int frac{x^2}{(sqrt{4-x^2})^3} dx = | vedclass

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(B) Let $I = \int \frac{x^2}{(\sqrt{4-x^2})^3} dx$. Substitute $x = 2 \sin 	heta$,then $dx = 2 \cos 	heta d	heta$. Substituting these into the inte

Vedclass · Accepted Answer

$\frac{x}{\sqrt{4-x^2}}-	an ^{-1}(\frac{x}{\sqrt{4-x^2}})+C$

Vedclass · Answer

(B) Let $I = \int \frac{x^2}{(\sqrt{4-x^2})^3} dx$. Substitute $x = 2 \sin 	heta$,then $dx = 2 \cos 	heta d	heta$. Substituting these into the integral: $I = \int \frac{(2 \sin 	heta)^2}{(\sqrt{4-4 \sin^2 	heta})^3} (2 \cos 	heta) d	heta$ $I = \int \frac{4 \sin^2 	heta \cdot 2 \cos 	heta}{(2 \cos 	heta)^3} d	heta$ $I = \int \frac{8 \sin^2 	heta \cos 	heta}{8 \cos^3 	heta} d	heta = \int 	an^2 	heta d	heta$ $I = \int (\sec^2 	heta - 1) d	heta = 	an 	heta - 	heta + C$. Since $x = 2 \sin 	heta$,we have $\sin 	heta = \frac{x}{2}$. Then $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{x/2}{\sqrt{1-(x/2)^2}} = \frac{x}{\sqrt{4-x^2}}$. Also,$	heta = \sin^{-1}(\frac{x}{2}) = 	an^{-1}(\frac{x}{\sqrt{4-x^2}})$. Thus,$I = \frac{x}{\sqrt{4-x^2}} - 	an^{-1}(\frac{x}{\sqrt{4-x^2}}) + C$.

$\int \frac{x^2}{(\sqrt{4-x^2})^3} dx =$

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