Problem 86 - 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 \(\tan \theta \leq 0\) ). Evaluate for \(\int \frac{\sqrt{x^{2}-1}}{x^{3}} d x,\) for \(x>1\) and for \(x<-1\)

Step-by-Step Solution

Answer

Question: Evaluate the integral of \(\frac{\sqrt{x^{2}-1}}{x^{3}}\) for \(x > 1\) and \(x < -1\). Answer: For \(x > 1\): \(\frac{\sqrt{x^{2}-1}}{x} - \sec^{-1}(x) + C1\) For \(x < -1\): \(-\frac{\sqrt{x^{2}-1}}{x} + \sec^{-1}(-x) + C2\)

Step 1: Case 1: x > 1 (x = secθ)

: For the case where \(x > 1\), we'll use the substitution \(x = \sec\theta\). First, we'll derive the differential, \(dx\), in terms of \(d\theta\). We have: $$\frac{dx}{d\theta} = \frac{d(\sec\theta)}{d\theta} = \sec\theta\tan\theta$$ So, $$dx = \sec\theta\tan\theta d\theta$$ Now, we'll substitute \(x\) and \(dx\) in the integral: $$\int \frac{\sqrt{x^{2}-1}}{x^{3}} dx = \int \frac{\sqrt{\sec^{2}\theta-1}}{(\sec\theta)^{3}}(\sec\theta\tan\theta d\theta)$$ Next, we'll simplify the integral: $$\int \frac{\sqrt{\sec^{2}\theta-1}}{(\sec\theta)^{3}}(\sec\theta\tan\theta d\theta) = \int \frac{\tan\theta}{\sec^{2}\theta}(\sec\theta\tan\theta d\theta)$$ $$= \int \tan^{2}\theta d\theta$$ Now, we'll integrate: $$\int \tan^{2}\theta d\theta = \int (\sec^{2}\theta - 1) d\theta$$ $$= \int \sec^{2}\theta d\theta - \int 1 d\theta$$ $$= \tan\theta - \theta + C$$ Finally, we'll use the substitution \(x = \sec\theta\) to express the result in terms of \(x\): $$\tan\theta - \theta + C = \frac{\sqrt{x^{2}-1}}{x} - \sec^{-1}(x) + C1$$ So, for \(x > 1\), the integral evaluates to: $$\int \frac{\sqrt{x^{2}-1}}{x^{3}} dx = \frac{\sqrt{x^{2}-1}}{x} - \sec^{-1}(x) + C1$$

Step 2: Case 2: x < -1 (x = -secθ)

: For the case where \(x < -1\), we'll use the substitution \(x = -\sec\theta\). Similarly to the previous case, we'll derive the differential, \(dx\), in terms of \(d\theta\): $$\frac{dx}{d\theta} = \frac{d(-\sec\theta)}{d\theta} = -\sec\theta\tan\theta$$ So, $$dx = -\sec\theta\tan\theta d\theta$$ We'll substitute \(x\) and \(dx\) in the integral: $$\int \frac{\sqrt{x^{2}-1}}{x^{3}} dx = \int \frac{\sqrt{(-\sec\theta)^{2}-1}}{(-\sec\theta)^{3}}(-\sec\theta\tan\theta d\theta)$$ The integral becomes: $$= \int \tan^{2}\theta d\theta$$ The integral is the same as in Case 1. Therefore, we can directly write down the result in terms of \(x\): $$\int \frac{\sqrt{x^{2}-1}}{x^{3}} dx = -\left(\frac{\sqrt{x^{2}-1}}{x} - \sec^{-1}(-x)\right) + C2$$ So, for \(x < -1\), the integral evaluates to: $$\int \frac{\sqrt{x^{2}-1}}{x^{3}} dx = -\frac{\sqrt{x^{2}-1}}{x} + \sec^{-1}(-x) + C2$$ To sum up, the integral \(\int \frac{\sqrt{x^{2}-1}}{x^{3}} dx\) evaluates to: For \(x > 1\): $$\frac{\sqrt{x^{2}-1}}{x} - \sec^{-1}(x) + C1$$ For \(x < -1\): $$-\frac{\sqrt{x^{2}-1}}{x} + \sec^{-1}(-x) + C2$$

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in calculus to simplify complicated integrals. It involves substituting a trigonometric function for a variable in the integral. This method is particularly useful for integrals that involve square roots of the form \(\sqrt{a^2-x^2}\), \(\sqrt{x^2-a^2}\), or \(\sqrt{x^2+a^2}\). In this exercise, we used the substitution \(x=a\sec\theta\) since the expression under the square root \(\sqrt{x^2-1}\) fits the \(\sec\theta\) trigonometric identity.

• The choice \(x=\sec\theta\) stems from the identity \(\sec^2\theta = 1 + \tan^2\theta\). This helps transform \(\sqrt{x^2 - 1}\) into \(\tan\theta\), making integration easier.
• The differential \(dx = \sec\theta\tan\theta d\theta\) helps convert the entire expression in terms of \(\theta\), which is easier to handle with integration.
• Deducing the correct limits of integration requires understanding the constraints on \(\theta\) imposed by the substitution \(x=a\sec\theta\).

Always remember: the key to successful trigonometric substitution is recognizing which identity simplifies the process the most. Be mindful of the domain restrictions for \(\theta\) to ensure accurate calculations.

Definite Integrals

Definite integrals are a fundamental concept in calculus, providing the area under a curve over a given interval. Unlike indefinite integrals, they yield a specific numerical value. The problem presented involves evaluating a definite integral using a trigonometric substitution.

• Definite integrals involve limits of integration which are the boundaries for the calculation. However, with trigonometric substitution, these limits might change temporarily as the integral is expressed in terms of \(\theta\). Always remember to convert back to the original variable or carefully adjust the limits.
• In our exercise, after performing trigonometric substitution, we find the integral in terms of \(\theta\) but ultimately adjust it to express the result back in terms of \(x\), ensuring continuity and correctness of limits.
• To master definite integrals, practice the art of adjusting limits of integration when substitutions are made and always verify by differentiating the result if possible.

Calculating definite integrals using such substitutions provides insight into a deeper understanding of calculus as it connects algebraic and trigonometric perspectives.

Calculus Problem Solving

Calculus problem-solving skills are built on understanding techniques such as differentiation, integration, limits, and their applications. In this exercise, we focus on solving integrals using specific techniques.

• The first approach was to assess the problem and recognize that trigonometric substitution could simplify the integration process. Recognition of patterns within the integrand is crucial.
• Perform the needed substitution: \(x = \sec\theta\) or \(x = -\sec\theta\) based on the range of \(x\) defined in the problem, ensuring that we solve within the appropriate contexts for \(x>1\) and \(x<-1\).
• Effectively solve the integral by breaking it down into manageable pieces, simplifying where possible, and then accurately reversing the substitution to present the final answer.

Developing a structured approach to complex calculus problems increases efficiency and accuracy. Engaging with problems consistently aids in mastering calculus concepts and techniques.