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Question Number 85669 by john santu last updated on 23/Mar/20
∫ ((√(1+x))/( (√(1−x)))) dx
$$\int\:\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 24/Mar/20
I =∫ ((√(1+x))/( (√(1−x))))dx ⇒ I =∫(√((1+x)/(1−x)))dx we use the changement (√((1+x)/(1−x)))=t ⇒((1+x)/(1−x))=t^2 ⇒1+x =t^2 −t^2 x ⇒(1+t^2 )x =t^2 −1 ⇒ x =((t^2 −1)/(t^2 +1)) =((t^2 +1−2)/(t^2 +1)) =1−(2/(t^2 +1)) ⇒dx =(−2)(−((2t)/((t^2 +1)^2 ))) =((4t)/((t^2 +1)^2 )) ⇒ I =∫ t((4t)/((t^2 +1)^2 ))dt =4 ∫ (t^2 /((t^2 +1)^2 ))dt =4 ∫ ((t^2 +1−1)/((t^2 +1)^2 ))dt =4 ∫ (dt/(t^2 +1)) −4 ∫ (dt/((t^2 +1)^2 )) =4arctan(t)−4∫ (dt/((t^2 +1)^2 )) we have ∫ (dt/((t^2 +1)^2 )) =_(t=tanu) ∫ (((1+tan^2 u)du)/((1+tan^2 u)^2 )) =∫ (du/(1+tan^2 u)) =∫ cos^2 u du =∫ ((1+cos(2u))/2)du =(u/2) +(1/4)sin(2u) =((arctan(t))/2) +(1/4)sin(2 arctant) +c (1/2)arctan((√((1+x)/(1−x))))+(1/4)sin(2 arctan((√((1+x)/(1−x)))))+C ⇒ I =4arctan((√((1+x)/(1−x))))−2 arctan((√((1+x)/(1−x))))−sin(2arctan((√(((1+x)/(1−x))))))) +C
$${I}\:=\int\:\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{\mathrm{1}−{x}}}{dx}\:\Rightarrow\:{I}\:=\int\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{dx}\:{we}\:{use}\:{the}\:{changement}\: \\ $$$$\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}={t}\:\Rightarrow\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}={t}^{\mathrm{2}} \:\Rightarrow\mathrm{1}+{x}\:={t}^{\mathrm{2}} −{t}^{\mathrm{2}} {x}\:\Rightarrow\left(\mathrm{1}+{t}^{\mathrm{2}} \right){x}\:={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow \\ $$$${x}\:=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\mathrm{1}−\frac{\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow{dx}\:=\left(−\mathrm{2}\right)\left(−\frac{\mathrm{2}{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:{I}\:=\int\:\:{t}\frac{\mathrm{4}{t}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\mathrm{4}\:\int\:\:\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{4}\:\int\:\:\frac{{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=\mathrm{4}\:\int\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:−\mathrm{4}\:\int\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{4}{arctan}\left({t}\right)−\mathrm{4}\int\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:{we}\:{have} \\ $$$$\int\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=_{{t}={tanu}} \:\:\int\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right){du}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {u}\right)^{\mathrm{2}} }\:=\int\:\frac{{du}}{\mathrm{1}+{tan}^{\mathrm{2}} {u}}\:=\int\:{cos}^{\mathrm{2}} {u}\:{du} \\ $$$$=\int\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{u}\right)}{\mathrm{2}}{du}\:=\frac{{u}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{u}\right) \\ $$$$=\frac{{arctan}\left({t}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}\:{arctant}\right)\:+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right)+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}\:{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right)\right)+{C}\:\Rightarrow \\ $$$${I}\:=\mathrm{4}{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right)−\mathrm{2}\:{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right)−{sin}\left(\mathrm{2}{arctan}\left(\sqrt{\left.\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)}\right)\right)\:+{C} \\ $$
Commented by mathmax by abdo last updated on 24/Mar/20
⇒ I =2arctan((√((1+x)/(1−x))))−sin(2arctan((√((1+x)/(1−x))))) +C
$$\Rightarrow\:{I}\:=\mathrm{2}{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right)−{sin}\left(\mathrm{2}{arctan}\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\right)\right)\:+{C} \\ $$
Commented by jagoll last updated on 24/Mar/20
mister, your solution uses a lot of hyperbolic functions
$$\mathrm{mister},\:\mathrm{your}\:\mathrm{solution}\:\mathrm{uses}\:\mathrm{a}\:\mathrm{lot} \\ $$$$\mathrm{of}\:\mathrm{hyperbolic}\:\mathrm{functions} \\ $$
Commented by mathmax by abdo last updated on 24/Mar/20
yes but the answer is correct...
$${yes}\:{but}\:{the}\:{answer}\:{is}\:{correct}… \\ $$
Answered by john santu last updated on 24/Mar/20
(√(1−x)) = t⇒x= 1−t^2 (√(1+x )) = (√(2−t^2 )) I= ∫ (((√(2−t^2 )) (−2t dt))/t) = I= −∫ (√(2−t^2 )) dt [ t = (√2) sin θ ] I = −∫ (√2) cos θ (√2) cos θ dθ I = −2 ∫ ((1/2)+ cos 2θ) dθ I = −θ −sin 2θ + c I = −sin^(−1) (((√(1−x))/( (√2))))−2(((√(1−x))/( (√2))))(((√(1+x))/( (√2))))+c I= −sin^(−1) ((√((1−x)/2)))−(√(1−x^2 )) +c
$$\sqrt{\mathrm{1}−{x}}\:=\:{t}\Rightarrow{x}=\:\mathrm{1}−{t}^{\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}+{x}\:}\:=\:\sqrt{\mathrm{2}−{t}^{\mathrm{2}} } \\ $$$${I}=\:\int\:\frac{\sqrt{\mathrm{2}−{t}^{\mathrm{2}} \:}\:\left(−\mathrm{2}{t}\:{dt}\right)}{{t}}\:=\: \\ $$$${I}=\:−\int\:\sqrt{\mathrm{2}−{t}^{\mathrm{2}} }\:{dt}\:\:\left[\:{t}\:=\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\theta\:\right] \\ $$$${I}\:=\:−\int\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\theta\:{d}\theta \\ $$$${I}\:=\:−\mathrm{2}\:\int\:\left(\frac{\mathrm{1}}{\mathrm{2}}+\:\mathrm{cos}\:\mathrm{2}\theta\right)\:{d}\theta \\ $$$${I}\:=\:−\theta\:−\mathrm{sin}\:\mathrm{2}\theta\:+\:{c} \\ $$$${I}\:=\:−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{2}}}\right)−\mathrm{2}\left(\frac{\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{2}}}\right)\left(\frac{\sqrt{\mathrm{1}+{x}}}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$$${I}=\:−\mathrm{sin}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{2}}}\right)−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+{c} \\ $$

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