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