Differentiation Questions

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

$$\\$$$$\Omega\::=\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{{x}}\:\sqrt{\mathrm{1}−\sqrt{{x}}\:}\:\sqrt{\mathrm{1}−\sqrt{\mathrm{1}−\sqrt{{x}}\:}}\:{dx}=? \\$$

Answered by mindispower last updated on 09/Oct/21

$$\sqrt{\mathrm{1}−\sqrt{{x}}}={u} \\$$$$\Rightarrow{x}=\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{u}^{\mathrm{2}} \right).\mathrm{2}{u}\left(\mathrm{1}−{u}^{\mathrm{2}} \right).{u}\sqrt{\mathrm{1}−{u}\:}.{du} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{u}\left({u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\mathrm{6}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} {du}−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\mathrm{4}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} {du} \\$$$$+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\mathrm{2}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}} {du} \\$$$$=\mathrm{2}\left(\beta\left(\mathrm{6},\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{4}\beta\left(\mathrm{4},\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{2}\beta\left(\mathrm{3},\frac{\mathrm{3}}{\mathrm{2}}\right)\right) \\$$$$=\mathrm{2}\left(\frac{\Gamma\left(\mathrm{6}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{6}+\frac{\mathrm{3}}{\mathrm{2}}\right)}−\mathrm{2}\frac{\Gamma\left(\mathrm{4}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{4}+\frac{\mathrm{3}}{\mathrm{2}}\right)}+\frac{\Gamma\left(\mathrm{3}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{3}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\right) \\$$$$=\mathrm{2}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right).\frac{\Gamma\left(\mathrm{3}\right)}{\Gamma\left(\mathrm{3}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\left(\frac{\mathrm{5}.\mathrm{4}.\mathrm{3}}{\left(\mathrm{5}+\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{4}+\frac{\mathrm{3}}{\mathrm{2}}\right)\left(\mathrm{3}+\frac{\mathrm{3}}{\mathrm{2}}\right)}−\frac{\mathrm{2}.\mathrm{3}}{\left(\mathrm{3}+\frac{\mathrm{3}}{\mathrm{2}}\right)}+\mathrm{1}\right) \\$$$$\\$$

Commented by mnjuly1970 last updated on 09/Oct/21

$${grateful}\:{sir}\:{power}\:... \\$$

Commented by mindispower last updated on 09/Oct/21

$${withe}\:{pleasur}\:{sir}\: \\$$