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Question Number 156206 by mnjuly1970 last updated on 09/Oct/21

$$$$$$\mathrm{\Omega }:={\int }_0^1\sqrt{x}\sqrt{1-\sqrt{x}}\sqrt{1-\sqrt{1-\sqrt{x}}}dx=?$$

Answered by mindispower last updated on 09/Oct/21

$$\sqrt{1-\sqrt{x}}=u$$$$\Rightarrow x={(1-u^2)}^2$$$$={\int }_0^1(1-u^2).2u(1-u^2).u\sqrt{1-u}.du$$$$={\int }_0^{1}2u(u^4-2u^2+1){(1-u)}^{\frac{1}{2}}$$$$=2{\int }_0^1{u}^{6-1}{(1-u)}^{\frac{3}{2}-1}du-4{\int }_0^1{u}^{4-1}{(1-u)}^{\frac{3}{2}-1}du$$$$+2{\int }_0^1{u}^{2-1}{(1-u)}^{\frac{3}{2}-1}du$$$$=2(\beta (6,\frac{3}{2})-4\beta (4,\frac{3}{2})+2\beta (3,\frac{3}{2}))$$$$=2(\frac{\mathrm{\Gamma }\left(6\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(6+\frac{3}{2})}-2\frac{\mathrm{\Gamma }\left(4\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(4+\frac{3}{2})}+\frac{\mathrm{\Gamma }\left(3\right)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(3+\frac{3}{2})})$$$$=2\mathrm{\Gamma }\left(\frac{3}{2}\right).\frac{\mathrm{\Gamma }\left(3\right)}{\mathrm{\Gamma }(3+\frac{3}{2})}(\frac{5.4.3}{(5+\frac{3}{2})(4+\frac{3}{2})(3+\frac{3}{2})}-\frac{2.3}{(3+\frac{3}{2})}+1)$$$$$$

Commented by mnjuly1970 last updated on 09/Oct/21

$$gratefulsirpower...$$

Commented by mindispower last updated on 09/Oct/21

$$withepleasursir$$

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