Problem 88 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{2 / \sqrt{3}}^{2} \frac{\cos \left(\sec ^{-1} x\right) d x}{x \sqrt{x^{2}-1}}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{12} + \frac{\sqrt{3}}{8} - \frac{1}{8} \).

Step 1: Change of Variables

Let \( \theta = \sec^{-1}(x) \). Then, \( x = \sec(\theta) \), and the differential \( dx = \sec(\theta) \tan(\theta) d\theta \). The limits of integration change accordingly: when \( x = 2/\sqrt{3} \), \( \theta = \sec^{-1}(2/\sqrt{3}) = \pi/6 \); and when \( x = 2 \), \( \theta = \sec^{-1}(2) = \pi/3 \).

Step 2: Simplify the Integral using Trigonometric Identities

Substitute \( x = \sec(\theta) \) and \( dx = \sec(\theta) \tan(\theta) d\theta \) into the integral. The integrand becomes: \( \cos(\theta) / (\sec(\theta) \sqrt{\sec^2(\theta)-1}) \). Simplify \( \sqrt{\sec^2(\theta)-1} = \sqrt{\tan^2(\theta)} = \tan(\theta) \). Substitute to get: \( \int_{\pi/6}^{\pi/3} \cos(\theta) / \sec(\theta) d\theta \).

Step 3: Evaluate the Simplified Integral

The expression simplifies to: \( \int_{\pi/6}^{\pi/3} \cos(\theta) \cos(\theta) d\theta = \int_{\pi/6}^{\pi/3} \cos^2(\theta) d\theta \). Use the identity \( \cos^2(\theta) = (1 + \cos(2\theta))/2 \) to further simplify. This gives: \[ \frac{1}{2} \int_{\pi/6}^{\pi/3} (1 + \cos(2\theta)) d\theta \].

Step 4: Perform the Integration

Integrate the expression: \[ \frac{1}{2} \left( \int_{\pi/6}^{\pi/3} 1 \,d\theta + \int_{\pi/6}^{\pi/3} \cos(2\theta) \,d\theta \right) \]. The first integral is \((\pi/3 - \pi/6) = \pi/6\). The second integral \( \int \cos(2\theta) \,d\theta = \frac{1}{2} \sin(2\theta)\). Compute it over the limits \( \pi/6 \) to \( \pi/3 \): \( \frac{1}{2}(\sin(\pi/3) - \sin(\pi/6)) = \frac{1}{2}(\sqrt{3}/2 - 1/2)\).

Step 5: Combine the Results

Combine the results from Step 4 to get: \( \frac{1}{2} \left( \frac{\pi}{6} + \frac{1}{2} (\frac{\sqrt{3}}{2} - \frac{1}{2}) \right) \). Simplify to get: \( \frac{1}{2} \left( \frac{\pi}{6} + \frac{\sqrt{3}}{4} - \frac{1}{4} \right) \). This simplifies further to: \( \frac{\pi}{12} + \frac{\sqrt{3}}{8} - \frac{1}{8} \).

Key Concepts

Integration Techniques

Integration is a cornerstone of calculus, essential for finding areas under curves and solving differential equations.
Various techniques are used to simplify the process. One significant method is substitution, where you replace variables with trigonometric or algebraic expressions to make the integral more manageable. Trigonometric substitution, in particular, leverages relationships in trigonometric functions to address integrands that involve square roots or quadratic expressions.
Another fundamental technique is integration by parts, which is particularly useful for products of functions. Selecting the right technique requires recognizing patterns in the integrand. Mastery of these methods comes from practice and understanding how and when to apply each tool effectively.

Trigonometric Substitution

Trigonometric substitution is a specific kind of substitution technique used to evaluate certain integrals, especially those involving square root expressions. The idea is to use trigonometric identities to simplify or eliminate these root expressions.
In our exercise, we start by letting \( \theta = \sec^{-1}(x) \), which transforms the variable \( x \) into a trigonometric form via \( x = \sec(\theta) \). This change of variables allows us to manipulate the integral into something more straightforward. Here are the key trigonometric identities applied:

• \( \sec^2(\theta) - 1 = \tan^2(\theta) \)
• \( \cos(\theta) = 1/\sec(\theta) \)

Such substitutions and simplifications are crucial for managing the intricacies found in integrals that do not yield easily to standard techniques.

Definite Integrals

Definite integrals compute the area under a curve from one point to another along the x-axis, providing significant geometric and physical interpretation.
A definite integral is different from an indefinite integral by having limits of integration, indicating the start and end points of the area to be calculated. The process begins with finding the antiderivative of the function, and then evaluating this antiderivative at both the upper and lower limits of integration.
The result is the net area between the curve and the x-axis. It's essential to correctly change the limits of integration when substitution is applied, as demonstrated in the exercise with new limits \( \pi/6 \) and \( \pi/3 \). Understanding how these integrals work is fundamental for analyzing continuous systems and solving real-world problems.