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

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