Question Number 108776 by bobhans last updated on 19/Aug/20
$$\:\overset{{bobhans}} {\vdots} \\ $$$$\int\:\frac{{dx}}{\:\sqrt{{x}\sqrt{{x}}\:−{x}^{\mathrm{2}} }}\:=\:? \\ $$
Answered by john santu last updated on 19/Aug/20
![((⊸JS⊸)/♥) I=∫ (dx/( (√(x^(3/2) −x^2 )))) = ∫ (dx/( (√(x^(3/2) (1−x^(1/2) ))))) I= ∫(dx/(x^(3/4) (√(1−x^(1/2) )))) . [ let x^(1/4) = z ] I= ∫ ((4z^3 dz)/(z^3 (√(1−z^2 )))) = 4∫ (dz/( (√(1−z^2 )))) I= 4 arc sin (z) + c I = 4 arc sin (((x ))^(1/4) ) + c](https://www.tinkutara.com/question/Q108777.png)
$$\:\:\:\:\:\frac{\multimap\mathcal{JS}\multimap}{\heartsuit} \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{x}^{\mathrm{2}} }}\:=\:\int\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)}} \\ $$$$\mathrm{I}=\:\int\frac{\mathrm{dx}}{\mathrm{x}^{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{\mathrm{1}−\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }}\:.\:\left[\:\mathrm{let}\:\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{4}}} =\:\mathrm{z}\:\right] \\ $$$$\mathrm{I}=\:\int\:\frac{\mathrm{4z}^{\mathrm{3}} \:\mathrm{dz}}{\mathrm{z}^{\mathrm{3}} \:\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\:=\:\mathrm{4}\int\:\frac{\mathrm{dz}}{\:\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }} \\ $$$$\mathrm{I}=\:\mathrm{4}\:\mathrm{arc}\:\mathrm{sin}\:\left(\mathrm{z}\right)\:+\:\mathrm{c} \\ $$$$\mathrm{I}\:=\:\mathrm{4}\:\mathrm{arc}\:\mathrm{sin}\:\left(\sqrt[{\mathrm{4}}]{\mathrm{x}\:}\right)\:+\:\mathrm{c}\: \\ $$