# Solved by the radical integration. \int\frac{\sqrt{1-x^2}}{x^4}dx

$$\int\frac{\sqrt{1-x^2}}{x^4}dx$$

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Arham Warner

Solution:
$$\int\frac{\sqrt{1-x^2}}{x^4}dx$$
Let $$x=\sin\theta\Rightarrow dx=\cos\theta\cdot d\theta$$
$$=\int\frac{\sqrt{1-\sin^2\theta}}{\sin^4\theta}\cos\theta d\theta$$
$$=\int\frac{\cos^2\theta}{\sin^4\theta}d\theta=\int\cot^2\theta\cdot\csc^2\theta d\theta$$
Let $$u=\cot\theta\Rightarrow du=-\csc^2\theta d\theta$$
$$=-\int u^2 du$$
$$=-\frac{u^3}{3}+c$$
$$=-\frac{\cot^3\theta}{3}+c$$
$$=-\frac{(1-x^2)\sqrt{1-x^2}}{3\times3}+c$$
$$\int\frac{\sqrt{1-x^2}}{x^4}dx=-\frac{(1-x^2)\sqrt{1-x^2}}{3\times3}+c$$