Question Number 222607 by MrGaster last updated on 01/Jul/25
$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{\mathrm{5}−{x}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}−\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\pi \\ $$
Answered by MrGaster last updated on 01/Jul/25
![t=x^3 dx=(1/3)t^(−(2/3)) dt ∫_0 ^1 (x^3 /(5−x^3 ))∙(1/( ((1−x^3 ))^(1/3) ))dx=∫_0 ^1 (t/((5−t)(1−t)))∙(1/3)t^(−(2/3)) dt=(1/3)∫_0 ^1 t^(1/3) (1−t)^(−(1/3)) (5−t)^(−1) dt (5−t)^(−1) =(1/5)Σ_(k=0) ^∞ ((t/5))^k (1/3)∙(1/5)Σ_(k=0) ^∞ (1/5^k )∫_0 ^1 t^(k+(1/3)) (1−t)^(−(1/3)) dt=(1/(15))Σ_(k=0) ^∞ 5^(−k) B(k+(4/3),(2/3)) B(k+(4/3),(2/3))=((Γ(k+(4/3))Γ((2/3)))/(Γ(k+2))) Σ_(k=0) ^∞ ((((4/3))_k )/((2)_k ))((1/5))^k =∫_0 ^1 (1−(s/5))^(−(4/3)) ds ∫_(0 ) ^1 (1−(s/5))^(−(4/3)) ds=∫_0 ^1 5^(4/3) (5−s)^(−(4/3)) ds=5^(4/3) ∫_4 ^5 u^(−(4/3)) du=5^(4/3) [−3u^(−(1/3)) ]_4 ^5 =5^(4/3) ∙3(4^(−(1/3)) −5^(−(1/3)) ) (1/(15))∙((2π)/(3(√3)))∙5^(4/3) ∙3(4^(−(1/3)) −5^(−(1/3)) )=(1/(15))∙((2π)/(3(√3)))∙3∙5^(4/3) (4^(−(1/3)) −5^(−(1/3)) ) =(1/(15))∙((2π)/( (√3)))∙5^(1/3) (4^(−(1/3)) −5^(−(1/3)) ) =((2π)/(15(√3)))∙5^(4/3) ∙4^(−(1/3)) −((2π)/(15(√3)))∙5 =((((10))^(1/3) −2)/(3(√3)))π](https://www.tinkutara.com/question/Q222608.png)
$${t}={x}^{\mathrm{3}} \\ $$$${dx}=\frac{\mathrm{1}}{\mathrm{3}}{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{\mathrm{5}−{x}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}}{\left(\mathrm{5}−{t}\right)\left(\mathrm{1}−{t}\right)}\centerdot\frac{\mathrm{1}}{\mathrm{3}}{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} {dt}=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{t}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{5}−{t}\right)^{−\mathrm{1}} {dt} \\ $$$$\left(\mathrm{5}−{t}\right)^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{5}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{t}}{\mathrm{5}}\right)^{{k}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\centerdot\frac{\mathrm{1}}{\mathrm{5}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{5}^{{k}} }\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{k}+\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{t}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} {dt}=\frac{\mathrm{1}}{\mathrm{15}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{5}^{−{k}} {B}\left({k}+\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$${B}\left({k}+\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\Gamma\left({k}+\frac{\mathrm{4}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left({k}+\mathrm{2}\right)} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{4}}{\mathrm{3}}\right)_{{k}} }{\left(\mathrm{2}\right)_{{k}} }\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{k}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\frac{{s}}{\mathrm{5}}\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} {ds} \\ $$$$\int_{\mathrm{0}\:} ^{\mathrm{1}} \left(\mathrm{1}−\frac{{s}}{\mathrm{5}}\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} {ds}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{5}^{\frac{\mathrm{4}}{\mathrm{3}}} \left(\mathrm{5}−{s}\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} {ds}=\mathrm{5}^{\frac{\mathrm{4}}{\mathrm{3}}} \int_{\mathrm{4}} ^{\mathrm{5}} {u}^{−\frac{\mathrm{4}}{\mathrm{3}}} {du}=\mathrm{5}^{\frac{\mathrm{4}}{\mathrm{3}}} \left[−\mathrm{3}{u}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right]_{\mathrm{4}} ^{\mathrm{5}} =\mathrm{5}^{\frac{\mathrm{4}}{\mathrm{3}}} \centerdot\mathrm{3}\left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{5}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$\frac{\mathrm{1}}{\mathrm{15}}\centerdot\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\centerdot\mathrm{5}^{\frac{\mathrm{4}}{\mathrm{3}}} \centerdot\mathrm{3}\left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{5}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right)=\frac{\mathrm{1}}{\mathrm{15}}\centerdot\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\centerdot\mathrm{3}\centerdot\mathrm{5}^{\frac{\mathrm{4}}{\mathrm{3}}} \left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{5}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{15}}\centerdot\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}}\centerdot\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{5}^{−\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{15}\sqrt{\mathrm{3}}}\centerdot\mathrm{5}^{\frac{\mathrm{4}}{\mathrm{3}}} \centerdot\mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{3}}} −\frac{\mathrm{2}\pi}{\mathrm{15}\sqrt{\mathrm{3}}}\centerdot\mathrm{5} \\ $$$$=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}−\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\pi \\ $$
Commented by Tawa11 last updated on 02/Jul/25
