If $\int \frac{\sqrt{1-x^2}}{x^4} \mathrm{~d} x=A(x)\left(\sqrt{1-x^2}\right)^m+C$ for a suitable chosen integer $m$ and a function $A(x)$, where $C$ is a constant of integration, then $(A(x))^m$ equals

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Aicharya · Accepted Answer

$\int \frac{\sqrt{1-x^2}}{x^4} \mathrm{~d} x=A(x)\left(\sqrt{1-x^2}ight)^m+C$ Let $x=\cos 	heta d x=-\sin 	heta d 	heta$ $\begin{aligned} & \Rightarrow \int \frac{\sqrt{1-\cos ^2 	heta}}{\cos ^4 	heta} \cdot(-\sin 	heta \mathrm{d} 	heta)=-\int \sec ^4 	heta \cdot \sin ^2 	heta \mathrm{d} 	heta \ & =-\int 	an ^2 	heta \cdot \sec ^2 	heta \mathrm{d} 	heta \ & =-\frac{	an ^3 	heta}{3}+C \ & =-\frac{1}{3}\left(\frac{\sqrt{1-\cos ^2 	heta}}{\cos 	heta}ight)^3+C \ & =-\frac{1}{3}\left(\frac{\sqrt{1-x^2}}{x}ight)^3+C \ & =\frac{-1}{3 x^3} \cdot\left(\sqrt{1-x^2}ight)^3+C \ & \Rightarrow A(x)=-\frac{1}{3 x^3} 	ext { and } m=3 \end{aligned}$ $\Rightarrow\{A(x)\}^m=\left(\frac{-1}{3 x^3}ight)^3=\frac{-1}{27 x^9}$

If $\int \frac{\sqrt{1-x^2}}{x^4} \mathrm{~d} x=A(x)\left(\sqrt{1-x^2}\right)^m+C$ for a suitable chosen…

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