∫ (√((x−a)/(b−x))) dx = ? where a <x


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 115656 by bobhans last updated on 27/Sep/20

$\int \sqrt{\frac{x - a}{b - x}} d x = ?$ $w h e r e a < x < b$
Commented bybemath last updated on 27/Sep/20

$I = \int \sqrt{\frac{x - a}{- x + b}} d x$ $I = - \sqrt{(x - a) (b - x)} - (a + b) \sin^{- 1} (\sqrt{\frac{b - x}{a + b}}) + C$
	Answered by bobhans last updated on 27/Sep/20

	$l e t t i n g x = a \cos^{2} s + b \sin^{2} s$ $d x = 2 (b - a) \cos s \sin s d s$ $I = 2 (b - a) \int \sqrt{\frac{\sin^{2} s}{\cos^{2} s}} \sin s \cos s d s$ $I = 2 (b - a) \int (\frac{1}{2} - \frac{1}{2} \cos 2 s) d s$ $I = (b - a) (s - \frac{1}{2} \sin 2 s) + c$ $I = (b - a) (s - \sin s \cos s) + c$ $I = (b - a) (arc \sin (\sqrt{\frac{x - a}{b - a}}) - \sqrt{(x - a) (b - x)}) + c$
	Answered by TANMAY PANACEA last updated on 27/Sep/20
	$∫((x−a)/( (√((x−a)(b−x)))))dx (x−a)(b−x)=xb−x^2 −ab+ax =x(a+b)−x^2 −ab =−ab−{x^2 −2.x.((a+b)/2)+(((a+b)/2))^2 −(((a+b)/2))^2 } =(((a+b)/2))^2 −ab−(x−((a+b)/2))^2 =(((a−b)/2))^2 −{x−((a+b)/2)}^2 →=(((b−a)/2))^2 −{x−((a+b)/2)}^2 now.. t^2 =x(a+b)−x^2 −ab 2tdt={(a+b)−2x}dx tdt=(((a+b)/2)−x)dx ∫((−(((a+b)/2)−x−((a+b)/2))−a)/( (√((x−a)(b−x)))))dx (−1)∫((((a+b)/2)−x)/( (√((x−a)(b−x)))))dx+∫((((a+b)/2)−a)/( (√((x−a)(b−x)))))dx (−1)∫((tdt)/t)+((b−a)/2)∫(dx/( (√((((b−a)/2))^2 −{x−((a+b)/2)}^2 )))) (−t)+((b−a)/2)×sin^(−1) (((x−((a+b)/2))/((b−a)/2)))+c −1×(√((x−a)(b−x))) +((b−a)/2)sin^(−1) (((x−((a+b)/2))/((b−a)/2)))+c pls check mistake if any$
	$\int \frac{x - a}{\sqrt{(x - a) (b - x)}} d x$ $(x - a) (b - x) = x b - x^{2} - a b + a x$ $= x (a + b) - x^{2} - a b$ $= - a b - {x^{2} - 2 . x . \frac{a + b}{2} + {(\frac{a + b}{2})}^{2} - {(\frac{a + b}{2})}^{2}}$ $= {(\frac{a + b}{2})}^{2} - a b - {(x - \frac{a + b}{2})}^{2}$ $= {(\frac{a - b}{2})}^{2} - {x - \frac{a + b}{2}}^{2} \to= {(\frac{b - a}{2})}^{2} - {x - \frac{a + b}{2}}^{2}$ $n o w . .$ $t^{2} = x (a + b) - x^{2} - a b$ $2 t d t = {(a + b) - 2 x} d x$ $t d t = (\frac{a + b}{2} - x) d x$ $\int \frac{- (\frac{a + b}{2} - x - \frac{a + b}{2}) - a}{\sqrt{(x - a) (b - x)}} d x$ $(- 1) \int \frac{\frac{a + b}{2} - x}{\sqrt{(x - a) (b - x)}} d x + \int \frac{\frac{a + b}{2} - a}{\sqrt{(x - a) (b - x)}} d x$ $(- 1) \int \frac{t d t}{t} + \frac{b - a}{2} \int \frac{d x}{\sqrt{{(\frac{b - a}{2})}^{2} - {x - \frac{a + b}{2}}^{2}}}$ $(- t) + \frac{b - a}{2} \times {s i n}^{- 1} (\frac{x - \frac{a + b}{2}}{\frac{b - a}{2}}) + c$ $- 1 \times \sqrt{(x - a) (b - x)} + \frac{b - a}{2} {s i n}^{- 1} (\frac{x - \frac{a + b}{2}}{\frac{b - a}{2}}) + c$ $p l s c h e c k m i s t a k e i f a n y$
	Answered by Bird last updated on 27/Sep/20
	$I =∫ (√((x−a)/(b−x)))dx we do the ch. (√((x−a)/(b−x)))=t ⇒((x−a)/(b−x))=t^2 ⇒ x−a =t^2 b−t^2 x ⇒(1+t^2 )x=bt^2 +a ⇒x =((bt^2 +a)/(t^2 +1)) ⇒(dx/dt)=((2bt(t^2 +1)−2t(bt^2 +a))/((t^2 +1)^2 )) =((2bt−2at)/((t^2 +1)^2 )) ⇒ I =(2b−2a)∫ (t^2 /((t^2 +1)^2 ))dt =(2b−2a)∫ ((t^2 +1−1)/((t^2 +1)^2 ))dt =(2b−2a){ arctant +∫ (dt/((t^2 +1)^2 ))} but ∫ (dt/((1+t^2 )^2 )) =_(t=tamθ) ∫ (((1+tan^2 θ)dθ)/((1+tan^2 θ)^2 )) =∫ (dθ/(1+tan^2 θ)) =∫ cos^2 θ dθ =(1/2)∫(1+cos(2θ))dθ =(θ/2) +(1/4)sin(2θ) +c =(1/2)arctant +(1/4)×((2t)/(1+t^2 )) +c =(1/2)arctan((√((x−a)/(b−x))))+((√((x−a)/(b−x)))/(2(1+((x−a)/(b−x)))))+c ⇒ I =2(b−a){ (3/2)arctan(√((x−a)/(b−x))) +((√((x−a)/(b−x)))/(2(1+((x−a)/(b−x)))))}+c$
	$I = \int \sqrt{\frac{x - a}{b - x}} d x w e d o t h e c h .$ $\sqrt{\frac{x - a}{b - x}} = t \Rightarrow \frac{x - a}{b - x} = t^{2} \Rightarrow$ $x - a = t^{2} b - t^{2} x \Rightarrow (1 + t^{2}) x = {b t}^{2} + a$ $\Rightarrow x = \frac{{b t}^{2} + a}{t^{2} + 1} \Rightarrow \frac{d x}{d t} = \frac{2 b t (t^{2} + 1) - 2 t ({b t}^{2} + a)}{{(t^{2} + 1)}^{2}}$ $= \frac{2 b t - 2 a t}{{(t^{2} + 1)}^{2}} \Rightarrow$ $I = (2 b - 2 a) \int \frac{t^{2}}{{(t^{2} + 1)}^{2}} d t$ $= (2 b - 2 a) \int \frac{t^{2} + 1 - 1}{{(t^{2} + 1)}^{2}} d t$ $= (2 b - 2 a) {a r c t a n t + \int \frac{d t}{{(t^{2} + 1)}^{2}}}$ $b u t \int \frac{d t}{{(1 + t^{2})}^{2}} =_{t = t a m θ} \int \frac{(1 + {t a n}^{2} θ) d θ}{{(1 + {t a n}^{2} θ)}^{2}}$ $= \int \frac{d θ}{1 + {t a n}^{2} θ} = \int {c o s}^{2} θ d θ$ $= \frac{1}{2} \int (1 + c o s (2 θ)) d θ$ $= \frac{θ}{2} + \frac{1}{4} s i n (2 θ) + c$ $= \frac{1}{2} a r c t a n t + \frac{1}{4} \times \frac{2 t}{1 + t^{2}} + c$ $= \frac{1}{2} a r c t a n (\sqrt{\frac{x - a}{b - x}}) + \frac{\sqrt{\frac{x - a}{b - x}}}{2 (1 + \frac{x - a}{b - x})} + c \Rightarrow$ $I = 2 (b - a) {\frac{3}{2} a r c t a n \sqrt{\frac{x - a}{b - x}}$ $+ \frac{\sqrt{\frac{x - a}{b - x}}}{2 (1 + \frac{x - a}{b - x})}} + c$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum