nice-calculus-evaluate-0-1-dx-x-x-3-1-2-

Question Number 129988 by mnjuly1970 last updated on 21/Jan/21

\dots n i c e c a l c u l u s \dots

e v a l u a t e :

ϕ = \int_{0}^{1} \frac{d x}{{(x - x^{3})}^{\frac{1}{2}}} = ?

Answered by Dwaipayan Shikari last updated on 21/Jan/21

\int_{0}^{1} \frac{1}{\sqrt{x} . \sqrt{1 + x^{2}}} x^{2} = u \Rightarrow 1 = \frac{1}{2 x} . \frac{d u}{d x}

= \frac{1}{2} \int_{0}^{1} u^{- \frac{3}{4}} {(1 + u)}^{- \frac{1}{2}} {(1 - u)}^{0} d u

_{2} F_{1} (a, b; c; - 1) = \frac{Γ (c)}{Γ (c - b) Γ (b)} \int_{0}^{1} t^{b - 1} {(1 - t)}^{c - b - 1} {(1 + t)}^{- a} d t

c - b = 1 (H e r e)

a = \frac{1}{2} b = \frac{1}{4} c = \frac{5}{4}

S o t h e i n t e g r a l i s

4_{2} F_{1} (\frac{1}{2}, \frac{1}{4}; \frac{5}{4}, - 1)

Answered by Dwaipayan Shikari last updated on 21/Jan/21

\int_{0}^{1} \frac{1}{{(x - x^{3})}^{\frac{1}{2}}} d x = \frac{1}{2} \int_{0}^{1} u^{- \frac{3}{4}} {(1 - u)}^{- \frac{1}{2}} d u = \frac{Γ (\frac{1}{4}) Γ (\frac{1}{2})}{2 Γ (\frac{3}{4})}

= \frac{\sqrt{\frac{π^{3}}{2}}}{Γ^{2} (\frac{3}{4})}

Commented by mnjuly1970 last updated on 22/Jan/21

t h a n k s a l o t \dots

Answered by mathmax by abdo last updated on 21/Jan/21

Φ = \int_{0}^{1} \frac{dx}{{(x - x^{3})}^{\frac{1}{2}}} \Rightarrow Φ = \int_{0}^{1} \frac{dx}{\sqrt{x} {(1 - x^{2})}^{\frac{1}{2}}} dx =_{\sqrt{x} = t} \int_{0}^{1} \frac{2 tdt}{t {(1 - t^{4})}^{\frac{1}{2}}}

= 2 \int_{0}^{1} \frac{dt}{{(1 - t^{4})}^{\frac{1}{2}}} =_{t = u^{\frac{1}{4}}} 2 \times \frac{1}{4} \int_{0}^{1} {(1 - u)}^{- \frac{1}{2}} u^{\frac{1}{4} - 1} du

= \frac{1}{2} \int_{0}^{1} u^{\frac{1}{4} - 1} {(1 - u)}^{\frac{1}{2} - 1} du = \frac{1}{2} B (\frac{1}{4}, \frac{1}{2})

= \frac{1}{2} \frac{Γ (\frac{1}{4}) . Γ (\frac{1}{2})}{Γ (\frac{1}{4} + \frac{1}{2})} = \frac{\sqrt{π}}{2} \frac{Γ (\frac{1}{4})}{Γ (\frac{3}{4})}

Commented by mnjuly1970 last updated on 22/Jan/21

g r a t e f u l s i r m a x \dots