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

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

2. **Rewrite the integral**: Substitute \( x = \sec(\theta) \) and \( dx = \sec(\theta) \tan(\theta) \, d\theta \):
   \[
   \int \sqrt{\sec^2(\theta) - 1} \sec(\theta) \tan(\theta) \, d\theta
   \]

3. **Simplify the expression**: Since \(\sec^2(\theta) - 1 = \tan^2(\theta)\), the integral becomes:
   \[
   \int \sqrt{\tan^2(\theta)} \sec(\theta) \tan(\theta) \, d\theta
   \]

   \[
   \int |\tan(\theta)| \sec(\theta) \tan(\theta) \, d\theta
   \]

4. **Consider the absolute value**: For \(\theta\) in the range where \(\sec(\theta) \geq 0\), \(\tan(\theta) \geq 0\). Thus, \(|\tan(\theta)| = \tan(\theta)\):
   \[
   \int \tan^2(\theta) \sec(\theta) \, d\theta
   \]

5. **Use trigonometric identities**: Recall that \(\tan^2(\theta) = \sec^2(\theta) - 1\):
   \[
   \int (\sec^2(\theta) - 1) \sec(\theta) \, d\theta
   \]

6. **Distribute**:
   \[
   \int \sec^3(\theta) \, d\theta - \int \sec(\theta) \, d\theta
   \]

7. **Solve each integral**:
   - The integral \(\int \sec(\theta) \, d\theta = \ln |\sec(\theta) + \tan(\theta)| + C\).

   - The integral \(\int \sec^3(\theta) \, d\theta\) can be solved using integration by parts or a reduction formula. Using the reduction formula, we get:
     \[
     \int \sec^3(\theta) \, d\theta = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln |\sec(\theta) + \tan(\theta)| + C
     \]

8. **Combine the results**:
   \[
   \int \sqrt{x^2 - 1} \, dx = \frac{1}{2} \sec(\theta) \tan(\theta) + \frac{1}{2} \ln |\sec(\theta) + \tan(\theta)| - \ln |\sec(\theta) + \tan(\theta)| + C
   \]

   \[
   \int \sqrt{x^2 - 1} \, dx = \frac{1}{2} \sec(\theta) \tan(\theta) - \frac{1}{2} \ln |\sec(\theta) + \tan(\theta)| + C
   \]

9. **Substitute back**: Since \( x = \sec(\theta) \), we have:
   \[
   \sec(\theta) = x \quad \text{and} \quad \tan(\theta) = \sqrt{x^2 - 1}
   \]

   Therefore, the integral becomes:
   \[
   \int \sqrt{x^2 - 1} \, dx = \frac{1}{2} x \sqrt{x^2 - 1} - \frac{1}{2} \ln \left| x + \sqrt{x^2 - 1} \right| + C
   \]

So, the final answer is:
\[
\int \sqrt{x^2 - 1} \, dx = \frac{1}{2} x \sqrt{x^2 - 1} - \frac{1}{2} \ln \left| x + \sqrt{x^2 - 1} \right| + C
\]

Would you like further details on any step or have any questions?

Here are some questions you might consider asking next:

1. How do you use integration by parts?
2. What is a reduction formula for trigonometric integrals?
3. How can we handle absolute values in integrals?
4. Can you explain trigonometric substitutions in integrals?
5. What are some common integrals involving square roots?

**Tip:** When dealing with integrals involving square roots of quadratic expressions, trigonometric substitutions can simplify the integral significantly.

Explore the step-by-step solution for the integral of sqrt(x^2 - 1). Learn how to apply trigonometric substitution to solve this integral with detailed explanations and formulas. Suitable for advanced mathematics enthusiasts and students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation