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Question Number 208335 by Shrodinger last updated on 12/Jun/24

$${\int }_{-1}^1\sqrt{1-t^4}dt$$

Answered by Frix last updated on 12/Jun/24

$$\underset{-1}{\stackrel{1}{\int }}\sqrt{1-t^4}dt=2\underset{0}{\stackrel{1}{\int }}\sqrt{1-t^4}dt\stackrel{u={\sin }^{-1}{t}^2}{=}$$$$=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}{\left(\sin u\right)}^{-\frac{1}{2}}{\left(\cos u\right)}^{2}du=$$$$=\frac{1}{2}B(\frac{1}{4},\frac{3}{2})=\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{2\mathrm{\Gamma }\left(\frac{7}{4}\right)}=\frac{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{5}{4}\right)}{\mathrm{\Gamma }\left(\frac{7}{4}\right)}=$$$$=\frac{1}{3\sqrt{2\pi }}{\left(\mathrm{\Gamma }\left(\frac{1}{4}\right)\right)}^2\approx 1.74803837$$

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