Paul Marsh
11/17/2023 · Elementary School
Use trigonometric substitution to evaluate \[ \begin{array}{l}I=\frac{\sqrt{9-x^{2}}}{x^{2}} d x \\ \text { A. }-\frac{\sqrt{9-x^{2}}}{x}+3 \sin ^{-1}\left(\frac{x}{3}\right)+C \\ \text { B. None of the options } \\ \text { C. } \frac{\sqrt{9-x^{2}}}{x}+\sin ^{-1}\left(\frac{x}{3}\right)+C \\ \text { D. }-\frac{\sqrt{9-x^{2}}}{x}-\sin ^{-1}\left(\frac{x}{3}\right)+C \\ \text { O. }-\frac{\sqrt{0-x^{2}}}{x^{2}}-\sin ^{-1}\left(\frac{x}{3}\right)+C\end{array} \]
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Step-by-step Solution
为了求解积分 \( I = \frac{\sqrt{9-x^2}}{x^2} \, dx \),我们可以使用三角代换法。具体步骤如下:
1. **选择合适的三角代换**:
我们注意到被积函数中有 \( \sqrt{9-x^2} \),这提示我们可以使用三角代换 \( x = 3 \sin \theta \)。这样,\( \sqrt{9-x^2} \) 可以表示为 \( \sqrt{9 - (3 \sin \theta)^2} = \sqrt{9 - 9 \sin^2 \theta} = 3 \cos \theta \)。
2. **计算微分**:
对 \( x = 3 \sin \theta \) 求导,得到 \( dx = 3 \cos \theta \, d\theta \)。
3. **代入原积分**:
将 \( x = 3 \sin \theta \) 和 \( dx = 3 \cos \theta \, d\theta \) 代入原积分,得到:
\[
I = \frac{\sqrt{9-x^2}}{x^2} \, dx = \frac{3 \cos \theta}{(3 \sin \theta)^2} \cdot 3 \cos \theta \, d\theta = \frac{3 \cos \theta}{9 \sin^2 \theta} \cdot 3 \cos \theta \, d\theta = \frac{\cos^2 \theta}{3 \sin^2 \theta} \, d\theta
\]
4. **简化积分**:
利用三角恒等式 \( \cos^2 \theta = 1 - \sin^2 \theta \),我们可以进一步简化积分:
\[
I = \frac{1 - \sin^2 \theta}{3 \sin^2 \theta} \, d\theta = \frac{1}{3} \left( \frac{1}{\sin^2 \theta} - \sin^2 \theta \right) \, d\theta
\]
利用三角恒等式 \( \csc^2 \theta = \frac{1}{\sin^2 \theta} \),得到:
\[
I = \frac{1}{3} \left( \csc^2 \theta - \sin^2 \theta \right) \, d\theta
\]
5. **计算积分**:
分别对 \( \csc^2 \theta \) 和 \( \sin^2 \theta \) 进行积分:
\[
\int \csc^2 \theta \, d\theta = -\cot \theta
\]
\[
\int \sin^2 \theta \, d\theta = \frac{1}{2} \theta - \frac{1}{4} \sin 2\theta
\]
因此,积分结果为:
\[
I = \frac{1}{3} \left( -\cot \theta - \left( \frac{1}{2} \theta - \frac{1}{4} \sin 2\theta \right) \right) + C
\]
6. **回代**:
由于 \( x = 3 \sin \theta \),所以 \( \theta = \sin^{-1} \left( \frac{x}{3} \right) \),并且 \( \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{9-x^2}}{x} \)。
因此,积分结果为:
\[
I = -\frac{\sqrt{9-x^2}}{x} + \frac{1}{3} \left( \sin^{-1} \left( \frac{x}{3} \right) - \frac{1}{2} \sin^{-1} \left( \frac{x}{3} \right) \right) + C
\]
\[
I = -\frac{\sqrt{9-x^2}}{x} + \frac{1}{3} \sin^{-1} \left( \frac{x}{3} \right) + C
\]
综上所述,正确答案是:
\[
\boxed{-\frac{\sqrt{9-x^2}}{x} + 3 \sin^{-1} \left( \frac{x}{3} \right) + C}
\]
即选项 A。
Quick Answer
The correct answer is A. \(-\frac{\sqrt{9-x^2}}{x} + 3 \sin^{-1} \left( \frac{x}{3} \right) + C\).
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