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

$$caniaskforsomehelp?$$ $$howtoprovethis?$$ $$\frac{1}{2}<\underset{0}{\stackrel{\frac{1}{2}}{\int }}\frac{dx}{\sqrt{1-x^3}}<\frac{\pi }{6}$$

Commented byDwaipayan Shikari last updated on 01/Mar/21

$$\frac{1}{\sqrt{1-x^3}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{x}^{3n}{\left(a\right)}_n=a(a+1)(a+2)...(a+n-1)$$ $${\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^3}}dx=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!(3n+1)}{\left(\frac{1}{2}\right)}^{3n+1}$$ $$=\frac{1}{6}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!(n+\frac{1}{3})}{\left(\frac{1}{8}\right)}^n=\frac{1}{6}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n{\left(\frac{1}{3}\right)}_n}{n!{\left(\frac{4}{3}\right)}_n}{\left(\frac{1}{8}\right)}^n$$ $$=\frac{1}{6}{}_2{F}_1(\frac{1}{2},\frac{1}{3};\frac{4}{3};\frac{1}{8})$$ $$So\frac{1}{2}<\frac{1}{6}{}_2{F}_1(\frac{1}{2},\frac{1}{3};\frac{4}{3};\frac{1}{8})<\frac{\pi }{6}$$ $$3{<}_2{F}_1(\frac{1}{2},\frac{1}{3};\frac{4}{3};\frac{1}{8})<\pi$$

Commented byruwedkabeh last updated on 01/Mar/21

$$thankyousomuch$$

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

$$for0⩽x⩽\frac{1}{2}:$$ $$0<x^3<x^2$$ $$1-x^2<1-x^3<1$$ $$\sqrt{1-x^2}<\sqrt{1-x^3}<1$$ $$1<\frac{1}{\sqrt{1-x^3}}<\frac{1}{\sqrt{1-x^2}}$$ $${\int }_0^{\frac{1}{2}}1dx<{\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^3}}dx<{\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}dx$$ $$1\times \frac{1}{2}<{\int }_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^3}}dx<{\left[{\sin }^{-1}x\right]}_0^{\frac{1}{2}}$$ $$\frac{1}{2}<{\int }_0^{\frac{1}{2}}\frac{dx}{\sqrt{1-x^3}}<\frac{\pi }{6}$$

Commented byruwedkabeh last updated on 01/Mar/21

$$thankyousomuch$$

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