find ∫_0 ^1 (dx/((√x) +(√(1−x))))


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Question Number 42435 by maxmathsup by imad last updated on 25/Aug/18

$f i n d \int_{0}^{1} \frac{d x}{\sqrt{x} + \sqrt{1 - x}}$
Commented by maxmathsup by imad last updated on 25/Aug/18

$l e t I = \int_{0}^{1} \frac{d x}{\sqrt{x} + \sqrt{1 - x}} c h a n g e m e n t x = {s i n}^{2} t g i v e$ $I = \int_{0}^{\frac{π}{2}} \frac{2 s i n t c o s t d t}{s i n t + c o s t} = \int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} t + {c o s}^{2} t + 2 s i n t c o s t - 1}{s i n t + c o s t} d t$ $= \int_{0}^{\frac{π}{2}} \frac{{(s i n t + c o s t)}^{2} - 1}{s i n t + c o s t} d t = \int_{0}^{\frac{π}{2}} (s i n t + c o s t) d t - \int_{0}^{\frac{π}{2}} \frac{d t}{s i n t + c o s t}$ $= {[s i n t - c o s t]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{d t}{s i n t + c o s t}$ $= 2 - \int_{0}^{\frac{π}{2}} \frac{d t}{s i n t + c o s t} c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e$ $\int_{0}^{\frac{π}{2}} \frac{d t}{s i n t + c o s t} = \int_{0}^{1} \frac{1}{\frac{2 u}{1 + u^{2}} + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = 2 \int_{0}^{1} \frac{d u}{2 u + 1 - u^{2}}$ $= - 2 \int_{0}^{1} \frac{d u}{u^{2} - 2 u - 1} = - 2 \int_{0}^{1} \frac{d u}{{(u - 1)}^{2} - 1} =_{u - 1 = - α} - 2 \int_{1}^{0} \frac{- d α}{α^{2} - 1}$ $= - 2 \int_{0}^{1} \frac{d α}{α^{2} - 1} = - \int_{0}^{1} (\frac{1}{α - 1} - \frac{1}{α + 1}) = - [l n ∣ \frac{α - 1}{α + 1} ∣_{0}^{1} = + \infty s o t h i s$ $i n t e g r a l d i v e r g e s! . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18
	$I= x=sin^2 α dx=2sinαcosα ∫_0 ^(Π/2) ((2sinαcosα)/(sinα+cosα))dα ∫_0 ^(Π/2) ((sin^2 α+cos^2 α+2sinαcosα−1)/(sinα+cosα))dα ∫_0 ^(Π/2) sinα+cosα dα−∫_0 ^(Π/2) (dα/(sinα+cosα)) ∫_0 ^(Π/2) sinα+cosα dα−∫_0 ^(Π/2) ((sec^2 (α/2))/(2tan(α/2)+1−tan^2 (α/2)))dα ∫_0 ^(Π/2) sinα+cosα dα+2∫_0 ^(Π/2) ((d(tan(α/2)))/(tan^2 (α/2)−2tan(α/2)+1−2)) ∫_0 ^(Π/2) sinα+cosα dα+2∫_0 ^(Π/2) ((d(tan(α/2)))/((tan(α/2)−1)^2 −1)) ∫_0 ^(Π/2) sinα+cosα dα−2∫_0 ^(Π/2) ((d(tan(α/2)))/(1−(tan(α/2)−1)^2 )) ∫_0 ^(Π/2) sinα+cosα dα−2∫_0 ^(Π/2) ((d(tan(α/2) −1))/((tan(α/2)−1)^2 −1))→∫(dx/(x^2 −a^2 )) =∣−cosα+sinα∣_0 ^(Π/2) −2×(1/2)ln∣((tan(α/2)−1−1)/(tan(α/2)−1+1))∣_0 ^(Π/2) ={(−0+1)−(−1+0)}−{ln∣((1−1−1)/(1−1+1))∣−ln∣((−2)/0)∣} =2−{ln∣−1∣−ln∣−∞∣}$
	$I =$ $x = {s i n}^{2} α d x = 2 s i n α c o s α$ $\int_{0}^{\frac{Π}{2}} \frac{2 s i n α c o s α}{s i n α + c o s α} d α$ $\int_{0}^{\frac{Π}{2}} \frac{{s i n}^{2} α + {c o s}^{2} α + 2 s i n α c o s α - 1}{s i n α + c o s α} d α$ $\int_{0}^{\frac{Π}{2}} s i n α + c o s α d α - \int_{0}^{\frac{Π}{2}} \frac{d α}{s i n α + c o s α}$ $\int_{0}^{\frac{Π}{2}} s i n α + c o s α d α - \int_{0}^{\frac{Π}{2}} \frac{{s e c}^{2} \frac{α}{2}}{2 t a n \frac{α}{2} + 1 - {t a n}^{2} \frac{α}{2}} d α$ $\int_{0}^{\frac{Π}{2}} s i n α + c o s α d α + 2 \int_{0}^{\frac{Π}{2}} \frac{d (t a n \frac{α}{2})}{{t a n}^{2} \frac{α}{2} - 2 t a n \frac{α}{2} + 1 - 2}$ $\int_{0}^{\frac{Π}{2}} s i n α + c o s α d α + 2 \int_{0}^{\frac{Π}{2}} \frac{d (t a n \frac{α}{2})}{{(t a n \frac{α}{2} - 1)}^{2} - 1}$ $\int_{0}^{\frac{Π}{2}} s i n α + c o s α d α - 2 \int_{0}^{\frac{Π}{2}} \frac{d (t a n \frac{α}{2})}{1 - {(t a n \frac{α}{2} - 1)}^{2}}$ $\int_{0}^{\frac{Π}{2}} s i n α + c o s α d α - 2 \int_{0}^{\frac{Π}{2}} \frac{d (t a n \frac{α}{2} - 1)}{{(t a n \frac{α}{2} - 1)}^{2} - 1} \to \int \frac{d x}{x^{2} - a^{2}}$ $=∣ - c o s α + s i n α ∣_{0}^{\frac{Π}{2}} - 2 \times \frac{1}{2} l n ∣ \frac{t a n \frac{α}{2} - 1 - 1}{t a n \frac{α}{2} - 1 + 1} ∣_{0}^{\frac{Π}{2}}$ $= {(- 0 + 1) - (- 1 + 0)} - {l n ∣ \frac{1 - 1 - 1}{1 - 1 + 1} ∣ - l n ∣ \frac{- 2}{0} ∣}$ $= 2 - {l n ∣ - 1 ∣ - l n ∣ - \infty ∣}$
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