Question Number 193411 by cortano12 last updated on 13/Jun/23
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Commented by BaliramKumar last updated on 13/Jun/23
$$\mathrm{put}\:\:\:{x}\:=\:{cos}\mathrm{2}\theta \\ $$
Answered by Frix last updated on 13/Jun/23
![∫_0 ^1 (√((1−x)/(1+x)))dx =^(t=(√((1+x)/(1−x)))) =4∫_1 ^∞ (dt/((t^2 +1)^2 ))=[((2t)/(t^2 +1))+2tan^(−1) t]_1 ^∞ = =(π/2)−1](https://www.tinkutara.com/question/Q193418.png)
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx}\:\overset{{t}=\sqrt{\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}}} {=}\: \\ $$$$=\mathrm{4}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\left[\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2tan}^{−\mathrm{1}} \:{t}\right]_{\mathrm{1}} ^{\infty} = \\ $$$$=\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$
Answered by Subhi last updated on 13/Jun/23
![put x=cos(θ) ⇛ dx=−sin(θ).dθ ∫_0 ^1 (√((1−cos(θ))/(1+cos(θ)))).−sin(θ)dθ −∫_0 ^1 tan((θ/2)).sin(θ)dθ ⇛ −2∫((sin((θ/2)))/(cos((θ/2)))).sin((θ/2)).cos((θ/2))dθ −2∫sin^2 ((θ/2))dθ ⇛ −∫1−cos(θ)dθ ⇛sin(θ)−θ x=(√(1−sin^2 (θ))) ⇛ sin(θ)=(√(1−x^2 )) [(√(1−x^2 ))−cos^(−1) (x)]_0 ^1 = 0−(1−(π/2))=(π/2)−1](https://www.tinkutara.com/question/Q193431.png)
$${put}\:{x}={cos}\left(\theta\right)\:\Rrightarrow\:{dx}=−{sin}\left(\theta\right).{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−{cos}\left(\theta\right)}{\mathrm{1}+{cos}\left(\theta\right)}}.−{sin}\left(\theta\right){d}\theta \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} {tan}\left(\frac{\theta}{\mathrm{2}}\right).{sin}\left(\theta\right){d}\theta\:\Rrightarrow\:−\mathrm{2}\int\frac{{sin}\left(\frac{\theta}{\mathrm{2}}\right)}{{cos}\left(\frac{\theta}{\mathrm{2}}\right)}.{sin}\left(\frac{\theta}{\mathrm{2}}\right).{cos}\left(\frac{\theta}{\mathrm{2}}\right){d}\theta \\ $$$$−\mathrm{2}\int{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right){d}\theta\:\Rrightarrow\:−\int\mathrm{1}−{cos}\left(\theta\right){d}\theta\:\Rrightarrow{sin}\left(\theta\right)−\theta \\ $$$${x}=\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left(\theta\right)}\:\Rrightarrow\:{sin}\left(\theta\right)=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left[\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }−{cos}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\mathrm{0}−\left(\mathrm{1}−\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$$$ \\ $$