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 ⇒ I = ∫_0 ^(π/2) arctan(cost) cost dt let integrate by parts u=arctan(cost) and v^′ =cost I = [sint arctan(cost)]_0 ^(π/2) −∫_0 ^(π/2) ((−sint)/(1+cos^2 t)) sint dt = ∫_0 ^(π/2) ((1−cos^2 t)/(1+cos^2 t)) dt = ∫_0 ^(π/2) ((1−((1+cos(2t))/2))/(1 +((1+cos(2t))/2))) dt = ∫_0 ^(π/2) ((1−cos(2t))/(3+cos(2t)))dt =_(2t=u) ∫_0 ^π ((1−cosu)/(3 +cosu)) (du/2) =(1/2) ∫_0 ^π ((1−cosu)/(3 +cosu))du ch.tan((u/2))=x give I = (1/2) ∫_0 ^∞ ((1− ((1−x^2 )/(1+x^2 )))/(3 +((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) =∫_0 ^∞ ((1+x^2 −1+x^2 )/((3 +3x^2 +1−x^2 )(1+x^2 ))) dx =∫_0 ^∞ ((2x^2 )/((4+2x^2 )(1+x^2 )))dx = ∫_0 ^∞ (x^2 /((x^2 +1)(x^2 +2)))dx =(1/2) ∫_(−∞) ^(+∞) (x^2 /((x^2 +1)(x^2 +2)))dx let consider ϕ(z) = (z^2 /((z^2 +1)(z^2 +2))) the poles of ϕ are i ,−i, i(√(2 )) ,−i(√2) and all are simples ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( Res(ϕ,i) +Res(ϕ,i(√2))) ϕ(z) = (z^2 /((z−i)(z+i)(z−i(√2))(z +i(√2)))) Res(ϕ,i) =lim_(z→i) (z−i)ϕ(z)= ((−1)/((2i) (1))) =((−1)/(2i)) Res(ϕ,i(√2)) =lim_(z→i(√2)) (z−i(√2))ϕ(z) = ((−2)/((−1)(2i(√2)))) =(1/(i(√2))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ( ((−1)/(2i)) +(1/(i(√2)))) = −π +((2π)/( (√2))) =−π +π(√2) =((√2) −1)π .](https://www.tinkutara.com/question/Q32514.png)
$${ch}.{x}={sint}\:\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{arctan}\left({cost}\right)\:{cost}\:{dt}\:{let}\:{integrate} \\ $$$${by}\:{parts}\:{u}={arctan}\left({cost}\right)\:{and}\:{v}^{'} \:={cost} \\ $$$${I}\:=\:\left[{sint}\:{arctan}\left({cost}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{−{sint}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:{sint}\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}−{cos}^{\mathrm{2}} {t}}{\mathrm{1}+{cos}^{\mathrm{2}} {t}}\:{dt}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}−\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}{\mathrm{1}\:+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{3}+{cos}\left(\mathrm{2}{t}\right)}{dt}\:\:=_{\mathrm{2}{t}={u}} \:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\mathrm{1}−{cosu}}{\mathrm{3}\:+{cosu}}\:\frac{{du}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\mathrm{1}−{cosu}}{\mathrm{3}\:+{cosu}}{du}\:\:{ch}.{tan}\left(\frac{{u}}{\mathrm{2}}\right)={x}\:{give} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{3}\:+\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}+{x}^{\mathrm{2}} }{\left(\mathrm{3}\:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{4}+\mathrm{2}{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)}{dx}\:{let}\:{consider} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+\mathrm{2}\right)}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\:,−{i},\:{i}\sqrt{\mathrm{2}\:}\:,−{i}\sqrt{\mathrm{2}}\:\:{and}\:{all}\:{are}\:{simples} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{i}\sqrt{\mathrm{2}}\right)\right)\:\: \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{{z}^{\mathrm{2}} }{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{i}\sqrt{\mathrm{2}}\right)\left({z}\:+{i}\sqrt{\mathrm{2}}\right)} \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)=\:\frac{−\mathrm{1}}{\left(\mathrm{2}{i}\right)\:\left(\mathrm{1}\right)}\:=\frac{−\mathrm{1}}{\mathrm{2}{i}} \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{2}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \left({z}−{i}\sqrt{\mathrm{2}}\right)\varphi\left({z}\right)\:=\:\:\frac{−\mathrm{2}}{\left(−\mathrm{1}\right)\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{{i}\sqrt{\mathrm{2}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\:\frac{−\mathrm{1}}{\mathrm{2}{i}}\:+\frac{\mathrm{1}}{{i}\sqrt{\mathrm{2}}}\right)\:=\:−\pi\:\:\:+\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}}}\:=−\pi\:+\pi\sqrt{\mathrm{2}} \\ $$$$=\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)\pi\:. \\ $$
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=∫_0 ^1 arctan((√(1−x^2 )))dx by parts u=arctan((√(1−x^2 )))⇒u′=((−x)/((2−x^2 )(√(1−x^2 )))) v′=1⇒v=x I=[xarctan((√(1−x^2 )))]_0 ^1 +∫_0 ^1 (x^2 /((2−x^2 )(√(1−x^2 ))))dx =−∫_0 ^1 ((2−x^2 −2)/((2−x^2 )(√(1−x^2 ))))dx=2∫_0 ^1 (dx/((2−x^2 )(√(1−x^2 ))))−∫_0 ^1 (dx/( (√(1−x^2 )))) let x=sint⇒dx=costdt I=2∫_0 ^(π/2) (dt/(2−sin^2 t))−(π/2) u=tant⇒dt=(du/(1+u^2 )) 2−sin^2 t=2−((tan^2 t)/(1+tan^2 t))=((2+tan^2 t)/(1+tan^2 t))=((2+u^2 )/(1+u^2 )) I=2∫_0 ^∞ (du/(2+u^2 ))−(π/2)=∫_0 ^∞ (du/(1+((u/( (√2))))^2 ))−(π/2)=(√2)[arctan((u/( (√2))))]_0 ^∞ −(π/2) I=((π(√2))/2)−(π/2)=(π/2)((√2)−1)](https://www.tinkutara.com/question/Q32466.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){dx} \\ $$$${by}\:{parts} \\ $$$${u}={arctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\Rightarrow{u}'=\frac{−{x}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${v}'=\mathrm{1}\Rightarrow{v}={x} \\ $$$${I}=\left[{xarctan}\left(\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}−{x}^{\mathrm{2}} −\mathrm{2}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${let}\:\:{x}={sint}\Rightarrow{dx}={costdt} \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dt}}{\mathrm{2}−{sin}^{\mathrm{2}} {t}}−\frac{\pi}{\mathrm{2}} \\ $$$${u}={tant}\Rightarrow{dt}=\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\mathrm{2}−{sin}^{\mathrm{2}} {t}=\mathrm{2}−\frac{{tan}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}=\frac{\mathrm{2}+{tan}^{\mathrm{2}} {t}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}=\frac{\mathrm{2}+{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{du}}{\mathrm{2}+{u}^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}}=\int_{\mathrm{0}} ^{\infty} \frac{{du}}{\mathrm{1}+\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }−\frac{\pi}{\mathrm{2}}=\sqrt{\mathrm{2}}\left[{arctan}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\infty} −\frac{\pi}{\mathrm{2}} \\ $$$${I}=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$