solve-0-1-dx-x-1-x-

Question Number 118566 by benjo_mathlover last updated on 18/Oct/20

s o l v e \underset{0}{\int^{1}} \frac{d x}{\sqrt{x} \sqrt{1 - x}} .

Commented by benjo_mathlover last updated on 18/Oct/20

i n o t h e r w a y b y s u b s t i t u t e t r i g o n o m e t r i

l e t t i n g x = \sin^{2} θ \to d x = 2 \sin θ \cos θ d θ

f o r x = 0 \to θ = 0, x = 1 \to θ = \frac{π}{2}

t h e n \sqrt{x} = ∣ \sin θ ∣ = \sin θ a n d \sqrt{1 - x} = ∣ \cos θ ∣ = \cos θ

s o t h e i n t e g r a l b e c o m e s

\int_{0}^{π / 2} \frac{2 \sin θ \cos θ}{\sin θ \cos θ} d θ = 2 \int_{0}^{π / 2} d θ = 2 θ ∣_{0}^{\frac{π}{2}} = π

Commented by Dwaipayan Shikari last updated on 18/Oct/20

\int_{0}^{1} {(x)}^{- \frac{1}{2}} {(1 - x)}^{- \frac{1}{2}} d x

= \int_{0}^{1} {(x)}^{\frac{1}{2} - 1} {(1 - x)}^{\frac{1}{2} - 1} d x

= β (\frac{1}{2}, \frac{1}{2}) = \frac{Γ (\frac{1}{2}) . Γ (\frac{1}{2})}{Γ (1)} = \frac{\sqrt{π} . \sqrt{π}}{1} = π ⊛

A n o t h e r w a y ★

Answered by TANMAY PANACEA last updated on 18/Oct/20

t^{2} = 1 - x

\int_{1}^{0} \frac{- 2 t d t}{\sqrt{1 - t^{2}} \times t}

2 \int_{0}^{1} \frac{d t}{\sqrt{1 - t^{2}}} \to 2 ∣ {s i n}^{- 1} t ∣_{0}^{1} = 2 \times \frac{π}{2} = π

Answered by Dwaipayan Shikari last updated on 18/Oct/20

\int_{0}^{1} \frac{d x}{\sqrt{x} . \sqrt{1 - x}} 1 - x = t^{2} \Rightarrow - 1 = 2 t \frac{d t}{d x}

- \int_{1}^{0} \frac{2 t d t}{t \sqrt{1 - t^{2}}} = 2 \int_{0}^{1} \frac{1}{\sqrt{1 - t^{2}}} d t

= π

Answered by MJS_new last updated on 18/Oct/20

\underset{0}{\int^{1}} \frac{d x}{\sqrt{x} \sqrt{1 - x}} =

[t = \frac{\sqrt{1 - x}}{\sqrt{x}} \to d x = - 2 \sqrt{x^{3}} \sqrt{1 - x} d t]

= - 2 \underset{+ \infty}{\int^{0}} \frac{d t}{t^{2} + 1} = 2 \underset{0}{\int^{+ \infty}} \frac{d t}{t^{2} + 1} =

= 2 {[\arctan t]}_{0}^{+ \infty} = π

Answered by 1549442205PVT last updated on 18/Oct/20

Put \frac{1}{x} - 1 = t^{2} \Rightarrow 2 tdt = - \frac{1}{x^{2}} dx

t = \frac{\sqrt{1 - x}}{\sqrt{x}}, dx = - \frac{2 t}{{(t^{2} + 1)}^{2}} dt, x = \frac{1}{t^{2} + 1}

\underset{0}{\int^{1}} \frac{d x}{\sqrt{x} \sqrt{1 - x}} = - \int_{\infty}^{0} \frac{t^{2} + 1}{t} \times \frac{2 tdt}{{(t^{2} + 1)}^{2}}

= 2 \int_{0}^{\infty} \frac{dt}{t^{2} + 1} = {2 \tan}^{- 1} (t) = π

Commented by MJS_new last updated on 18/Oct/20

is it just coincidence or do you repost my

answers for a special reason ?

Commented by MJS_new last updated on 18/Oct/20

sorry I asked . you posted many good answers .

Answered by Bird last updated on 18/Oct/20

I = \int_{0}^{1} \frac{d x}{\sqrt{x} \sqrt{1 - x}} w e d o t h e c h a n g .

\sqrt{x} = t \Rightarrow I = \int_{0}^{1} \frac{2 t d t}{t \sqrt{1 - t^{2}}}

= 2 \int_{0}^{1} \frac{d t}{\sqrt{1 - t^{2}}} = 2 {[a r c s i n t]}_{0}^{1}

= 2 \times \frac{π}{2} = π

Answered by mindispower last updated on 18/Oct/20

= 2 \int_{0}^{1} . \frac{1}{2 \sqrt{x}} . \frac{1}{\sqrt{1 - {(\sqrt{x})}^{2}}} = 2 {[a r c s i n (\sqrt{x})]}_{0}^{1} = 2 . \frac{π}{2} = 1 π