Problem 33 - Thomas Calculus

Question

In Exercises \(29-36\) , use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$ \int \frac{d x}{x \sqrt{x^{2}-1}} $$

Step-by-Step Solution

Answer

\(\sec^{-1}(x) + C\)

Step 1: Identify the Substitution

The integral involves the expression \(x\sqrt{x^2 - 1}\). Recognize that this expression is suitable for a trigonometric substitution. First, perform a substitution: let \( x = \sec(\theta) \). This substitution is chosen because in the expression \( \sec^2(\theta) - 1 = \tan^2(\theta) \), which simplifies the square root.

Step 2: Calculate the Derivatives and Simplify

From \( x = \sec(\theta) \), we have \( dx = \sec(\theta)\tan(\theta)d\theta \). Substitute \(x = \sec(\theta)\) in the integral, it becomes: \[ \int \frac{\sec(\theta)\tan(\theta)d\theta}{\sec(\theta)\sqrt{\sec^2(\theta) - 1}}. \]Simplify the square root: \( \sqrt{\sec^2(\theta) - 1} = \tan(\theta) \). Thus, the integral simplifies to: \[ \int \frac{\sec(\theta)\tan(\theta)}{\sec(\theta)\tan(\theta)}d\theta = \int d\theta. \]

Step 3: Integrate with Respect to \(\theta\)

The integral \(\int d\theta\) is straightforward. The result is simply \( \theta + C \), where \( C \) is the integration constant.

Step 4: Back-Substitute \( \theta \) to \( x \)

We back-substitute \( \theta \) using our substitution, \( x = \sec(\theta) \). From \( x = \sec(\theta) \), it follows that \( \theta = \sec^{-1}(x) \). Hence, the solution to the integral is:\[ \sec^{-1}(x) + C. \]

Key Concepts

Definite Integral

The concept of a definite integral revolves around finding the area under a curve within a specific interval. Unlike indefinite integrals, which focus on finding a general antiderivative, definite integrals provide a precise numerical value. This numerical value is the accumulated sum of infinitely small quantities over a range, often visualized as the area under a function graph.
When dealing with definite integrals, you must evaluate the antiderivative at the upper and lower limits of integration and then find their difference. Definite integrals are essential in many real-world applications, such as calculating distances, areas, and volumes. Understanding definite integrals requires comfort with fundamental calculus concepts like limits, continuity, and antiderivatives.

Substitution Method

The substitution method is a powerful technique to simplify integrals, making them easier to solve. The key idea is to transform a complicated integral into a more manageable form using a change of variables. This often involves substituting part of the integral with a single variable, thereby simplifying the expression.
To perform a substitution:

• Choose a substitution variable (often denoted as different letters like \( u \) or \( v \)).
• Express the original variable and its differential in terms of this new variable.
• Rewrite the integral using these new terms, which often reduces complexity.
• Integrate with respect to the substitution variable.
• Finally, revert back to the original variable using your substitution formula.

The substitution method is particularly useful in trigonometric substitution, where complex algebraic expressions can be simplified using trigonometric identities. This technique is highlighted in the step-by-step solution, transforming the original problem into an elementary integral.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential when reverting from trigonometric expressions back to algebraic forms. Common inverse trigonometric functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \). They are crucial in calculus, especially in integrals involving trigonometric substitutions.In the given problem, the substitution \( x = \sec(\theta) \) is used. After simplifying and integrating, the result is expressed in terms of \( \theta \), an angle. To return to the original variable \( x \), inverse trigonometric functions are utilized. Specifically, since \( x = \sec(\theta) \), we use the inverse secant function \( \sec^{-1}(x) \) to express the solution.These functions help bridge the gap between trigonometric angles and their corresponding algebraic representations. They allow us to not only solve integrals but also to represent our solutions in terms that align with the original variables in a problem.