Problem 13 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \sec \frac{t}{3} d t $$

Step-by-Step Solution

Answer

\( 3 \ln \left| \sec \frac{t}{3} + \tan \frac{t}{3} \right| + C \)

Step 1: Choose a substitution

To evaluate \( \int \sec \frac{t}{3} \, dt \), let's use the substitution \( u = \frac{t}{3} \). This means that \( du = \frac{1}{3} \, dt \), or equivalently \( dt = 3 \, du \).

Step 2: Change the variable of integration

Substitute \( u \) and \( du \) in the integral. The integral becomes \( \int \sec u \, (3 \, du) \). Simplify this to \( 3 \int \sec u \, du \).

Step 3: Evaluate the standard integral

The integral \( \int \sec u \, du \) is a standard integral with the result \( \ln |\sec u + \tan u| + C \). Thus, \( 3 \int \sec u \, du = 3 \ln |\sec u + \tan u| + C \).

Step 4: Back-substitute the original variable

Replace \( u \) with \( \frac{t}{3} \) to express the integral in terms of the original variable. The integral evaluates to \( 3 \ln \left| \sec \frac{t}{3} + \tan \frac{t}{3} \right| + C \).

Key Concepts

Substitution Method

The substitution method is a popular technique in integral calculus used to simplify the integration process. It's like the reverse chain rule for derivatives. By changing the variable of integration, we can transform a complex integral into a simpler form. Here's how it generally works:

• Identify a part of the integrand that can be substituted with a simpler variable, often denoted as "u".
• Express the new variable "u" in terms of the original variable. For our exercise, we substitute with \( u = \frac{t}{3} \).
• Compute the derivative of your substitution, \( du \), and express the original differential (in this case, \( dt \)) in terms of \( du \). From \( u = \frac{t}{3} \), we get \( du = \frac{1}{3} \, dt \) or \( dt = 3 \, du \).
• Replace all instances of the original variable and differential in the integral with their corresponding expressions in terms of \( u \) and \( du \). This simplification often results in a standard integral form, which is easier to solve.

After substitution, integrating becomes a straightforward task much like solving a puzzle with fewer pieces. However, after integrating the simpler expression, don't forget to back-substitute the original variable.

Standard Integral Forms

Standard integral forms are the essential building blocks of calculus. They are well-recognized integrals whose solutions are known and commonly used as shortcuts. In our problem, after substitution, the integral was transformed into \( \int \sec u \, du \), a standard form.These standard integrals have known results that are derived from fundamental calculus concepts. Some common examples include:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for \(n eq -1\)
• \( \int e^x \, dx = e^x + C \)
• \( \int \sec u \, du = \ln |\sec u + \tan u| + C \)

A good grasp of these forms is crucial as it allows us to quickly identify and compute integrals without having to derive each result from scratch. They are like pieces of a mathematical toolkit that facilitate solving complex problems.

Trigonometric Integrals

Trigonometric integrals, as the name suggests, involve the integration of trigonometric functions. These can sometimes be complex due to the cyclic nature of trig functions. Nevertheless, understanding them is essential for tackling problems involving periodic phenomena.When evaluating these integrals, some useful strategies include:

• Simplifying the integral using trigonometric identities, such as \( \sin^2 x + \cos^2 x = 1 \).
• Using substitution, as demonstrated in the original exercise, to transform the integral into a standard form.
• Remembering the standard integral results for basic trigonometric functions. For example, \( \int \sec x \, dx \) transforms using the identity \( \int \sec x \, dx = \ln |\sec x + \tan x| + C \).

These methods help break down even the most intricate expressions into simpler parts, making it easier to integrate. With practice, solving trigonometric integrals becomes an intuitive process.