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Question Number 16703 by Tinkutara last updated on 25/Jun/17

Evaluate: ∫_0 ^(1/2) (dx/((1 + x^2 )(√(1 − x^2 ))))

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

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17

x=tgϕ⇒dx=(1+tg^2 ϕ)dϕ  1+x^2 =(1/(cos^2 ϕ))⇒cos^2 ϕ=(1/(1+x^2 ))⇒  sin^2 ϕ=1−(1/(1+x^2 ))=(x^2 /(1+x^2 ))⇒sinϕ=(x/(√(1+x^2 )))  ⇒I=∫(((1+tg^2 ϕ)dϕ)/((1+tg^2 ϕ)(√(1−tg^2 ϕ))))=∫(dϕ/(√(1−tg^2 ϕ)))=  =∫((cosϕdϕ)/(√(1−2sin^2 ϕ)))=∫(du/(√(1−2u^2 )))=(1/(√2))∫(du/(√((1/2)−u^2 )))=    =(1/(√2))×sin^(−1) (√2)u+C=(1/(√2))sin^(−1) ((√2)sinϕ)+C=  =(1/(√2))sin^(−1) (((x(√2))/(√(1+x^2 ))))+C  ⇒I=F((1/2))−F(0)=(1/(√2))sin^(−1) (((2×(1/2)(√2))/(√5)))−0=  =(1/(√2))sin^(−1) ((√(2/5)))=0.483924  .■

$${x}={tg}\varphi\Rightarrow{dx}=\left(\mathrm{1}+{tg}^{\mathrm{2}} \varphi\right){d}\varphi \\ $$$$\mathrm{1}+{x}^{\mathrm{2}} =\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \varphi}\Rightarrow{cos}^{\mathrm{2}} \varphi=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\Rightarrow \\ $$$${sin}^{\mathrm{2}} \varphi=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\Rightarrow{sin}\varphi=\frac{{x}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow{I}=\int\frac{\left(\mathrm{1}+{tg}^{\mathrm{2}} \varphi\right){d}\varphi}{\left(\mathrm{1}+{tg}^{\mathrm{2}} \varphi\right)\sqrt{\mathrm{1}−{tg}^{\mathrm{2}} \varphi}}=\int\frac{{d}\varphi}{\sqrt{\mathrm{1}−{tg}^{\mathrm{2}} \varphi}}= \\ $$$$=\int\frac{{cos}\varphi{d}\varphi}{\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \varphi}}=\int\frac{{du}}{\sqrt{\mathrm{1}−\mathrm{2}{u}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\int\frac{{du}}{\sqrt{\frac{\mathrm{1}}{\mathrm{2}}−{u}^{\mathrm{2}} }}= \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}×{sin}^{−\mathrm{1}} \sqrt{\mathrm{2}}{u}+{C}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}{sin}\varphi\right)+{C}= \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\frac{{x}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)+{C} \\ $$$$\Rightarrow{I}={F}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−{F}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{5}}}\right)−\mathrm{0}= \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}\right)=\mathrm{0}.\mathrm{483924}\:\:.\blacksquare \\ $$

Commented by Tinkutara last updated on 26/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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.

$${you}\:{are}\:{right}.{it}\:{is}\:{correct}\:{now}. \\ $$

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))

$$\mathrm{Let}\:{x}\:=\:\frac{\mathrm{1}}{{t}}\:\Rightarrow\:{t}\:=\:\frac{\mathrm{1}}{{x}}\:\mathrm{and}\:{dx}\:=\:\frac{−{dt}}{{t}^{\mathrm{2}} } \\ $$$$\therefore\:\int\frac{{dx}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\:=\:\int\frac{\frac{−{dt}}{{t}^{\mathrm{2}} }}{\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)\sqrt{\mathrm{1}\:−\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}} \\ $$$$=\:−\int\frac{{tdt}}{\left({t}^{\mathrm{2}} \:+\:\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} \:−\:\mathrm{1}}} \\ $$$$\mathrm{Now}\:\mathrm{let}\:{t}^{\mathrm{2}} \:−\:\mathrm{1}\:=\:{z}^{\mathrm{2}} .\:\mathrm{So}\:{tdt}\:=\:{zdz}. \\ $$$$\mathrm{Integral}\:\mathrm{becomes}\:−\int\frac{{zdz}}{\left({z}^{\mathrm{2}} \:+\:\mathrm{2}\right){z}} \\ $$$$=\:−\int\frac{{dz}}{{z}^{\mathrm{2}} \:+\:\mathrm{2}}\:=\:\frac{−\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \:\frac{{z}}{\sqrt{\mathrm{2}}} \\ $$$$\mathrm{Now}\:{z}\:=\:\sqrt{{t}^{\mathrm{2}} \:−\:\mathrm{1}}\:=\:\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\mathrm{1}} \\ $$$$\mathrm{As}\:{x}\rightarrow\mathrm{0},\:{z}\rightarrow\infty\:\mathrm{and}\:{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}},\:{z}\rightarrow\sqrt{\mathrm{3}}. \\ $$$$\therefore\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dx}}{\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}\:−\:{x}^{\mathrm{2}} }}\:=\:\frac{−\mathrm{1}}{\sqrt{\mathrm{2}}}\:\left[\mathrm{tan}^{−\mathrm{1}} \:\frac{{z}}{\sqrt{\mathrm{2}}}\right]_{\infty} ^{\sqrt{\mathrm{3}}} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\left[\frac{\pi}{\mathrm{2}}\:−\:\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\right] \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \:\sqrt{\frac{\mathrm{2}}{\mathrm{3}}} \\ $$

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