Problem 38 - Thomas Calculus

Question

Use the table of integrals at the back of the book to evaluate the integrals. \(\int \cos \frac{\theta}{2} \cos 7 \theta d \theta\)

Step-by-Step Solution

Answer

\( \frac{1}{15} \sin \frac{15\theta}{2} + \frac{1}{13} \sin \frac{13\theta}{2} + C \).

Step 1: Identify the integral form

In this problem, we need to evaluate the integral \( \int \cos \frac{\theta}{2} \cos 7\theta \, d\theta \). Look for similar forms in the table of integrals at the back of the book. There might be a formula for the product of cosines.

Step 2: Use Trigonometric Identity

According to trigonometric identities, the product of cosines can be transformed using the formula \( \cos A \cos B = \frac{1}{2} [ \cos(A+B) + \cos(A-B) ] \). Apply this to the given integral: \( \cos \frac{\theta}{2} \cos 7\theta = \frac{1}{2} [ \cos(\frac{\theta}{2} + 7\theta) + \cos(\frac{\theta}{2} - 7\theta) ] \).

Step 3: Simplify the expression

Simplify the arguments of the cosine terms: \( \cos(\frac{\theta}{2} + 7\theta) = \cos(\frac{15\theta}{2}) \) and \( \cos(\frac{\theta}{2} - 7\theta) = \cos(-\frac{13\theta}{2}) = \cos(\frac{13\theta}{2}) \). Hence, \( \cos \frac{\theta}{2} \cos 7\theta = \frac{1}{2} [ \cos \frac{15\theta}{2} + \cos \frac{13\theta}{2} ] \).

Step 4: Split the integral

The integral becomes: \( \int \cos \frac{\theta}{2} \cos 7\theta d\theta = \frac{1}{2} \int [ \cos \frac{15\theta}{2} + \cos \frac{13\theta}{2} ] d\theta \). This can be split into two separate integrals: \( \frac{1}{2} \left( \int \cos \frac{15\theta}{2} \, d\theta + \int \cos \frac{13\theta}{2} \, d\theta \right) \).

Step 5: Integrate each term

Using the standard integral \( \int \cos k\theta \, d\theta = \frac{1}{k} \sin k\theta + C \), integrate each term:- For \( \int \cos \frac{15\theta}{2} \, d\theta \), we have \( \frac{2}{15} \sin \frac{15\theta}{2} + C_1 \).- For \( \int \cos \frac{13\theta}{2} \, d\theta \), we have \( \frac{2}{13} \sin \frac{13\theta}{2} + C_2 \).

Step 6: Combine and simplify the result

Combine the results from the integration: \( \frac{1}{2} \left( \frac{2}{15} \sin \frac{15\theta}{2} + \frac{2}{13} \sin \frac{13\theta}{2} \right) = \frac{1}{15} \sin \frac{15\theta}{2} + \frac{1}{13} \sin \frac{13\theta}{2} + C \), where \( C \) is the integration constant.

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical equations that express one trigonometric function in terms of others. These identities help simplify complex expressions and solve equations involving trigonometric functions.
The exercise used the trigonometric product-to-sum identity:

• \[\cos A \cos B = \frac{1}{2} [ \cos(A+B) + \cos(A-B) ]\]

This identity is particularly helpful in calculus when integrating products of cosines. Without such identities, directly integrating a product of cosines could be quite difficult.
By simplifying the expression with the identity, the integral becomes a sum of simpler terms that are easier to integrate. This transformation is crucial in making complex integrals manageable. Integrating these simpler trigonometric functions is typically straightforward, relying on basic integral formulas available in calculus tables.

Product of Cosines

The product of cosines, as seen in the integral \( \int \cos \frac{\theta}{2} \cos 7\theta \, d\theta \), involves multiplying two cosine functions together.
This product can often be a perfect candidate for simplification through trigonometric identities, like the product-to-sum identity. By rewriting the product of cosines as the sum of two cosines, you transform a challenging problem into a simpler one involving basic trigonometric integrals.
Here's how this process works:

• First, express the original product \( \cos \frac{\theta}{2} \cos 7\theta \) with the identity to get: \[\frac{1}{2} [ \cos(\frac{15\theta}{2}) + \cos(\frac{13\theta}{2}) ]\]
• Next, solve separate integrals for each of these cosine terms.

Thus, by breaking down the product into a sum, each integral becomes more approachable, and calculus tables provide an effective route to finding overall solutions.

Definite Integrals

Definite integrals calculate the area under a curve over a specific interval. However, the given exercise focuses on an indefinite integral, which represents a family of functions as a result of the integration process.
When solving indefinite integrals:

• You must integrate each term separately.
• Apply the rules and formulas from integral calculus to find an antiderivative.

Once each integral term is individually solved, combine them to obtain the final result. Don't forget the integration constant \( C \) since it represents a family of potential solutions.
This constant is crucial in indefinite integrals because it encompasses all the possible vertical shifts of a function – necessary because differentiating a constant results in zero, indicating no change in slope.
Understanding the nuances between definite and indefinite integrals enhances comprehension of calculus concepts, allowing for more robust problem-solving skills.