To evaluate the integral $\int \sqrt{x^2 - 1} \, dx$, we can use trigonometric substitution. Here are the steps:

1. Substitute: Let $x = \sec(\theta)$, hence $dx = \sec(\theta) \tan(\theta) \, d\theta$.

2. Simplify the Integrand: $\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \sqrt{\tan^2(\theta)} = |\tan(\theta)|$ Since $\sec(\theta) \geq 1$, $\theta$ ranges from $0$ to $\pi/2$ or from $\pi$ to $3\pi/2$, and $\tan(\theta) \geq 0$, we can write $|\tan(\theta)| = \tan(\theta)$.

3. Rewrite the Integral: $\int \sqrt{x^2 - 1} \, dx = \int \tan(\theta) \sec(\theta) \tan(\theta) \, d\theta = \int \tan^2(\theta) \sec(\theta) \, d\theta$

4. Simplify: Use the identity $\tan^2(\theta) = \sec^2(\theta) - 1$: $\int \tan^2(\theta) \sec(\theta) \, d\theta = \int (\sec^2(\theta) - 1) \sec(\theta) \, d\theta = \int (\sec^3(\theta) - \sec(\theta)) \, d\theta$

5. Integrate: The integral can be split into two parts: $\int \sec^3(\theta) \, d\theta - \int \sec(\theta) \, d\theta$

   The integrals are: $\int \sec(\theta) \, d\theta = \ln |\sec(\theta) + \tan(\theta)| + C_1$ $\int \sec^3(\theta) \, d\theta = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln |\sec(\theta) + \tan(\theta)| + C_2$

   Combining these: $\int \sec^3(\theta) \, d\theta - \int \sec(\theta) \, d\theta = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln |\sec(\theta) + \tan(\theta)| - \ln |\sec(\theta) + \tan(\theta)| + C$ $= \frac{1}{2} \sec(\theta) \tan(\theta) - \frac{1}{2} \ln |\sec(\theta) + \tan(\theta)| + C$

6. Substitute Back: Recall that $x = \sec(\theta)$, so $\sec(\theta) = x$ and $\tan(\theta) = \sqrt{x^2 - 1}$: $\int \sqrt{x^2 - 1} \, dx = \frac{1}{2} x \sqrt{x^2 - 1} - \frac{1}{2} \ln |x + \sqrt{x^2 - 1}| + C$

Therefore, the final result is: $\int \sqrt{x^2 - 1} \, dx = \frac{1}{2} x \sqrt{x^2 - 1} - \frac{1}{2} \ln |x + \sqrt{x^2 - 1}| + C$

Would you like more details or have any questions?

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Tip: For integrals involving square roots of quadratic expressions, trigonometric substitutions are often very useful.