Question Number 205245 by pticantor last updated on 13/Mar/24
$$\boldsymbol{{calculate}} \\ $$$$\:\:\: \\ $$$$\int\frac{\mathrm{3}\boldsymbol{{xdx}}}{\:\sqrt{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{4}} }} \\ $$$$\boldsymbol{{plsssssss}} \\ $$
Answered by mr W last updated on 13/Mar/24
$$=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}\:\:{with}\:{u}={x}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{d}\left(\mathrm{sin}\:\theta\right)}{\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta}}\:\:{with}\:{u}=\mathrm{sin}\:\theta \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{cos}\:\theta\:{d}\theta}{\:\mathrm{cos}\:\theta} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int{d}\theta \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\theta+{C} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} {u}+{C} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} {x}^{\mathrm{2}} +{C} \\ $$