We will go through the necessary trigonometric substitution and reduction steps systematically.

Step 1: Trigonometric Substitution

We recognize that the integral
$I = \int \frac{\sqrt{4 - x^2}}{x} \,dx$ suggests a trigonometric substitution based on the form $4 - x^2$. The standard substitution is:

$x = 2\sin\theta$

Step 2: Expressing Square Root and dx in Terms of $\theta$

Since $x = 2\sin\theta$, we compute:

$\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4(1 - \sin^2\theta)} = \sqrt{4\cos^2\theta} = 2\cos\theta$

The derivative is:

$dx = 2\cos\theta d\theta$

Step 3: Substituting in the Integral

Rewriting the given integral:

$I = \int \frac{2\cos\theta}{2\sin\theta} \cdot 2\cos\theta d\theta$

$= \int \frac{2\cos^2\theta}{\sin\theta} d\theta$

This is now of the form:

$k \int \sin^m\theta \cos^n\theta d\theta$

where we compare terms and find:

• $k = 2$
• $m = -1$ (since $\sin\theta$ is in the denominator)
• $n = 2$

Step 4: Another Substitution

We introduce:

$u = \sin\theta$

Then,

$du = \cos\theta d\theta$

Rewriting:

$I = 2 \int \frac{\cos^2\theta}{\sin\theta} d\theta$

Using $\cos^2\theta = 1 - \sin^2\theta = 1 - u^2$, we get:

$I = 2 \int \frac{1 - u^2}{u} du$

Step 5: Expressing as a Rational Function

$R(u) = \frac{1 - u^2}{u} = \frac{1}{u} - u$

Step 6: Partial Fraction Expansion

$\frac{1}{u} - u$ does not require further expansion.

Step 7: Integrating

$\int R(u) du = \int \frac{1}{u} du - \int u du$

$= \ln |u| - \frac{u^2}{2}$

Since $u = \sin\theta$, we substitute back:

$I = 2 \left(\ln |\sin\theta| - \frac{\sin^2\theta}{2} \right)$

Step 8: Final Expression in Terms of $x$

Since $\sin\theta = \frac{x}{2}$, we substitute:

$\ln |\sin\theta| = \ln \left|\frac{x}{2}\right|$

$\sin^2\theta = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4}$

Thus, the final result is:

$\int \frac{\sqrt{4 - x^2}}{x} dx = 2 \ln \left|\frac{x}{2}\right| - \frac{x^2}{4} + C$

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Final Answers to Blanks:

1. $x(\theta) = 2\sin\theta$
2. $\sqrt{4 - x^2} = 2\cos\theta$
3. $k = 2$
4. $m = -1$
5. $n = 2$
6. $u(\theta) = \sin\theta$
7. $R(u) = \frac{1 - u^2}{u}$
8. Partial fraction expansion: $\frac{1}{u} - u$
9. $\int R(u) du = \ln |u| - \frac{u^2}{2}$
10. $\int \frac{\sqrt{4 - x^2}}{x} dx = 2 \ln \left|\frac{x}{2}\right| - \frac{x^2}{4} + C$

Would you like any step to be explained in more detail?

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Related Questions:

1. How do we choose the correct trigonometric substitution?
2. What are other common substitutions for integrals involving square roots?
3. How do we solve similar integrals involving $\sqrt{a^2 - x^2}$?
4. How do we apply partial fractions in more complex cases?
5. What is the geometric interpretation of this integral?

Tip:
Trigonometric substitutions are useful for integrals containing $\sqrt{a^2 - x^2}$, $\sqrt{x^2 - a^2}$, and $\sqrt{x^2 + a^2}$. Identifying the right substitution makes integration much easier!