Question Number 32352 by abdo imad last updated on 23/Mar/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$
Commented by abdo imad last updated on 26/Mar/18
$${ch}.{x}={sint}\:\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{arctan}\left({cost}\right)\:{cost}\:{dt}\:{let}\:{integrate} \\ $$$${by}\:{parts}\:{u}={arctan}\left({cost}\right)\:{and}\:{v}^{'} \:={cost} \\ $$$${I}\:=\:\left[{sint}\:{arctan}\left({cost}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{−{sint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:{sint}\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}−{cos}^{\mathrm{2}} {t}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:{dt}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}−\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}{\mathrm{1}\:+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}{dt}\:\:=_{\mathrm{2}{t}={u}} \:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\mathrm{1}−{cosu}}{\mathrm{3}\:+{cosu}}\:\frac{{du}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\mathrm{1}−{cosu}}{\mathrm{3}\:+{cosu}}{du}\:\:{ch}.{tan}\left(\frac{{u}}{\mathrm{2}}\right)={x}\:{give} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{3}\:+\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}+{x}^{\mathrm{2}} }{\left(\mathrm{3}\:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{4}+\mathrm{2}{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)}{dx}\:{let}\:{consider} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+\mathrm{2}\right)}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\:,−{i},\:{i}\sqrt{\mathrm{2}\:}\:,−{i}\sqrt{\mathrm{2}}\:\:{and}\:{all}\:{are}\:{simples} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{i}\sqrt{\mathrm{2}}\right)\right)\:\: \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{{z}^{\mathrm{2}} }{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{i}\sqrt{\mathrm{2}}\right)\left({z}\:+{i}\sqrt{\mathrm{2}}\right)} \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)=\:\frac{−\mathrm{1}}{\left(\mathrm{2}{i}\right)\:\left(\mathrm{1}\right)}\:=\frac{−\mathrm{1}}{\mathrm{2}{i}} \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{2}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \left({z}−{i}\sqrt{\mathrm{2}}\right)\varphi\left({z}\right)\:=\:\:\frac{−\mathrm{2}}{\left(−\mathrm{1}\right)\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{{i}\sqrt{\mathrm{2}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\frac{−\mathrm{1}}{\mathrm{2}{i}}\:+\frac{\mathrm{1}}{{i}\sqrt{\mathrm{2}}}\right)\:=\:−\pi\:\:\:+\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}}}\:=−\pi\:+\pi\sqrt{\mathrm{2}} \\ $$$$=\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)\pi\:. \\ $$
Commented by abdo imad last updated on 26/Mar/18
$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\frac{\pi}{\mathrm{2}}\left(\sqrt{\mathrm{2}}\:\:−\mathrm{1}\right)\:. \\ $$
Answered by sma3l2996 last updated on 25/Mar/18
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$$${by}\:{parts} \\ $$$${u}={arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\Rightarrow{u}'=\frac{−{x}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${v}'=\mathrm{1}\Rightarrow{v}={x} \\ $$$${I}=\left[{xarctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}−{x}^{\mathrm{2}} −\mathrm{2}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${let}\:\:{x}={sint}\Rightarrow{dx}={costdt} \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dt}}{\mathrm{2}−{sin}^{\mathrm{2}} {t}}−\frac{\pi}{\mathrm{2}} \\ $$$${u}={tant}\Rightarrow{dt}=\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\mathrm{2}−{sin}^{\mathrm{2}} {t}=\mathrm{2}−\frac{{tan}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}=\frac{\mathrm{2}+{tan}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}=\frac{\mathrm{2}+{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{du}}{\mathrm{2}+{u}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}}=\int_{\mathrm{0}} ^{\infty} \frac{{du}}{\mathrm{1}+\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}}=\sqrt{\mathrm{2}}\left[{arctan}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\infty} −\frac{\pi}{\mathrm{2}} \\ $$$${I}=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$