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Question Number 134239 by ruwedkabeh last updated on 01/Mar/21

c a n i a s k f o r s o m e h e l p ?

h o w t o p r o v e t h i s ?

\frac{1}{2} < \underset{0}{\int^{\frac{1}{2}}} \frac{d x}{\sqrt{1 - x^{3}}} < \frac{π}{6}

Commented by Dwaipayan Shikari last updated on 01/Mar/21

\frac{1}{\sqrt{1 - x^{3}}} = \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n!} x^{3 n} {(a)}_{n} = a (a + 1) (a + 2) \dots (a + n - 1)

\int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{3}}} d x = \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n! (3 n + 1)} {(\frac{1}{2})}^{3 n + 1}

= \frac{1}{6} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n}}{n! (n + \frac{1}{3})} {(\frac{1}{8})}^{n} = \frac{1}{6} \underset{n = 0}{\sum^{\infty}} \frac{{(\frac{1}{2})}_{n} {(\frac{1}{3})}_{n}}{n! {(\frac{4}{3})}_{n}} {(\frac{1}{8})}^{n}

= \frac{1}{6}_{2} F_{1} (\frac{1}{2}, \frac{1}{3}; \frac{4}{3}; \frac{1}{8})

S o \frac{1}{2} < \frac{1}{6}_{2} F_{1} (\frac{1}{2}, \frac{1}{3}; \frac{4}{3}; \frac{1}{8}) < \frac{π}{6}

3 <_{2} F_{1} (\frac{1}{2}, \frac{1}{3}; \frac{4}{3}; \frac{1}{8}) < π

Commented by ruwedkabeh last updated on 01/Mar/21

t h a n k y o u s o m u c h

Answered by mr W last updated on 01/Mar/21

f o r 0 ⩽ x ⩽ \frac{1}{2} :

0 < x^{3} < x^{2}

1 - x^{2} < 1 - x^{3} < 1

\sqrt{1 - x^{2}} < \sqrt{1 - x^{3}} < 1

1 < \frac{1}{\sqrt{1 - x^{3}}} < \frac{1}{\sqrt{1 - x^{2}}}

\int_{0}^{\frac{1}{2}} 1 d x < \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{3}}} d x < \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{2}}} d x

1 \times \frac{1}{2} < \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{3}}} d x < {[\sin^{- 1} x]}_{0}^{\frac{1}{2}}

\frac{1}{2} < \int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{1 - x^{3}}} < \frac{π}{6}

Commented by ruwedkabeh last updated on 01/Mar/21

t h a n k y o u s o m u c h