Question Number 16703 by Tinkutara last updated on 25/Jun/17
$$\mathrm{Evaluate}:\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dx}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }} \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17
$${x}={tg}\varphi\Rightarrow{dx}=\left(\mathrm{1}+{tg}^{\mathrm{2}} \varphi\right){d}\varphi \\ $$$$\mathrm{1}+{x}^{\mathrm{2}} =\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \varphi}\Rightarrow{cos}^{\mathrm{2}} \varphi=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\Rightarrow \\ $$$${sin}^{\mathrm{2}} \varphi=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\Rightarrow{sin}\varphi=\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow{I}=\int\frac{\left(\mathrm{1}+{tg}^{\mathrm{2}} \varphi\right){d}\varphi}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \varphi\right)\sqrt{\mathrm{1}−{tg}^{\mathrm{2}} \varphi}}=\int\frac{{d}\varphi}{\:\sqrt{\mathrm{1}−{tg}^{\mathrm{2}} \varphi}}= \\ $$$$=\int\frac{{cos}\varphi{d}\varphi}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \varphi}}=\int\frac{{du}}{\:\sqrt{\mathrm{1}−\mathrm{2}{u}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{du}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}−{u}^{\mathrm{2}} }}= \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×{sin}^{−\mathrm{1}} \sqrt{\mathrm{2}}{u}+{C}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}{sin}\varphi\right)+{C}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\frac{{x}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)+{C} \\ $$$$\Rightarrow{I}={F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{F}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{5}}}\right)−\mathrm{0}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\right)=\mathrm{0}.\mathrm{483924}\:\:.\blacksquare \\ $$
Commented by Tinkutara last updated on 26/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17
$${you}\:{are}\:{right}.{it}\:{is}\:{correct}\:{now}. \\ $$
Answered by Tinkutara last updated on 05/Jul/17
![Let x = (1/t) ⇒ t = (1/x) and dx = ((−dt)/t^2 ) ∴ ∫(dx/((1 + x^2 )(√(1 − x^2 )))) = ∫(((−dt)/t^2 )/((1 + (1/t^2 ))(√(1 − (1/t^2 ))))) = −∫((tdt)/((t^2 + 1)(√(t^2 − 1)))) Now let t^2 − 1 = z^2 . So tdt = zdz. Integral becomes −∫((zdz)/((z^2 + 2)z)) = −∫(dz/(z^2 + 2)) = ((−1)/( (√2))) tan^(−1) (z/( (√2))) Now z = (√(t^2 − 1)) = (√((1/x^2 ) − 1)) As x→0, z→∞ and x→(1/2), z→(√3). ∴ ∫_0 ^(1/2) (dx/((1 + x^2 )(√(1 − x^2 )))) = ((−1)/( (√2))) [tan^(−1) (z/( (√2)))]_∞ ^(√3) = (1/( (√2))) [(π/2) − tan^(−1) (√(3/2))] = (1/( (√2))) tan^(−1) (√(2/3))](https://www.tinkutara.com/question/Q17442.png)
$$\mathrm{Let}\:{x}\:=\:\frac{\mathrm{1}}{{t}}\:\Rightarrow\:{t}\:=\:\frac{\mathrm{1}}{{x}}\:\mathrm{and}\:{dx}\:=\:\frac{−{dt}}{{t}^{\mathrm{2}} } \\ $$$$\therefore\:\int\frac{{dx}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\:=\:\int\frac{\frac{−{dt}}{{t}^{\mathrm{2}} }}{\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\sqrt{\mathrm{1}\:−\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}} \\ $$$$=\:−\int\frac{{tdt}}{\left({t}^{\mathrm{2}} \:+\:\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} \:−\:\mathrm{1}}} \\ $$$$\mathrm{Now}\:\mathrm{let}\:{t}^{\mathrm{2}} \:−\:\mathrm{1}\:=\:{z}^{\mathrm{2}} .\:\mathrm{So}\:{tdt}\:=\:{zdz}. \\ $$$$\mathrm{Integral}\:\mathrm{becomes}\:−\int\frac{{zdz}}{\left({z}^{\mathrm{2}} \:+\:\mathrm{2}\right){z}} \\ $$$$=\:−\int\frac{{dz}}{{z}^{\mathrm{2}} \:+\:\mathrm{2}}\:=\:\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \:\frac{{z}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{Now}\:{z}\:=\:\sqrt{{t}^{\mathrm{2}} \:−\:\mathrm{1}}\:=\:\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\mathrm{1}} \\ $$$$\mathrm{As}\:{x}\rightarrow\mathrm{0},\:{z}\rightarrow\infty\:\mathrm{and}\:{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}},\:{z}\rightarrow\sqrt{\mathrm{3}}. \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dx}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\:=\:\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\mathrm{tan}^{−\mathrm{1}} \:\frac{{z}}{\:\sqrt{\mathrm{2}}}\right]_{\infty} ^{\sqrt{\mathrm{3}}} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\frac{\pi}{\mathrm{2}}\:−\:\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\right] \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$