x-a-b-x-dx-where-a-lt-x-lt-b-

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 by bemath 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

\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 = \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

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