cos x}{\sin x+\cos x}dx$$$$\mathrm{then}I=??$$

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

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

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

$${\int }_0^{\frac{\pi }{2}}\frac{\mathrm{sinx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}={\int }_0^{\frac{\pi }{2}}\frac{\sin (\frac{\pi }{2}-\mathrm{x})}{\sin (\frac{\pi }{2}-\mathrm{x})+\cos (\frac{\pi }{2}-\mathrm{x})}\mathrm{dx}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\mathrm{I}$$$$\mathrm{so},2\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{sinx}+\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\frac{\pi }{2}$$$$\mathrm{So},\mathrm{I}=\frac{\pi }{4}$$$$$$

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

$$\{\begin{array}{l}\mathcal{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{sinx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}..........\left(1\right)\\ \mathcal{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}...........\left(2\right)\end{array}\Rightarrow \mathcal{I}+\mathcal{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{sinx}+\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}$$$$\Rightarrow 2\mathcal{I}={\int }_0^{\frac{\pi }{2}}\mathrm{dx}=\frac{\pi }{2}\Rightarrow \mathcal{I}=\frac{\pi }{4}$$

Commented by Coronavirus last updated on 25/Jun/20

$$clean$$

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

Ouais

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

$$\mathrm{thank}{\mathrm{you}}^{\prime }\mathrm{ll}$$

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