int frac{x^2}{(9 - x^2)^{3/2}} , dx = | vedclass

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Question

(A) Let $x = 3\sin 	heta$,then $dx = 3\cos 	heta \, d	heta$.Substituting these into the integral:$\int \frac{9\sin^2 	heta}{(9 - 9\sin^2 	heta)^{

Vedclass · Accepted Answer

$\frac{x}{\sqrt{9 - x^2}} - \sin^{-1}\left(\frac{x}{3}\right) + c$

Vedclass · Answer

(A) Let $x = 3\sin 	heta$,then $dx = 3\cos 	heta \, d	heta$.Substituting these into the integral:$\int \frac{9\sin^2 	heta}{(9 - 9\sin^2 	heta)^{3/2}} \cdot 3\cos 	heta \, d	heta = \int \frac{27\sin^2 	heta \cos 	heta}{(9\cos^2 	heta)^{3/2}} \, d	heta$$= \int \frac{27\sin^2 	heta \cos 	heta}{27\cos^3 	heta} \, d	heta = \int 	an^2 	heta \, d	heta$$= \int (\sec^2 	heta - 1) \, d	heta = 	an 	heta - 	heta + c$Since $\sin 	heta = \frac{x}{3}$,we have $	heta = \sin^{-1}\left(\frac{x}{3}ight)$ and $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{x/3}{\sqrt{1 - (x/3)^2}} = \frac{x}{\sqrt{9 - x^2}}$.Thus,the integral is $\frac{x}{\sqrt{9 - x^2}} - \sin^{-1}\left(\frac{x}{3}ight) + c$.

$\int \frac{x^2}{(9 - x^2)^{3/2}} \, dx = $

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