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 151894 by peter frank last updated on 24/Aug/21

$${\int }_0^{\mathrm{a}}\mathrm{x}\sqrt{\frac{{\mathrm{a}}^2-{\mathrm{x}}^2}{{\mathrm{a}}^2+{\mathrm{x}}^2}}$$

Commented by peter frank last updated on 24/Aug/21

$$\mathrm{limit}\mathrm{from}\mathrm{a}\to 0$$

Answered by Ar Brandon last updated on 24/Aug/21

$$I={\int }_0^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}dx=\frac{1}{2}{\int }_0^{a^2}\sqrt{\frac{a^2-u}{a^2+u}}du$$$$=\frac{a^2}{2}{\int }_0^1\sqrt{\frac{1-v}{1+v}}dv=\frac{a^2}{2}{\int }_0^1\frac{1-v}{\sqrt{1-v^2}}dv$$$$=\frac{a^2}{2}{[\arcsin \left(v\right)+\sqrt{1-v^2}]}_0^1=\frac{a^2}{2}(\frac{\pi }{2}-1)$$$$-a⩽x⩽a$$

Commented by peter frank last updated on 24/Aug/21

$$\mathrm{more}\mathrm{clarification}\mathrm{please}$$$$\mathrm{step}\mathrm{one}\mathrm{and}\mathrm{two}$$

Commented by Ar Brandon last updated on 24/Aug/21

$$\mathrm{Change}\mathrm{of}\mathrm{variable}u=x^2=a^{2}v$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com