To find the integral of $\sqrt{x^2 - 1}$, we can use trigonometric substitution.

Step-by-step process:

1. Substitute: We recognize that $x^2 - 1$ suggests a trigonometric identity. A good substitution is: $x = \sec(\theta) \quad \text{because} \quad \sec^2(\theta) - 1 = \tan^2(\theta)$

2. Differentiating $x = \sec(\theta)$: $dx = \sec(\theta) \tan(\theta) \, d\theta$

3. Rewrite the integrand: Now, express the square root in terms of $\theta$: $\sqrt{x^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \sqrt{\tan^2(\theta)} = \tan(\theta)$

4. Substitute into the integral: $\int \sqrt{x^2 - 1} \, dx = \int \tan(\theta) \cdot \sec(\theta) \tan(\theta) \, d\theta$ Simplifying: $= \int \sec(\theta) \tan^2(\theta) \, d\theta$

5. Use a trigonometric identity: To simplify, use the identity $\tan^2(\theta) = \sec^2(\theta) - 1$: $\int \sec(\theta) (\sec^2(\theta) - 1) \, d\theta$ $= \int (\sec^3(\theta) - \sec(\theta)) \, d\theta$

6. Integrate: The two terms can be integrated separately. The integral of $\sec^3(\theta)$ can be solved using a standard reduction formula, and the integral of $\sec(\theta)$ is a well-known result: $\int \sec(\theta) \, d\theta = \ln|\sec(\theta) + \tan(\theta)| + C$ The integral of $\sec^3(\theta)$ is: $\int \sec^3(\theta) \, d\theta = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| + C$

7. Substitute back: Finally, substitute $\theta$ back in terms of $x$ using $x = \sec(\theta)$ and $\tan(\theta) = \sqrt{x^2 - 1}$.

Thus, the integral of $\sqrt{x^2 - 1} \, dx$ becomes: $\int \sqrt{x^2 - 1} \, dx = \frac{x}{2} \sqrt{x^2 - 1} - \frac{1}{2} \ln|x + \sqrt{x^2 - 1}| + C$

Final Answer:

$\boxed{\frac{x}{2} \sqrt{x^2 - 1} - \frac{1}{2} \ln|x + \sqrt{x^2 - 1}| + C}$

Would you like a detailed explanation of any step or the use of any specific formula?

Here are some related questions you might find interesting:

1. How do trigonometric substitution methods work for integrals?
2. What is the reduction formula for $\int \sec^3(\theta) \, d\theta$?
3. What other common integrals can be solved with trigonometric substitution?
4. How do you solve integrals involving $\sqrt{a^2 - x^2}$ using substitution?
5. What is the role of absolute values in logarithmic integrals?

Tip: Always check the range of the variables when working with trigonometric substitution to ensure the correct sign for the logarithmic term.