To find the integral of $\sqrt{x^2 - 1}$, we can use trigonometric substitution. Specifically, we can use the substitution $x = \sec(\theta)$, which implies that $dx = \sec(\theta) \tan(\theta) \, d\theta$. This substitution simplifies the integral as follows:

1. Substitute $x = \sec(\theta)$: $\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \sqrt{\tan^2(\theta)} = |\tan(\theta)|$ Since $\theta$ is in the range where $\sec(\theta) \geq 1$, $\tan(\theta) \geq 0$, and $|\tan(\theta)| = \tan(\theta)$.

2. The integral becomes: $\int \sqrt{x^2 - 1} \, dx = \int \tan(\theta) \cdot \sec(\theta) \tan(\theta) \, d\theta = \int \tan^2(\theta) \sec(\theta) \, d\theta$

3. We know that $\tan^2(\theta) = \sec^2(\theta) - 1$. So, $\int \tan^2(\theta) \sec(\theta) \, d\theta = \int (\sec^2(\theta) - 1) \sec(\theta) \, d\theta = \int \sec^3(\theta) \, d\theta - \int \sec(\theta) \, d\theta$

4. The integral of $\sec(\theta)$ is well-known: $\int \sec(\theta) \, d\theta = \ln|\sec(\theta) + \tan(\theta)| + C$

5. The integral of $\sec^3(\theta)$ requires a standard technique. One approach is integration by parts, but the result can be directly used: $\int \sec^3(\theta) \, d\theta = \frac{1}{2} (\sec(\theta) \tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|) + C$

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

7. Simplifying: $\frac{1}{2} \sec(\theta) \tan(\theta) - \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| + C$

8. Returning to the original variable $x$ where $x = \sec(\theta)$: $\sec(\theta) = x \quad \text{and} \quad \tan(\theta) = \sqrt{x^2 - 1}$

9. The final answer 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 about this solution? Here are some related questions you might find interesting:

1. How do you perform integration by parts?
2. What is the process for solving integrals using partial fractions?
3. How do you solve integrals involving exponential functions?
4. What is the method for solving trigonometric integrals?
5. How do you find the integral of $\sin^2(x)$ or $\cos^2(x)$?

Tip: When dealing with integrals that involve $\sqrt{x^2 - a^2}$, $\sqrt{a^2 - x^2}$, or $\sqrt{x^2 + a^2}$, trigonometric substitutions are often a powerful technique to simplify the integral.