Integration Questions

Question Number 85807 by jagoll last updated on 25/Mar/20

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{dx}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }} \\$$

Answered by Joel578 last updated on 25/Mar/20

$${I}\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}} \left(\mathrm{1}\:−\:{x}^{\mathrm{4}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{dx} \\$$ $$\:\:\:\:\:\:\:\:\:\:\left[{u}\:=\:{x}^{\mathrm{4}} \:\rightarrow\:{x}\:=\:{u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\rightarrow\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}}{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \:{du}\right] \\$$ $${I}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {u}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}\:−\:{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \:{du} \\$$ $$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {u}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\left(\mathrm{1}\:−\:{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{du} \\$$ $$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}{B}\left(\frac{\mathrm{3}}{\mathrm{4}},\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$

Commented byJoel578 last updated on 25/Mar/20

$${I}\:\approx\:\mathrm{0}.\mathrm{59907} \\$$

Commented byjagoll last updated on 25/Mar/20

$$\mathrm{betha}\:\mathrm{function} \\$$