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Question Number 152608 by ZiYangLee last updated on 30/Aug/21

By using the substitution x=cos 2θ, prove that ∫ (√((1+x)/(1−x))) dx = −sin 2θ−2θ+C

$$\mathrm{By}\:\mathrm{using}\:\mathrm{the}\:\mathrm{substitution}\:{x}=\mathrm{cos}\:\mathrm{2}\theta, \\ $$$$\mathrm{prove}\:\mathrm{that}\:\int\:\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx}\:=\:−\mathrm{sin}\:\mathrm{2}\theta−\mathrm{2}\theta+{C} \\ $$

Answered by Olaf_Thorendsen last updated on 30/Aug/21

F(x) = ∫(√( ((1+x)/(1−x)))) dx F(θ) = ∫(√( ((1+cos2θ)/(1−cos2θ)))) (−2sin2θdθ) F(θ) = ∫(√( ((2cos^2 θ)/(2sin^2 θ)))) (−4sinθcosθdθ) F(θ) = ∫∣cotθ∣ (−4sinθcosθdθ) F(θ) = −4∫cos^2 θdθ) F(θ) = −2∫(1+cos2θ)dθ) F(θ) = −2(θ+(1/2)sin2θ) F(θ) = −sin2θ−2θ+C

$$\mathrm{F}\left({x}\right)\:=\:\int\sqrt{\:\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}\:{dx} \\ $$$$\mathrm{F}\left(\theta\right)\:=\:\int\sqrt{\:\frac{\mathrm{1}+\mathrm{cos2}\theta}{\mathrm{1}−\mathrm{cos2}\theta}}\:\left(−\mathrm{2sin2}\theta{d}\theta\right) \\ $$$$\mathrm{F}\left(\theta\right)\:=\:\int\sqrt{\:\frac{\mathrm{2cos}^{\mathrm{2}} \theta}{\mathrm{2sin}^{\mathrm{2}} \theta}}\:\left(−\mathrm{4sin}\theta\mathrm{cos}\theta{d}\theta\right) \\ $$$$\mathrm{F}\left(\theta\right)\:=\:\int\mid\mathrm{cot}\theta\mid\:\left(−\mathrm{4sin}\theta\mathrm{cos}\theta{d}\theta\right) \\ $$$$\left.\mathrm{F}\left(\theta\right)\:=\:−\mathrm{4}\int\mathrm{cos}^{\mathrm{2}} \theta{d}\theta\right) \\ $$$$\left.\mathrm{F}\left(\theta\right)\:=\:−\mathrm{2}\int\left(\mathrm{1}+\mathrm{cos2}\theta\right){d}\theta\right) \\ $$$$\mathrm{F}\left(\theta\right)\:=\:−\mathrm{2}\left(\theta+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}\theta\right) \\ $$$$\mathrm{F}\left(\theta\right)\:=\:−\mathrm{sin2}\theta−\mathrm{2}\theta+\mathrm{C} \\ $$

Commented by puissant last updated on 30/Aug/21

Genial Mr !!!

$${Genial}\:{Mr}\:!!!\: \\ $$

Answered by Ar Brandon last updated on 30/Aug/21

I=∫(√((1+x)/(1−x)))dx, x=cos2ϑ =−2∫(√((1+cos2ϑ)/(1−cos2ϑ)))∙sin2ϑdϑ =−2∫((1+cos2ϑ)/( (√(1−cos^2 2ϑ))))∙sin2ϑdϑ =−2∫((1+cos2ϑ)/(∣sin2ϑ∣))∙sin2ϑdϑ =∓2∫(1+cos2ϑ)dϑ =∓(2ϑ+sin2ϑ)+C

$${I}=\int\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}{dx},\:{x}=\mathrm{cos2}\vartheta \\ $$$$\:\:=−\mathrm{2}\int\sqrt{\frac{\mathrm{1}+\mathrm{cos2}\vartheta}{\mathrm{1}−\mathrm{cos2}\vartheta}}\centerdot\mathrm{sin2}\vartheta{d}\vartheta \\ $$$$\:\:=−\mathrm{2}\int\frac{\mathrm{1}+\mathrm{cos2}\vartheta}{\:\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{2}\vartheta}}\centerdot\mathrm{sin2}\vartheta{d}\vartheta \\ $$$$\:\:=−\mathrm{2}\int\frac{\mathrm{1}+\mathrm{cos2}\vartheta}{\mid\mathrm{sin2}\vartheta\mid}\centerdot\mathrm{sin2}\vartheta{d}\vartheta \\ $$$$\:\:=\mp\mathrm{2}\int\left(\mathrm{1}+\mathrm{cos2}\vartheta\right){d}\vartheta \\ $$$$\:\:=\mp\left(\mathrm{2}\vartheta+\mathrm{sin2}\vartheta\right)+{C} \\ $$

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