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evaluate-1-0-dx-1-x-3-1-3-




Question Number 126344 by mnjuly1970 last updated on 19/Dec/20
          evaluate ::   ∫_(−1) ^(  0) (dx/( ((1+x^3 ))^(1/3) )) =?
$$\:\:\:\:\: \\ $$$$\:\:\:{evaluate}\:::\:\:\:\int_{−\mathrm{1}} ^{\:\:\mathrm{0}} \frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{x}^{\mathrm{3}} }}\:=? \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 19/Dec/20
∫_0 ^1 (dt/( ((1−t^3 ))^(1/3) ))        x=−t  ⇒(1/3)∫_0 ^1 u^(−(2/3)) (1−u)^(−(1/3)) du       t^3 =u    =(1/3)Γ((1/3))Γ((2/3))=((2π)/(3(√3)))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−{t}^{\mathrm{3}} }}\:\:\:\:\:\:\:\:{x}=−{t} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} {du}\:\:\:\:\:\:\:{t}^{\mathrm{3}} ={u}\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$
Commented by mnjuly1970 last updated on 19/Dec/20
thanks alot mr payan...
$${thanks}\:{alot}\:{mr}\:{payan}… \\ $$

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