Question Number 30512 by abdo imad last updated on 22/Feb/18
$${find}\:\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\sqrt{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}}\:{dt}\:. \\ $$
Commented by abdo imad last updated on 24/Feb/18
![the ch.t=cosθ give I= ∫_0 ^(π/2) (√((1−cosθ)/(1+cosθ))) sinθdθ = ∫_0 ^(π/2) tan((θ/2)) sinθ d.θ =∫_0 ^(π/2) ((sin((θ/2)))/(cos((θ/2))))2 sin((θ/2))cos((θ/2))dθ =∫_0 ^(π/2) 2 sin^2 ((θ/2))dθ= ∫_0 ^(π/2) (1−cosθ)dθ =(π/2) −[sinθ]_0 ^(π/2) =(π/2) −1.](https://www.tinkutara.com/question/Q30703.png)
$${the}\:{ch}.{t}={cos}\theta\:{give}\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\sqrt{\frac{\mathrm{1}−{cos}\theta}{\mathrm{1}+{cos}\theta}}\:{sin}\theta{d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{tan}\left(\frac{\theta}{\mathrm{2}}\right)\:{sin}\theta\:{d}.\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}\left(\frac{\theta}{\mathrm{2}}\right)}{{cos}\left(\frac{\theta}{\mathrm{2}}\right)}\mathrm{2}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right){cos}\left(\frac{\theta}{\mathrm{2}}\right){d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{2}\:{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right){d}\theta=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}−{cos}\theta\right){d}\theta\:=\frac{\pi}{\mathrm{2}}\:−\left[{sin}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{2}}\:−\mathrm{1}. \\ $$