To evaluate the integral

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

we'll use the Fundamental Theorem of Calculus. However, notice that the limits are reversed, as $\sqrt{2}$ is less than 2. To simplify, we’ll rewrite the integral with proper limits:

$\int_{2}^{\sqrt{2}} \frac{dx}{x \sqrt{x^2 - 1}} = -\int_{\sqrt{2}}^{2} \frac{dx}{x \sqrt{x^2 - 1}}$

Step 1: Substitute

We can approach this integral by using a trigonometric substitution. Set $x = \sec(\theta)$. Then,

1. $dx = \sec(\theta) \tan(\theta) \, d\theta$,
2. $\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \sqrt{\tan^2(\theta)} = |\tan(\theta)|$.

Since $x \geq 1$, we have $\tan(\theta) \geq 0$, so $\sqrt{x^2 - 1} = \tan(\theta)$.

Step 2: Substitute into the Integral

Now the integral becomes

$-\int_{\sqrt{2}}^{2} \frac{\sec(\theta) \tan(\theta)}{\sec(\theta) \tan(\theta)} \, d\theta = -\int_{\sqrt{2}}^{2} 1 \, d\theta$

This simplifies to:

$-\int_{\theta_1}^{\theta_2} 1 \, d\theta = -(\theta_2 - \theta_1)$

Step 3: Find the Limits in Terms of $\theta$

When $x = \sqrt{2}$, $\sec(\theta) = \sqrt{2} \Rightarrow \theta = \frac{\pi}{4}$.

When $x = 2$, $\sec(\theta) = 2 \Rightarrow \theta = \frac{\pi}{3}$.

So the integral now is:

$-(\frac{\pi}{3} - \frac{\pi}{4}) = -\left(\frac{4\pi - 3\pi}{12}\right) = -\frac{\pi}{12}$

Final Answer

$\int_{2}^{\sqrt{2}} \frac{dx}{x \sqrt{x^2 - 1}} = -\frac{\pi}{12}$

Would you like more details or have any questions?

Here are some related questions to deepen understanding:

1. How does the Fundamental Theorem of Calculus apply to definite integrals?
2. What other trigonometric substitutions work well with similar integrals?
3. How can we identify the correct trigonometric substitution for a given integral?
4. Why does reversing limits affect the sign of an integral?
5. Can integrals of this form be evaluated without trigonometric substitution?

Tip: When limits are reversed in a definite integral, you can change the order of integration by factoring out a negative sign.