Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$$\underset{0}{\stackrel{1}{\int }}\frac{{\mathrm{x}}^2\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^4}}$$

Answered by Joel578 last updated on 25/Mar/20

$$I={\int }_0^1{x}^2{(1-x^4)}^{-\frac{1}{2}}dx$$$$[u=x^4\to x=u^{\frac{1}{4}}\to dx=\frac{1}{4}{u}^{-\frac{3}{4}}du]$$$$I=\frac{1}{4}{\int }_0^1{u}^{\frac{1}{2}}{(1-u)}^{-\frac{1}{2}}{u}^{-\frac{3}{4}}du$$$$=\frac{1}{4}{\int }_0^1{u}^{-\frac{1}{4}}{(1-u)}^{-\frac{1}{2}}du$$$$=\frac{1}{4}B(\frac{3}{4},\frac{1}{2})$$

Commented by Joel578 last updated on 25/Mar/20

$$I\approx 0.59907$$

Commented by jagoll last updated on 25/Mar/20

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com