Question Number 134239 by ruwedkabeh last updated on 01/Mar/21
$${can}\:{i}\:{ask}\:{for}\:{some}\:{help}? \\ $$$${how}\:{to}\:{prove}\:{this}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}<\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}<\frac{\pi}{\mathrm{6}} \\ $$
Commented by Dwaipayan Shikari last updated on 01/Mar/21
$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{x}^{\mathrm{3}{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({a}\right)_{{n}} ={a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)…\left({a}+{n}−\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{3}{n}+\mathrm{1}\right)}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{3}{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left({n}+\frac{\mathrm{1}}{\mathrm{3}}\right)}\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{6}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)_{{n}} }{{n}!\left(\frac{\mathrm{4}}{\mathrm{3}}\right)_{{n}} }\left(\frac{\mathrm{1}}{\mathrm{8}}\right)^{{n}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{3}};\frac{\mathrm{4}}{\mathrm{3}};\frac{\mathrm{1}}{\mathrm{8}}\right) \\ $$$${So}\:\frac{\mathrm{1}}{\mathrm{2}}<\frac{\mathrm{1}}{\mathrm{6}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{3}};\frac{\mathrm{4}}{\mathrm{3}};\frac{\mathrm{1}}{\mathrm{8}}\right)<\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{3}<_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{3}};\frac{\mathrm{4}}{\mathrm{3}};\frac{\mathrm{1}}{\mathrm{8}}\right)<\pi \\ $$
Commented by ruwedkabeh last updated on 01/Mar/21
$${thank}\:{you}\:{so}\:{much} \\ $$
Answered by mr W last updated on 01/Mar/21
![for 0≤x≤(1/2): 0<x^3 <x^2 1−x^2 <1−x^3 <1 (√(1−x^2 ))<(√(1−x^3 ))<1 1<(1/( (√(1−x^3 ))))<(1/( (√(1−x^2 )))) ∫_0 ^(1/2) 1dx<∫_0 ^(1/2) (1/( (√(1−x^3 ))))dx<∫_0 ^(1/2) (1/( (√(1−x^2 ))))dx 1×(1/2)<∫_0 ^(1/2) (1/( (√(1−x^3 ))))dx<[sin^(−1) x]_0 ^(1/2) (1/2)<∫_0 ^(1/2) (dx/( (√(1−x^3 ))))<(π/6)](https://www.tinkutara.com/question/Q134242.png)
$${for}\:\mathrm{0}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$$\mathrm{0}<{x}^{\mathrm{3}} <{x}^{\mathrm{2}} \\ $$$$\mathrm{1}−{x}^{\mathrm{2}} <\mathrm{1}−{x}^{\mathrm{3}} <\mathrm{1} \\ $$$$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }<\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }<\mathrm{1} \\ $$$$\mathrm{1}<\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}<\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{1}{dx}<\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}<\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}<\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}<\left[\mathrm{sin}^{−\mathrm{1}} {x}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}<\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}<\frac{\pi}{\mathrm{6}} \\ $$
Commented by ruwedkabeh last updated on 01/Mar/21
$${thank}\:{you}\:{so}\:{much} \\ $$