Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\int_{2}^{3} \frac{d t}{t^{2} \sqrt{t^{2}-1}}\)

Step-by-Step Solution

Answer

\( \frac{4\sqrt{2} - 3\sqrt{3}}{6} \)

Step 1: Recognize the Integral Form

The integral given is \( \int \frac{1}{t^2 \sqrt{t^2 - 1}} \, dt \) from 2 to 3. Noticing the square root, this could potentially involve a trigonometric substitution like \( t = \sec(\theta) \), as it fits the form \( \sqrt{t^2 - 1} = \sqrt{\sec^2(\theta) - 1} = \tan(\theta) \).

Step 2: Apply Trigonometric Substitution

Substitute \( t = \sec(\theta) \), which implies \( dt = \sec(\theta)\tan(\theta)\,d\theta \). This transforms the integral into: \[ \int \frac{\sec(\theta)\tan(\theta)}{\sec^2(\theta)\tan(\theta)} \, d\theta = \int \frac{1}{\sec(\theta)} \, d\theta = \int \cos(\theta) \, d\theta \].

Step 3: Determine the New Limits of Integration

Initially, \( t = 2 \rightarrow \sec(\theta) = 2 \) hence \( \theta = \sec^{-1}(2) \). For \( t = 3 \rightarrow \sec(\theta) = 3 \), so \( \theta = \sec^{-1}(3) \). Thus, the limits change from 2 and 3 to \( \sec^{-1}(2) \) and \( \sec^{-1}(3) \) respectively.

Step 4: Integrate with respect to \( \theta \)

The integral \( \int \cos(\theta) \, d\theta \) can be evaluated as \( \sin(\theta) \). Hence, evaluate \( \sin(\theta) \) at the new limits: \[ \left[ \sin(\theta) \right]_{\sec^{-1}(2)}^{\sec^{-1}(3)} = \sin(\sec^{-1}(3)) - \sin(\sec^{-1}(2)) \].

Step 5: Express Trigonometric Functions in Terms of \( t \)

For \( \theta = \sec^{-1}(t) \), \( \cos(\theta) = \frac{1}{t} \) and \( \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \frac{1}{t^2} \). Thus, \( \sin(\theta) = \sqrt{1 - \frac{1}{t^2}} \). Applying this:\( \sin(\sec^{-1}(3)) = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \) and \( \sin(\sec^{-1}(2)) = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \).

Step 6: Calculate and Simplify the Result

Substitute the values from Step 5 to find the difference: \[ \frac{2\sqrt{2}}{3} - \frac{\sqrt{3}}{2} \].Find a common denominator (6):\[ \frac{4\sqrt{2}}{6} - \frac{3\sqrt{3}}{6} = \frac{4\sqrt{2} - 3\sqrt{3}}{6} \] as the final answer.

Key Concepts

Understanding Definite Integrals

A definite integral is a way of calculating the accumulation of quantities, much like finding the area under a curve. In mathematics, this is represented as a sum of an infinite number of infinitesimally small quantities. The notation for a definite integral usually includes upper and lower limits on the integral sign, denoting the interval over which the function is integrated. For instance, in our exercise, the definite integral is denoted as \[ \int_{2}^{3} \frac{dt}{t^2 \sqrt{t^2 - 1}} \]This tells us that we need to calculate the integral of the function from 2 to 3. These limits provide a boundary for the integration, resulting in a single number that represents the area or accumulated quantity between those two points.
Definite integrals are essential in calculus because they help solve problems related to area, volume, and physical concepts like work and energy. Understanding them is key to mastering the evaluation of accumulated values over an interval.

The Role of Trigonometric Functions in Integration

Trigonometric functions play a crucial role in solving integrals that involve square roots. When dealing with integrands involving expressions like \( \sqrt{t^2 - 1} \), a trigonometric substitution is often used to simplify the integral into a more manageable form.
In our case, we used the substitution \( t = \sec(\theta) \), which simplifies \( \sqrt{t^2 - 1} \) to \( \tan(\theta) \). This substitution transforms the integral's variables from \( t \) to \( \theta \), converting the original function into something simpler:

• The change in variable introduces a new expression in terms of \( \theta \).
• This variable change allows us to integrate trigonometric functions using standard results like \( \int \cos(\theta) \, d\theta = \sin(\theta) \).

This technique is part of a broader class called 'trigonometric substitution', which exploits the identities from trigonometry to make integration feasible. These substitutions are integral to handling integrals that include square roots and other complex forms.

Adjusting Integral Limits for Substitution

Another important aspect when using trigonometric substitution is adjusting the integral limits to correspond to the new variable. In definite integrals, changing the variable also changes the limits of integration. Originally, our limits were from 2 to 3 when integrated with respect to \( t \).
Upon substituting \( t = \sec(\theta) \), we calculated the new limits by determining what values of \( \theta \) correspond to \( t = 2 \) and \( t = 3 \). This results in:

• \( \theta = \sec^{-1}(2) \) when \( t = 2 \)
• \( \theta = \sec^{-1}(3) \) when \( t = 3 \)

Thus, the integration interval changes from [2, 3] to \([\sec^{-1}(2), \sec^{-1}(3)]\). It's essential to make this adjustment, as it ensures that the computation of the definite integral remains accurate, even with a change in variables. This conversion maintains the bounds consistent with the integral's initial setup, allowing us to compute the exact area or quantity being measured over the original interval in the transformed space.