To solve the integral $\int \sqrt{x^2 - 1} \, dx$, we can use a trigonometric substitution.

Steps:

1. Substitute $x = \sec(\theta)$, which implies $dx = \sec(\theta) \tan(\theta) \, d\theta$.

   • This substitution works because $\sec^2(\theta) - 1 = \tan^2(\theta)$, which simplifies the integrand.

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

3. Solve the integral: We can rewrite $\tan^2(\theta)$ as $\sec^2(\theta) - 1$: $\int \sec(\theta) (\sec^2(\theta) - 1) \, d\theta = \int \sec^3(\theta) \, d\theta - \int \sec(\theta) \, d\theta$

4. Handle the two integrals:

   • For $\int \sec(\theta) \, d\theta$, the result is $\ln|\sec(\theta) + \tan(\theta)|$.
   • For $\int \sec^3(\theta) \, d\theta$, we use the reduction formula: $\int \sec^3(\theta) \, d\theta = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)|$

5. Combine results: Substituting these results back gives: $\frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| - \ln|\sec(\theta) + \tan(\theta)| + C$ Simplifying: $\frac{1}{2} \sec(\theta) \tan(\theta) - \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| + C$

6. Substitute back $\theta = \sec^{-1}(x)$:

   • $\sec(\theta) = x$
   • $\tan(\theta) = \sqrt{x^2 - 1}$

Thus, the final answer is: $\frac{1}{2} x \sqrt{x^2 - 1} - \frac{1}{2} \ln|x + \sqrt{x^2 - 1}| + C$

Would you like further details on any of the steps?

Related Questions:

1. How do you derive the reduction formula for powers of secant?
2. What are other methods to evaluate integrals involving square roots?
3. How does trigonometric substitution work for other forms like $\sqrt{a^2 - x^2}$?
4. Why is the logarithmic term necessary in integrals involving secant?
5. What are hyperbolic function substitutions, and when can they be used instead?

Tip: For integrals involving $\sqrt{x^2 - a^2}$, always consider trigonometric substitutions like $x = a \sec(\theta)$.