Problem 85 - Calculus Early Transcendentals

Question

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0)\) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). $$\begin{aligned} &\text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}=\\\ &\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{aligned}$$

Step-by-Step Solution

Answer

In conclusion, by applying the substitution rule, solving the integral in terms of θ, and converting the solution back into terms of x, we have shown that the solution to the given integral takes the given forms for the two different intervals of x. For x > 1, the integral is equal to \(\tan^{-1}\sqrt{x^2-1} + C\), and for x < -1, the integral is equal to \(-\tan^{-1}\sqrt{x^2-1} + C\).

Step 1: Apply the Substitution Rule

Let us apply the substitution \(x=a\sec\theta\), where \(a=1\) (since the integral does not include 'a'). By this substitution, we get \(x=\sec\theta\). To find \(\frac{dx}{d\theta}\), we differentiate \(x\) with respect to \(\theta\): \(\frac{dx}{d\theta}=\frac{d(\sec\theta)}{d\theta}=\sec\theta\tan\theta\) Now, we substitute \(x=\sec\theta\) and \(dx=\sec\theta\tan\theta d\theta\) in the integral. So, $$\begin{aligned} \int \frac{dx}{x\sqrt{x^2-1}}=\int\frac{\sec\theta\tan\theta d\theta}{\sec\theta\sqrt{\sec^2\theta-1}} \end{aligned}$$

Step 2: Simplify the Integral in terms of \(\theta\)

Now, let's simplify the integral and solve it: $$\begin{aligned} \int\frac{\sec\theta\tan\theta d\theta}{\sec\theta\sqrt{\sec^2\theta-1}}&=\int\frac{\tan\theta d\theta}{\sqrt{\sec^2\theta-1}}\\ &=\int\frac{\sin\theta\cos\theta d\theta}{\cos\theta\sin\theta} \\ &=\int d\theta \end{aligned}$$ Now, let's find the integral of both sides: $$\begin{aligned} \int d\theta &= \theta + C \end{aligned}$$

Step 3: Convert back to terms of \(x\) and Verify the given solution

Now we need to convert this solution back into terms of \(x\). We have \(x=\sec\theta\), so \(\theta=\sec^{-1}x\). Thus, $$\begin{aligned} \int \frac{dx}{x\sqrt{x^2-1}} &= \sec^{-1}x + C \end{aligned}$$ Now let's split the solution based on the interval of \(x\): 1. If \(x > 1\): $$\theta = \sec^{-1}x = \tan^{-1}\sqrt{x^2-1}$$ So, the integral becomes: $$\int \frac{dx}{x\sqrt{x^2-1}} = \tan^{-1}\sqrt{x^2-1} + C$$ 2. If \(x < -1\): $$\theta = \sec^{-1}x = -\tan^{-1}\sqrt{x^2-1}$$ So, the integral becomes: $$\begin{aligned} \int \frac{dx}{x\sqrt{x^2-1}} &= -\tan^{-1}\sqrt{x^2-1} + C \end{aligned}$$ So, we have proved that the solution to the given integral matches the two forms given for the different intervals of \(x\).

Key Concepts

Integral Calculus

Integral calculus is a fundamental concept in mathematics that focuses on finding the antiderivative or the integral of a function. This is important for calculating areas under curves or solving problems involving accumulation. In the context of this problem, integral calculus is used to find the antiderivative of the function \(\frac{1}{x \sqrt{x^2 - 1}}\).
To solve integrals, techniques such as substitution are often employed. The substitution method, like the one used here, involves replacing variables with trigonometric functions to simplify the integration process. This particular problem uses the trigonometric substitution \(x = a \sec \theta\), where \(a = 1\).
The goal is to simplify the original integral so it becomes easier to solve using known integral forms or fundamental integration techniques.

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the action of the standard trigonometric functions. They are essential when solving integrals involving trigonometric identities. In this exercise, the inverse trigonometric functions \(\sec^{-1} x\) and \(\tan^{-1} x\) are highlighted.
During the substitution process, you often end up with inverse trigonometric forms as part of the antiderivative. For example, after transforming the integral with \(x = \sec \theta\), the simplification leads to terms that are solved through inverse tangent, \(\tan^{-1}\), and inverse secant, \(\sec^{-1}\).
These functions provide solutions to trigonometric equations where the values of certain angles need to be recovered, playing a key role in converting back from the substitution variable \(\theta\) to the original variable \(x\).

Definite Integrals

Definite integrals are used to calculate the net area between a function and the x-axis over a specified interval. While this exercise focused on an indefinite integral, understanding definite integrals is a logical progression. In definite integrals, limits of integration are set, transforming the antiderivative results into an actual number representative of accumulated area.
Consider the following properties of definite integrals:

• They provide a signed area, where the area below the x-axis is considered negative.
• In applications, they are used for calculating work, loads, probabilities, and other physical quantities.
• The Fundamental Theorem of Calculus ties together the concept of differentiation and integration, providing a systematic way to evaluate definite integrals.

Understanding definite integrals enhances the comprehension of the area function and establishes a deeper connection between the physical interpretation of integral calculus and real-world applications.