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Question Number 21810 by j.masanja06@gmail.com last updated on 04/Oct/17

integrate  ∫(x^2 /(√(1−x^2 )))dx

$${integrate} \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$

Answered by sma3l2996 last updated on 04/Oct/17

x=sint⇒dx=costdt=(√(1−sin^2 t))dt  dx=(√(1−x^2 ))dt⇔dt=(dx/(√(1−x^2 )))  ∫(x^2 /(√(1−x^2 )))dx=∫sin^2 (t)dt=∫((1−cos(2t))/2)dt  =(1/2)(t−(1/2)sin(2t))+C=(1/2)(t−sin(t)cos(t))+C  ∫(x^2 /(√(1−x^2 )))dx=(1/2)(sin^(−1) (x)−x(√(1−x^2 )))+C

$${x}={sint}\Rightarrow{dx}={costdt}=\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}{dt} \\ $$$${dx}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dt}\Leftrightarrow{dt}=\frac{{dx}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=\int{sin}^{\mathrm{2}} \left({t}\right){dt}=\int\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({t}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{t}\right)\right)+{C}=\frac{\mathrm{1}}{\mathrm{2}}\left({t}−{sin}\left({t}\right){cos}\left({t}\right)\right)+{C} \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left({sin}^{−\mathrm{1}} \left({x}\right)−{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)+{C} \\ $$

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