Integration Questions

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

$$\int_{\mathrm{0}} ^{\mathrm{a}} \:\:\:\mathrm{x}\:\sqrt{\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }} \\$$

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

$$\mathrm{limit}\:\mathrm{from}\:\mathrm{a}\rightarrow\mathrm{0} \\$$

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

$${I}=\int_{\mathrm{0}} ^{{a}} {x}\sqrt{\frac{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{a}^{\mathrm{2}} } \sqrt{\frac{{a}^{\mathrm{2}} −{u}}{{a}^{\mathrm{2}} +{u}}}{du} \\$$$$\:\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−{v}}{\mathrm{1}+{v}}}{dv}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{v}}{\:\sqrt{\mathrm{1}−{v}^{\mathrm{2}} }}{dv} \\$$$$\:\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left[\mathrm{arcsin}\left({v}\right)+\sqrt{\mathrm{1}−{v}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right) \\$$$$−{a}\leqslant{x}\leqslant{a} \\$$

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

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