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Question Number 131558 by liberty last updated on 06/Feb/21

$${\int }_0^1\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^4}}=?$$

Answered by Dwaipayan Shikari last updated on 06/Feb/21

$${\int }_0^1\frac{dx}{\sqrt{1-x^4}}{x}^4=u$$$$=\frac{1}{4}{\int }_0^1{u}^{-\frac{3}{4}}{(1-u)}^{-\frac{1}{2}}du=\frac{1}{4}.\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}=\frac{{\pi }^{\frac{3}{2}}}{2\sqrt{2}{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)}$$

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