Problem 40 - Essential Calculus

Question

Find the integral. $$ \int \cos 4 \theta \cos (-3 \theta) d \theta $$

Step-by-Step Solution

Answer

The integration results in \(\frac{1}{14}\sin(7\theta) + \frac{1}{2}\sin(\theta) + C\).

Step 1: Apply the Product-to-Sum Identity

To simplify the integration process, first, apply the product-to-sum identity to the integrand. The identity for products of cosine functions is \(\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]\). Applying this identity gives: \[\int \cos 4 \theta \cos (-3 \theta) d \theta = \frac{1}{2} \int [\cos(4\theta - -3\theta) + \cos(4\theta + -3\theta)] d\theta\] This simplifies to: \[\frac{1}{2} \int [\cos(7\theta) + \cos(\theta)] d\theta\]

Step 2: Integration

Now that the integrand is simplified, execute the integration. As you likely know, the integral of \(\cos A\theta\) is \(\frac{1}{A} \sin A\theta\). So completing the integration gives: \[\frac{1}{2} [\frac{1}{7} \sin(7\theta) + \sin(\theta) ] + C\]

Step 3: Simplify the Result

Simplify the final result by distributing the 1/2: \[= \frac{1}{14}\sin(7\theta) + \frac{1}{2}\sin(\theta) + C\]

Key Concepts

Product-to-Sum Identity

To tackle trigonometric integrals, one key tool is the Product-to-Sum Identity. This handy identity allows us to transform the product of two cosine functions into a sum, making calculations more manageable. In our problem, we initially face the integral of \(\int \cos 4\theta \cos (-3\theta) d\theta\). This expression is a product of cosine terms.

According to the Product-to-Sum Identity, \(\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]\). By applying this identity, complexities reduce, turning the original product into a simple sum:

• \(A = 4\theta\)
• \(B = -3\theta\)

The expression reduces to \(\frac{1}{2} [\cos(7\theta) + \cos(\theta)]\) after substituting and simplifying the identity. This transformation lays the groundwork for easier integration.

Using identities like Product-to-Sum not only aids in simplifying expressions but also makes subsequent mathematical operations much simpler by transforming them into known functions.

Trigonometric Integrals

Integrating trigonometric functions is a vital skill in calculus. Recognizing standard integral forms can greatly speed up this process. Once we simplify the expression \(\cos 4\theta \cos (-3\theta)\) into \(\frac{1}{2}[\cos(7\theta) + \cos(\theta)]\) using the Product-to-Sum Identity, we can focus on solving simpler, more manageable integrals.

The key here is remembering that the integral of \(\cos A\theta\) is \(\frac{1}{A} \sin A\theta\). Thus, integrate each term separately:

• The integral of \(\cos(7\theta)\) is \(\frac{1}{7} \sin(7\theta)\).
• The integral of \(\cos(\theta)\) is \(\sin(\theta)\).

These integrals yield straightforward results that can be combined using the linearity of integration.

After integrating, remember to simplify: \(\frac{1}{2}\left[\frac{1}{7} \sin(7\theta) + \sin(\theta)\right] + C\), where \(C\) represents the constant of integration. This final expression represents the solution to our integration problem.

Definite and Indefinite Integrals

Understanding the distinction between definite and indefinite integrals is fundamental in calculus. The problem we're addressing involves an indefinite integral, as indicated by the absence of upper and lower limits in the integral notation. Indefinite integrals are used to find the antiderivative or primitive function of a given function.

When approaching indefinite integrals, always remember:

• They include a constant of integration, \(C\), since there are infinitely many antiderivatives.
• The result is a family of functions differing by a constant.

Definite integrals, on the other hand, calculate the net area under the curve between given limits and do not include a \(C\). Each type of integral serves distinct purposes in mathematics and applications, such as calculating areas, accumulated quantities, or solving differential equations.

In our exercise, after finding the antiderivative using trigonometric simplifications, the general solution is represented as:\(\frac{1}{14}\sin(7\theta) + \frac{1}{2}\sin(\theta) + C\). This expression captures the essence of solving an indefinite integral.