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Question Number 31504 by abdo imad last updated on 09/Mar/18

calculate ∫_0 ^1     (dt/(t +(√(1−t^2 )))) .

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{{t}\:+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:. \\ $$

Commented by abdo imad last updated on 15/Mar/18

let put t=sinx ⇒ I=∫_0 ^(π/2)     ((cosx dx)/(sinx +cosx))  = ∫_0 ^(π/2)  ((cosx +sinx −sinx)/(sinx +cosx))dx=(π/2) −∫_0 ^(π/2)    ((sinx)/(sinx +csx))dx  ch.x=(π/2) −t give ∫_0 ^(π/2)  ((sinx)/(sinx +cosx))dx=∫_0 ^(π/2)   ((cosx)/(cosx +sinx))dx  =I ⇒ 2I= (π/2) ⇒ I=(π/4) .

$${let}\:{put}\:{t}={sinx}\:\Rightarrow\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{cosx}\:{dx}}{{sinx}\:+{cosx}} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cosx}\:+{sinx}\:−{sinx}}{{sinx}\:+{cosx}}{dx}=\frac{\pi}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sinx}}{{sinx}\:+{csx}}{dx} \\ $$$${ch}.{x}=\frac{\pi}{\mathrm{2}}\:−{t}\:{give}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sinx}}{{sinx}\:+{cosx}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cosx}}{{cosx}\:+{sinx}}{dx} \\ $$$$={I}\:\Rightarrow\:\mathrm{2}{I}=\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}=\frac{\pi}{\mathrm{4}}\:. \\ $$

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