Question

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

Integrals
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asked 2021-05-12
Solved by the radical integration.
\(\int\frac{\sqrt{1-x^2}}{x^4}dx\)

Answers (1)

2021-05-13

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 \(u=\cot\theta\Rightarrow du=-\csc^2\theta d\theta\)
\(=-\int u^2 du\)
\(=-\frac{u^3}{3}+c\)
\(=-\frac{\cot^3\theta}{3}+c\)
\(=-\frac{(1-x^2)\sqrt{1-x^2}}{3\times3}+c\)
\(\int\frac{\sqrt{1-x^2}}{x^4}dx=-\frac{(1-x^2)\sqrt{1-x^2}}{3\times3}+c\)

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