Question Number 122967 by bemath last updated on 21/Nov/20
$$\:\:\int\:\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }}\:{dx}\: \\ $$
Answered by bobhans last updated on 21/Nov/20
$${let}\:{x}^{\mathrm{2}} \:=\:\mathrm{sin}\:{t}\:\Rightarrow\:\mathrm{2}{x}\:{dx}\:=\:\mathrm{cos}\:{t}\:{dt} \\ $$$$\emptyset\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{cos}\:{t}\:{dt}}{\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:{dt}\: \\ $$$$\emptyset\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\:{c}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{arcsin}\:\left({x}^{\mathrm{2}} \right)\:+\:{c}. \\ $$
Answered by TANMAY PANACEA last updated on 21/Nov/20
$${t}={x}^{\mathrm{2}} \rightarrow{dt}=\mathrm{2}{xdx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left({t}\right)+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)+{C} \\ $$