Question

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

Short answer

Based on the step-by-step solution, provide a short answer to the problem: The integral can be evaluated as follows: $$\int \frac{d x}{x\sqrt{x^{2}-1}} = \left\\{\begin{array}{ll} \sec ^{-1} x+C & \text { if } x > 1 \\\ -\sec ^{-1} x+C & \text { if } x < -1 \end{array}\right.$$

Step-by-step solution

Substitution

Let \(x = \sec \theta\), then we have \(dx = \sec \theta \tan \theta d\theta\). Now, substitute in the integral and simplify: $$\int \frac{d x}{x\sqrt{x^{2}-1}} = \int\frac{\sec \theta \tan \theta d\theta}{\sec \theta \sqrt{\sec^2 \theta - 1}}$$

Simplification

Simplify the expression inside the integral: $$\int\frac{\sec \theta \tan \theta d\theta}{\sec \theta \sqrt{\sec^2 \theta - 1}} = \int\frac{\tan \theta d\theta}{\sqrt{\sec^2 \theta - 1}}$$ Use the identity \(\tan ^{2} \theta = \sec ^{2}\theta - 1\) to further simplify the integral: $$= \int\frac{\tan \theta d\theta}{\sqrt{\tan^2 \theta}}$$ Now, we have two cases: 1. \(x > 1\): In this case, \(0 \leq \theta < \frac{\pi}{2}\) and \(\tan \theta \geq 0\). Thus, the integral becomes: $$\int\frac{\tan \theta d\theta}{\tan \theta} = \int d\theta$$ 2. \(x < -1\): In this case, \(\frac{\pi}{2} < \theta \leq \pi\) and \(\tan \theta \leq 0\). Thus, the integral becomes: $$\int\frac{\tan \theta d\theta}{-\tan \theta} = -\int d\theta$$

Evaluate the Integrals

Now, evaluate the integrals for both cases: 1. \(x > 1\): $$\int d\theta = \theta + C = \sec^{-1} x + C$$ 2. \(x < -1\): $$-\int d\theta = -\theta + C = -\sec^{-1} x + C$$

Final Answer

Combine the answers from both cases: $$\int \frac{d x}{x\sqrt{x^{2}-1}} = \left\\{\begin{array}{ll} \sec ^{-1} x+C & \text { if } x > 1 \\\ -\sec ^{-1} x+C & \text { if } x < -1 \end{array}\right.$$

Key concepts

Secant Substitution

Secant substitution is a powerful technique often used to simplify integrals that involve expressions resembling the form \(\sqrt{x^2 - a^2}\) or \(\sqrt{a^2 - x^2}\). By substituting \(x = a \sec \theta\), we harness the trigonometrical identity \( \sec^2 \theta - 1 = \tan^2 \theta \), which makes the radical expression easier to handle.

Here's how it works:

• This substitution is particularly useful when dealing with expressions involving \(\sqrt{x^2 - 1}\). By setting \(x = \sec \theta\) and calculating the derivative \(dx = \sec \theta \tan \theta d\theta\), the complex expression simplifies substantially.
• The substitution rephrases the problem using trigonometric identities, allowing for easier integration.

In our original exercise, this method demonstrates how using \(x = \sec \theta\) condenses the integral into a manageable expression by recognizing \(\sqrt{\sec^2 \theta - 1} = \tan \theta\). This deftly replaces the root with a simpler trigonometric term, setting the stage for straightforward integration. It's a classic calculus maneuver that showcases the interplay between algebraic and trigonometric functions.

Inverse Trigonometric Functions

Inverse trigonometric functions offer a way to express angles whose trigonometric ratios are known. They "undo" the trigonometric function, providing an angle given a ratio. In calculus, they also offer solutions to certain integrals that do not easily fit a straightforward antiderivative approach.

The integral in the original problem involved \(\sec^{-1} x\), which is the inverse of the secant function. This function helps express angles in terms of the secant ratio instead of the familiar sine or cosine functions. When considering the inverse secant, it's important to note a few key details:

• \(\sec^{-1} x\) is defined for \(x \leq -1\) or \(x \geq 1\), reflecting the behavior of the secant function outside the interval where it traditionally flips back on itself (between \(-1\) and \(1\)).
• Its range spans from \([0, \pi/2) \cup (\pi/2, \pi]\). This helps dictate which part of the unit circle corresponds to particular values of \(x\).

Inverse trigonometric functions simplify integration by transforming integrals into more familiar forms, often facilitating the extraction of the antiderivative needed for a problem like the one in the exercise.

Trigonometric Identities

Trigonometric identities are foundational tools in mathematics that describe relationships between trigonometric functions. They allow various trigonometric expressions to be rewritten in a simpler or different form without changing their value. These identities are crucial when performing operations such as differentiating or integrating expressions involving trigonometric functions.

Let's discuss some important identities relevant to our original problem:

• The Pythagorean Identity: \(\tan^2 \theta = \sec^2 \theta - 1\). This identity was key in transforming the integral \(\sqrt{\sec^2 \theta - 1} = \tan \theta\), thus simplifying the expression under the integral sign.
• Using such identities often converts an otherwise complex problem into a straightforward integral calculation.
• These identities are not only helpful for simplification but also when attempting to solve equations involving trigonometric terms or when establishing additional relationships between different angles and their functions.

In summary, the power of trigonometric identities lies in their ability to bridge gaps between different mathematical expressions, enabling us to tackle complex integrals and equations with greater ease.