∫_( 0) ^1 (√((1−x)/(1+x))) dx =


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Question Number 54033 by qw last updated on 28/Jan/19

$\underset{0}{\int^{1}} \sqrt{\frac{1 - x}{1 + x}} d x =$
Commented by maxmathsup by imad last updated on 29/Jan/19
$let I =∫_0 ^1 (√((1−x)/(1+x)))dx let use the chan. (√((1−x)/(1+x)))=t ⇒((1−x)/(1+x)) =t^2 ⇒ 1−x =t^2 +t^2 x ⇒1−t^2 =(1+t^2 )x ⇒x =((1−t^2 )/(1+t^2 )) ⇒ (dx/dt) =((−2t(1+t^2 )−(1−t^2 )(2t))/((1+t^2 )^2 )) =((−2t−2t^3 −2t +2t^3 )/((1+t^2 )^2 )) =((−4t)/((1+t^2 )^2 )) ⇒ I = −∫_0 ^1 t(((−4t)/((1+t^2 )^2 )) )dt =4 ∫_0 ^1 (t^2 /((1+t^2 )^2 )) dt =4{ ∫_0 ^1 ((1+t^2 −1)/((1+t^2 )^2 ))dt}=4{∫_0 ^1 (dt/(1+t^2 )) −∫_0 ^1 (dt/((1+t^2 )^2 ))} but ∫_0 ^1 (dt/(1+t^2 )) =[arctant]_0 ^1 =(π/4) ∫_0 ^1 (dt/((1+t^2 )^2 )) =_(t =tanθ) ∫_0 ^(π/4) (((1+tan^2 θ))/((1+tan^2 θ)^2 ))dθ = ∫_0 ^(π/4) (dθ/(1+tan^2 θ)) =∫_0 ^(π/4) cos^2 θ dθ =∫_0 ^(π/4) ((1+cos(2θ))/2) dθ =(π/8) +(1/4)[sin(2θ)]_0 ^(π/4) =(π/8) +(1/4) ⇒ I =4{(π/4) −(π/8) −(1/4)} =π−(π/2) −1 =(π/2) −1 .$
$l e t I = \int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x l e t u s e t h e c h a n . \sqrt{\frac{1 - x}{1 + x}} = t \Rightarrow \frac{1 - x}{1 + x} = t^{2} \Rightarrow$ $1 - x = t^{2} + t^{2} x \Rightarrow 1 - t^{2} = (1 + t^{2}) x \Rightarrow x = \frac{1 - t^{2}}{1 + t^{2}} \Rightarrow \frac{d x}{d t} = \frac{- 2 t (1 + t^{2}) - (1 - t^{2}) (2 t)}{{(1 + t^{2})}^{2}}$ $= \frac{- 2 t - 2 t^{3} - 2 t + 2 t^{3}}{{(1 + t^{2})}^{2}} = \frac{- 4 t}{{(1 + t^{2})}^{2}} \Rightarrow I = - \int_{0}^{1} t (\frac{- 4 t}{{(1 + t^{2})}^{2}}) d t$ $= 4 \int_{0}^{1} \frac{t^{2}}{{(1 + t^{2})}^{2}} d t = 4 {\int_{0}^{1} \frac{1 + t^{2} - 1}{{(1 + t^{2})}^{2}} d t} = 4 {\int_{0}^{1} \frac{d t}{1 + t^{2}} - \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}}} b u t$ $\int_{0}^{1} \frac{d t}{1 + t^{2}} = {[a r c t a n t]}_{0}^{1} = \frac{π}{4}$ $\int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}} =_{t = t a n θ} \int_{0}^{\frac{π}{4}} \frac{(1 + {t a n}^{2} θ)}{{(1 + {t a n}^{2} θ)}^{2}} d θ = \int_{0}^{\frac{π}{4}} \frac{d θ}{1 + {t a n}^{2} θ}$ $= \int_{0}^{\frac{π}{4}} {c o s}^{2} θ d θ = \int_{0}^{\frac{π}{4}} \frac{1 + c o s (2 θ)}{2} d θ = \frac{π}{8} + \frac{1}{4} {[s i n (2 θ)]}_{0}^{\frac{π}{4}} = \frac{π}{8} + \frac{1}{4} \Rightarrow$ $I = 4 {\frac{π}{4} - \frac{π}{8} - \frac{1}{4}} = π - \frac{π}{2} - 1 = \frac{π}{2} - 1 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19

	$\int \frac{\sqrt{1 - x}}{\sqrt{1 + x}} d x$ $\int \frac{1 - x}{\sqrt{1 - x^{2}}} d x$ $\int \frac{d x}{\sqrt{1 - x^{2}}} + \frac{1}{2} \int \frac{d (1 - x^{2})}{\sqrt{1 - x^{2}}}$ ${s i n}^{- 1} (x) + \frac{1}{2} \times \frac{{(1 - x^{2})}^{\frac{- 1}{2} + 1}}{\frac{1}{2}} + c$ $s o r e q u i r e d a n s w e r i s$ $∣ {s i n}^{- 1} x + {(1 - x^{2})}^{\frac{1}{2}} ∣_{0}^{1}$ $= {s i n}^{- 1} (1) + 0 - {s i n}^{- 1} (0) - {(1 - 0)}^{\frac{1}{2}}$ $= \frac{π}{2} - 1$
	Answered by ajfour last updated on 28/Jan/19

	$l e t x = \cos 2 θ$ $\int_{π / 4}^{0} \tan θ (- 4 \sin θ \cos θ) d θ$ $= 2 \int_{0}^{π / 4} (1 - \cos 2 θ) d θ$ $= \frac{π}{2} - 1 .$
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