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 = ∫_0 ^( 1) x^2 (1 − x^4 )^(−(1/2)) dx [u = x^4 → x = u^(1/4) → dx = (1/4)u^(−(3/4)) du] I = (1/4)∫_0 ^( 1) u^(1/2) (1 − u)^(−(1/2)) u^(−(3/4)) du = (1/4) ∫_0 ^( 1) u^(−(1/4)) (1 − u)^(−(1/2)) du = (1/4)B((3/4), (1/2))](https://www.tinkutara.com/question/Q85808.png)
$${I}\:\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}} \left(\mathrm{1}\:−\:{x}^{\mathrm{4}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}\:=\:{x}^{\mathrm{4}} \:\rightarrow\:{x}\:=\:{u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\rightarrow\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{4}}{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \:{du}\right] \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {u}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}\:−\:{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \:{du} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {u}^{−\frac{\mathrm{1}}{\mathrm{4}}} \:\left(\mathrm{1}\:−\:{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{du} \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}{B}\left(\frac{\mathrm{3}}{\mathrm{4}},\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$
Commented by Joel578 last updated on 25/Mar/20
$${I}\:\approx\:\mathrm{0}.\mathrm{59907} \\ $$
Commented by jagoll last updated on 25/Mar/20
$$\mathrm{betha}\:\mathrm{function} \\ $$