Question Number 105755 by bemath last updated on 31/Jul/20
$$\int\:\frac{{dx}}{\:\sqrt{{x}\sqrt{{x}}\:−{x}^{\mathrm{2}} }}\:? \\ $$
Answered by john santu last updated on 31/Jul/20
$${let}\:{u}\:=\:\sqrt{{x}}\:\Rightarrow\:\int\:\frac{\mathrm{2}{u}\:{du}}{\:\sqrt{{u}^{\mathrm{3}} −{u}^{\mathrm{4}} }}\:=\: \\ $$$$\mathrm{2}\int\:\frac{{du}}{\:\sqrt{{u}−{u}^{\mathrm{2}} }}\:=\:\mathrm{2}\:\int\:\frac{{du}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$${set}\:{u}−\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{m}\: \\ $$$$=\int\frac{\mathrm{cos}\:{m}\:{dm}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} {m}}} \\ $$$$=\:\mathrm{2}\int\frac{\mathrm{cos}\:{m}\:{dm}}{\:\sqrt{\mathrm{cos}\:^{\mathrm{2}} {m}}}\:=\:\mathrm{2}{m}\:+\:{C} \\ $$$$=\:\mathrm{2}\:\mathrm{arc}\:\mathrm{sin}\:\left(\mathrm{2}{u}−\mathrm{1}\right)+\:{C} \\ $$$$=\:\mathrm{2}\:\mathrm{arc}\:\mathrm{sin}\:\left(\mathrm{2}\sqrt{{x}}\:−\mathrm{1}\right)\:+\:{C}\: \\ $$
Answered by Dwaipayan Shikari last updated on 31/Jul/20
$${x}={u}^{\mathrm{2}} \\ $$$$\int\frac{\mathrm{2}{udu}}{\:\sqrt{{u}^{\mathrm{2}} .{u}−{u}^{\mathrm{4}} }}=\int\frac{\mathrm{2}{u}}{{u}\sqrt{{u}−{u}^{\mathrm{2}} }}{du}=\mathrm{2}\int\frac{{du}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{4}}−\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$\mathrm{2}{sin}^{−\mathrm{1}} \frac{{u}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{2}{sin}^{−\mathrm{1}} \left(\mathrm{2}{u}−\mathrm{1}\right)+{C} \\ $$$$=\mathrm{2}{sin}^{−\mathrm{1}} \left(\mathrm{2}\sqrt{{x}}−\mathrm{1}\right)+{C} \\ $$