Question Number 31504 by abdo imad last updated on 09/Mar/18
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{{t}\:+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:. \\ $$
Commented by abdo imad last updated on 15/Mar/18
$${let}\:{put}\:{t}={sinx}\:\Rightarrow\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{cosx}\:{dx}}{{sinx}\:+{cosx}} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cosx}\:+{sinx}\:−{sinx}}{{sinx}\:+{cosx}}{dx}=\frac{\pi}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{sinx}}{{sinx}\:+{csx}}{dx} \\ $$$${ch}.{x}=\frac{\pi}{\mathrm{2}}\:−{t}\:{give}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sinx}}{{sinx}\:+{cosx}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cosx}}{{cosx}\:+{sinx}}{dx} \\ $$$$={I}\:\Rightarrow\:\mathrm{2}{I}=\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{I}=\frac{\pi}{\mathrm{4}}\:. \\ $$