1-x-1-x-dx-x-1-1-

Question Number 119920 by bramlexs22 last updated on 28/Oct/20

$$\:\int\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:{dx}\:=\:?\:,\:{x}\epsilon\left(−\mathrm{1},\mathrm{1}\right) \\ $$

Answered by mathmax by abdo last updated on 28/Oct/20

I =∫(√((1−x)/(1+x)))dx we do the changement x =cost ⇒ I =∫ (√((2sin^2 ((t/2)))/(2cos^2 ((t/(2 ))))))sint dt =∫ ((sin((t/2)))/(cos((t/2))))×2sin((t/2))cos((t/2))dt =2 ∫ sin^2 ((t/2))dt =2∫ ((1−cos(t))/2)dt =∫(1−cost)dt =t−sint +c =arcosx −(√(1−x^2 )) +c =arcsinx −(√(1−x^2 )) +C

$$\mathrm{I}\:=\int\sqrt{\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}\:=\mathrm{cost}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int\:\sqrt{\frac{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{2cos}^{\mathrm{2}} \left(\frac{\mathrm{t}}{\mathrm{2}\:}\right)}}\mathrm{sint}\:\mathrm{dt} \\ $$$$=\int\:\:\frac{\mathrm{sin}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{cos}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}×\mathrm{2sin}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\mathrm{dt} \\ $$$$=\mathrm{2}\:\int\:\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{t}}{\mathrm{2}}\right)\mathrm{dt}\:=\mathrm{2}\int\:\:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{t}\right)}{\mathrm{2}}\mathrm{dt}\:=\int\left(\mathrm{1}−\mathrm{cost}\right)\mathrm{dt} \\ $$$$=\mathrm{t}−\mathrm{sint}\:+\mathrm{c}\:=\mathrm{arcosx}\:−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\mathrm{c}\:\:=\mathrm{arcsinx}\:−\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\mathrm{C} \\ $$$$ \\ $$

Answered by 1549442205PVT last updated on 28/Oct/20

∫ (√((1−x)/(1+x))) dx = ? , xε(−1,1) Put x=cos ϕ⇒dx=−sinϕdϕ F=−∫(√(((1−cosϕ)/(1+cosϕ)) )).sinϕdϕ =−∫(√((2sin^2 (ϕ/2))/(2cos^2 (ϕ/2)))).sinϕdϕ =−∫tan(ϕ/2).2sin(ϕ/2)cos(ϕ/2)dϕ =−2∫sin^2 (ϕ/2)dϕ=−∫(1−cosϕ)dϕ =−ϕ+sinϕ=−cos^(−1) (x)+(√(1−x^2 ))+C

$$\:\int\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:{dx}\:=\:?\:,\:{x}\epsilon\left(−\mathrm{1},\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{x}=\mathrm{cos}\:\varphi\Rightarrow\mathrm{dx}=−\mathrm{sin}\varphi\mathrm{d}\varphi \\ $$$$\mathrm{F}=−\int\sqrt{\frac{\mathrm{1}−\mathrm{cos}\varphi}{\mathrm{1}+\mathrm{cos}\varphi}\:}.\mathrm{sin}\varphi\mathrm{d}\varphi \\ $$$$=−\int\sqrt{\frac{\mathrm{2sin}^{\mathrm{2}} \frac{\varphi}{\mathrm{2}}}{\mathrm{2cos}^{\mathrm{2}} \frac{\varphi}{\mathrm{2}}}}.\mathrm{sin}\varphi\mathrm{d}\varphi \\ $$$$=−\int\mathrm{tan}\frac{\varphi}{\mathrm{2}}.\mathrm{2sin}\frac{\varphi}{\mathrm{2}}\mathrm{cos}\frac{\varphi}{\mathrm{2}}\mathrm{d}\varphi \\ $$$$=−\mathrm{2}\int\mathrm{sin}^{\mathrm{2}} \frac{\varphi}{\mathrm{2}}\mathrm{d}\varphi=−\int\left(\mathrm{1}−\mathrm{cos}\varphi\right)\mathrm{d}\varphi \\ $$$$=−\varphi+\mathrm{sin}\varphi=−\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }+\mathrm{C} \\ $$

Answered by bobhans last updated on 28/Oct/20

∫ (√((1−x)/(1+x))) dx =? ,x∈(−1,1) K=∫(√((1−x)/(1+x))) dx [ let (√(1+x)) =1+ t ] 1+x = 1+2t+t^2 ⇒dx = (2+2t) dt K=∫ ((√(1−2t−t^2 ))/(1+t))(2+2t)dt = 2∫ (√(2−(1+t)^2 )) dt=2∫ (√(((√2))^2 −(1+t)^2 )) dt by substituting 1+t=(√2) sin w K=4∫cos^2 w dw = 4∫((1/2)+(1/2)cos 2w)dw K=2w+sin 2w + c K=2arc sin (((1+t)/( (√2))))+2(((1+t)/( (√2))))(((√(1−2t−t^2 ))/( (√2))))+c K=2arc sin ((√((1+x)/2)))+(√(1−x^2 )) + c

$$\:\int\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:{dx}\:=?\:,{x}\in\left(−\mathrm{1},\mathrm{1}\right) \\ $$$${K}=\int\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:{dx}\:\left[\:{let}\:\sqrt{\mathrm{1}+{x}}\:=\mathrm{1}+\:{t}\:\right] \\ $$$$\mathrm{1}+{x}\:=\:\mathrm{1}+\mathrm{2}{t}+{t}^{\mathrm{2}} \:\Rightarrow{dx}\:=\:\left(\mathrm{2}+\mathrm{2}{t}\right)\:{dt} \\ $$$${K}=\int\:\frac{\sqrt{\mathrm{1}−\mathrm{2}{t}−{t}^{\mathrm{2}} }}{\mathrm{1}+{t}}\left(\mathrm{2}+\mathrm{2}{t}\right){dt}\:= \\ $$$$\mathrm{2}\int\:\sqrt{\mathrm{2}−\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\:{dt}=\mathrm{2}\int\:\sqrt{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\:{dt} \\ $$$${by}\:{substituting}\:\mathrm{1}+{t}=\sqrt{\mathrm{2}}\:\mathrm{sin}\:{w}\: \\ $$$${K}=\mathrm{4}\int\mathrm{cos}^{\mathrm{2}} {w}\:{dw}\:=\:\mathrm{4}\int\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{w}\right){dw} \\ $$$${K}=\mathrm{2}{w}+\mathrm{sin}\:\mathrm{2}{w}\:+\:{c} \\ $$$${K}=\mathrm{2arc}\:\mathrm{sin}\:\left(\frac{\mathrm{1}+{t}}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{2}\left(\frac{\mathrm{1}+{t}}{\:\sqrt{\mathrm{2}}}\right)\left(\frac{\sqrt{\mathrm{1}−\mathrm{2}{t}−{t}^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$$${K}=\mathrm{2arc}\:\mathrm{sin}\:\left(\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{2}}}\right)+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:{c}\:\: \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 28/Oct/20

I=∫(√((1−x)/(1+x)))dx=∫((1−x)/( (√(1−x^2 ))))dx =∫(dx/( (√(1−x^2 ))))−∫(x/( (√(1−x^2 ))))dx =Arcsin(x)+(√(1−x^2 ))+C

$$\mathcal{I}=\int\sqrt{\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}\mathrm{dx}=\int\frac{\mathrm{1}−\mathrm{x}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$\:\:\:=\int\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}−\int\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx} \\ $$$$\:\:\:=\mathrm{Arcsin}\left(\mathrm{x}\right)+\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }+\mathcal{C} \\ $$

Answered by Dwaipayan Shikari last updated on 28/Oct/20

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

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

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