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Question Number 136765 by Ñï= last updated on 25/Mar/21

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\$$

Answered by snipers237 last updated on 25/Mar/21

$${let}\:{named}\:{it}\:{A} \\$$$${by}\:{stating}\:{x}={sint}\:\:\: \\$$$${A}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({sint}+{cost}\right)}{{tant}}{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{tant}\:.{ln}\left({sint}+{cost}\right){dt}\:\:{due}\:{to}\:\frac{\pi}{\mathrm{2}}−{t}\:{stability} \\$$$$\mathrm{2}{A}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({tant}+\frac{\mathrm{1}}{{tant}}\right){ln}\left({sint}+{cost}\right){dt} \\$$$$\mathrm{2}{A}=\int_{\mathrm{0}\:\:} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{ln}\left({sint}+{cost}\right)}{{sin}\mathrm{2}{t}}{dt}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left(\mathrm{1}+{sin}\mathrm{2}{t}\right)}{{sin}\left(\mathrm{2}{t}\right)}{dt} \\$$$$\mathrm{4}{A}=\int_{\mathrm{0}} ^{\pi} \frac{{ln}\left(\mathrm{1}+{sint}\right)}{{sint}}{dt} \\$$$${Let}\:\:{g}\left({a}\right)=\int_{\mathrm{0}} ^{\pi\:} \frac{{ln}\left(\mathrm{1}+{asint}\right)}{{sint}}{dt} \\$$$$\:{g}^{'} \left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\frac{{sint}\:{dt}}{{sint}\left(\mathrm{1}+{asint}\right)}\:=\int_{\mathrm{0}} ^{\pi} \frac{{a}^{−\mathrm{1}} }{{a}^{−\mathrm{1}} +{sint}}{dt} \\$$$${g}'\left({a}\right)=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{a}^{−\mathrm{1}} {du}}{{a}^{−\mathrm{1}} +{cosu}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{a}^{−\mathrm{1}} }{{a}^{−\mathrm{1}} +{cosu}}{du}\:=\:\mathrm{2}{a}^{−\mathrm{1}} .\frac{\mathrm{2}\pi}{\:\sqrt{{a}^{−\mathrm{2}} −\mathrm{1}}}\:\:\:\:\:\:\:\:\:\:\:\:\:{with}\:\:\:{u}=\frac{\pi}{\mathrm{2}}−{t} \\$$$${g}'\left({a}\right)=\frac{\mathrm{4}\pi}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }\:}\:.{So}\:\:\:{g}\left({a}\right)=\mathrm{4}\pi{arcsin}\left({a}\right)\:{cause}\:{g}\left(\mathrm{0}\right)=\mathrm{0} \\$$$${Then}\:\:{g}\left(\mathrm{1}\right)=\mathrm{4}\pi{arcsin}\left(\mathrm{1}\right)=\mathrm{2}\pi^{\mathrm{2}} \: \\$$$${and}\:\:\:{A}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\: \\$$$$\\$$

Answered by Ñï= last updated on 26/Mar/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}{dx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)}{\mathrm{tan}\:}{dx} \\$$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\:\left(\mathrm{1}+\mathrm{tan}\:{x}\right)}{\mathrm{tan}\:{x}}{dx}+\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\:\mathrm{cos}\:{x}}{\mathrm{tan}\:{x}}{dx} \\$$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{yln}\:{y}}{\mathrm{1}−{y}^{\mathrm{2}} }{dy} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\:\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{1}} ^{\infty} \frac{{ln}\:\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{\mathrm{2}{n}+\mathrm{1}} {ln}\:{ydy} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\:\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\:\left(\mathrm{1}+{x}\right)−{xln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\:\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \left[\frac{\mathrm{1}}{{x}}\left(\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\right)\right]{ln}\:\left(\mathrm{1}+{x}\right)−\frac{{xln}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}{dx}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} {lnxdx}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\$$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−\mathrm{2}} \right)\frac{\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\$$

Commented by mnjuly1970 last updated on 26/Mar/21

$$\:\:\:\:{very}\:{nice}\:..{grateful}... \\$$