Question-109378

Question Number 109378 by shahria14 last updated on 23/Aug/20

Answered by john santu last updated on 23/Aug/20

((≈JS ≈)/(■★■)) (√(((1+x)/(1−x)).((1+x)/(1+x)))) = ((1+x)/( (√(1−x^2 )))) ∫ (√((1+x)/(1−x))) dx = ∫ (1/( (√(1−x^2 )))) dx +∫ (x/( (√(1−x^2 )))) dx = arc sin (x) −(1/2)∫ ((d(1−x^2 ))/( (√(1−x^2 )))) = arc sin (x)−(1/2).2(√(1−x^2 )) + c = arc sin (x)−(√(1−x^2 )) + c

$$\:\:\:\:\:\frac{\approx{JS}\:\approx}{\blacksquare\bigstar\blacksquare} \\ $$$$\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}.\frac{\mathrm{1}+{x}}{\mathrm{1}+{x}}}\:=\:\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\: \\ $$$$\int\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx}\:=\:\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\:+\int\:\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx} \\ $$$$=\:\mathrm{arc}\:\mathrm{sin}\:\left({x}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{d}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$=\:\mathrm{arc}\:\mathrm{sin}\:\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:{c} \\ $$$$=\:\mathrm{arc}\:\mathrm{sin}\:\left({x}\right)−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:{c}\: \\ $$

Commented by shahria14 last updated on 23/Aug/20

thanks

Answered by 1549442205PVT last updated on 23/Aug/20

Put x=cosϕ,ϕ∈[0;π]⇒dx=−sinϕdϕ ∫(√((1+x)/(1−x))) dx=∫(√((1+cosϕ)/(1−cosϕ)))(−sinϕ)dϕ =−∫(√(((2cos^2 (ϕ/2))/(2sin^2 (ϕ/2))).))2sin(ϕ/2)cos(ϕ/2)dϕ =−∫2cos^2 (ϕ/2)dϕ=−∫(1+cosϕ)dϕ =−ϕ+sinϕ=−cos^(−1) (x)+(√(1−x^2 )) +C

$$\mathrm{Put}\:\mathrm{x}=\mathrm{cos}\varphi,\varphi\in\left[\mathrm{0};\pi\right]\Rightarrow\mathrm{dx}=−\mathrm{sin}\varphi\mathrm{d}\varphi \\ $$$$\int\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}\:\mathrm{dx}=\int\sqrt{\frac{\mathrm{1}+\mathrm{cos}\varphi}{\mathrm{1}−\mathrm{cos}\varphi}}\left(−\mathrm{sin}\varphi\right)\mathrm{d}\varphi \\ $$$$=−\int\sqrt{\frac{\mathrm{2cos}^{\mathrm{2}} \frac{\varphi}{\mathrm{2}}}{\mathrm{2sin}^{\mathrm{2}} \frac{\varphi}{\mathrm{2}}}.}\mathrm{2sin}\frac{\varphi}{\mathrm{2}}\mathrm{cos}\frac{\varphi}{\mathrm{2}}\mathrm{d}\varphi \\ $$$$=−\int\mathrm{2cos}^{\mathrm{2}} \frac{\varphi}{\mathrm{2}}\mathrm{d}\varphi=−\int\left(\mathrm{1}+\mathrm{cos}\varphi\right)\mathrm{d}\varphi \\ $$$$=−\varphi+\mathrm{sin}\varphi=−\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\mathrm{C} \\ $$

Answered by Dwaipayan Shikari last updated on 23/Aug/20

∫(√((1+x)/(1−x))) dx ∫(1/( (√(1−x^2 ))))+(x/( (√(1−x^2 ))))dx sin^(−1) x−(√(1−x^2 )) +C

$$\int\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$$$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${sin}^{−\mathrm{1}} {x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\:+{C} \\ $$

Answered by mathmax by abdo last updated on 23/Aug/20

let I =∫ (√((1+x)/(1−x)))dx we do the changement x =cos(θ) ⇒ I =∫(√((2cos^2 ((θ/2)))/(2sin^2 ((θ/2)))))(−sinθ)dθ =−∫((cos((θ/2)))/(sn((θ/2))))2cos((θ/2))sin((θ/2))dθ =−∫ 2cos^2 ((θ/2))dθ =−∫(1+cosθ)dθ =−θ−sinθ +C =−arcosx−(√(1−x^2 )) + C

$$\mathrm{let}\:\mathrm{I}\:=\int\:\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}\:=\mathrm{cos}\left(\theta\right)\:\Rightarrow \\ $$$$\mathrm{I}\:=\int\sqrt{\frac{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}}\left(−\mathrm{sin}\theta\right)\mathrm{d}\theta\:=−\int\frac{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{sn}\left(\frac{\theta}{\mathrm{2}}\right)}\mathrm{2cos}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{d}\theta \\ $$$$=−\int\:\mathrm{2cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)\mathrm{d}\theta\:=−\int\left(\mathrm{1}+\mathrm{cos}\theta\right)\mathrm{d}\theta\:=−\theta−\mathrm{sin}\theta\:\:+\mathrm{C} \\ $$$$=−\mathrm{arcosx}−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\:\mathrm{C} \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 23/Aug/20

another method I =∫ (√((1+x)/(1−x)))dx we do 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)) ⇒ (dx/dt) =((2t(t^2 +1)−2t(t^2 −1))/((t^2 +1)^2 )) =((4t)/((t^2 +1)^2 )) ⇒I =∫t((4t)/((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 )) we have ∫ (dt/(1+t^2 )) =arctan(t) +c_0 ∫ (dt/((1+t^2 )^2 )) =_(t =tanθ) ∫ ((1+tan^2 θ)/((1+tan^2 θ)^2 ))dθ =∫ (dθ/(1+tan^2 θ)) =∫ cos^2 θ dθ =(1/2)∫(1+cos(2θ))dθ =(θ/2) +(1/4)sin(2θ) =(1/2)arctan(t)+(1/4)×((2t)/(1+t^2 )) ⇒ I =4arctan((√((1+x)/(1−x))))−2arctan((√((1+x)/(1−x))))+(1/2)((√((1+x)/(1−x)))/(1+((1+x)/(1−x)))) +C =2arctan((√((1+x)/(1−x)))) +(1/4)(1−x)(√((1+x)/(1−x)))+C

$$\mathrm{another}\:\mathrm{method}\:\:\mathrm{I}\:=\int\:\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}=\mathrm{t} \\ $$$$\Rightarrow\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}\:=\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{1}+\mathrm{x}\:=\mathrm{t}^{\mathrm{2}} −\mathrm{t}^{\mathrm{2}} \mathrm{x}\:\Rightarrow\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{x}\:=\mathrm{t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow\mathrm{x}\:=\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\frac{\mathrm{dx}}{\mathrm{dt}}\:=\frac{\mathrm{2t}\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)−\mathrm{2t}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{4t}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\mathrm{I}\:=\int\mathrm{t}\frac{\mathrm{4t}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\mathrm{4}\:\int\:\:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dt}\:=\mathrm{4}\:\int\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}−\mathrm{4}\:\int\:\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{we}\:\mathrm{have} \\ $$$$\int\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:=\mathrm{arctan}\left(\mathrm{t}\right)\:+\mathrm{c}_{\mathrm{0}} \\ $$$$\int\:\frac{\mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=_{\mathrm{t}\:=\mathrm{tan}\theta} \:\:\:\int\:\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}{\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\mathrm{d}\theta\:=\int\:\frac{\mathrm{d}\theta}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\:=\int\:\mathrm{cos}^{\mathrm{2}} \theta\:\mathrm{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2}\theta\right)\right)\mathrm{d}\theta\:=\frac{\theta}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2}\theta\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{arctan}\left(\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\mathrm{I}\:=\mathrm{4arctan}\left(\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}\right)−\mathrm{2arctan}\left(\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\frac{\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}}{\mathrm{1}+\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}\:+\mathrm{C} \\ $$$$=\mathrm{2arctan}\left(\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{x}\right)\sqrt{\frac{\mathrm{1}+\mathrm{x}}{\mathrm{1}−\mathrm{x}}}+\mathrm{C} \\ $$$$ \\ $$

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