Problem 69 - Calculus Early Transcendentals

Question

Integrals of the form \(\int \sin m x \cos n x d x\) Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}(\cos ((m-n) x)-\cos ((m+n) x))\\\&\sin m x \cos n x=\frac{1}{2}(\sin ((m-n) x)+\sin ((m+n) x))\\\&\cos m x \cos n x=\frac{1}{2}(\cos ((m-n) x)+\cos ((m+n) x))\end{aligned}$$ $$\int \sin 3 x \sin 2 x d x$$

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int \sin 3x \sin 2x dx\). Answer: \(\int \sin 3x \sin 2x dx = \frac{1}{2}\sin x - \frac{1}{10}\sin 5x + C\)

Step 1: Apply the given identity to simplify the integrand

We are given the identity: $$\sin mx \sin nx =\frac{1}{2}(\cos ((m-n)x)-\cos ((m+n)x))$$ Applying this identity to \(\sin 3x \sin 2x\), let \(m = 3\) and \(n = 2\): $$\sin 3x \sin 2x = \frac{1}{2}(\cos ((3-2)x)-\cos ((3+2)x))$$ So, our integrand simplifies to: $$\sin 3x \sin 2x = \frac{1}{2}\cos x - \frac{1}{2}\cos 5x$$

Step 2: Integrate the simplified expression

Now, we have the integral: $$\int \sin 3x \sin 2x dx = \int \left(\frac{1}{2}\cos x - \frac{1}{2}\cos 5x\right) dx$$ Integrating term by term: $$\int \sin 3x \sin 2x dx = \frac{1}{2} \int \cos x dx - \frac{1}{2} \int \cos 5x dx$$

Step 3: Integrate cosine functions

Now, we will integrate the cosine functions: $$\int \cos x dx = \sin x + C_1$$ and, $$\int \cos 5x dx = \frac{1}{5}\sin 5x + C_2$$

Step 4: Substitute and combine constants

Substituting these results back into the integral, we get: $$\int \sin 3x \sin 2x dx = \frac{1}{2}(\sin x + C_1) - \frac{1}{2}\left(\frac{1}{5}\sin 5x + C_2\right)$$ Let \(C = C_1 - \frac{1}{2}C_2\). Then our final answer is: $$\int \sin 3x \sin 2x dx = \frac{1}{2}\sin x - \frac{1}{10}\sin 5x + C$$

Key Concepts

Trigonometric Identities

When it comes to integrating trigonometric functions, knowledge of trigonometric identities is invaluable. These identities are equations that relate the trigonometric functions to one another, allowing for simplification and manipulation of complex expressions.

Three particularly useful identities for integrals involving products of sine and cosine functions include:

• \( \text{sin}(mx)\text{sin}(nx) = \frac{1}{2}[\text{cos}((m-n)x) - \text{cos}((m+n)x)] \)
• \( \text{sin}(mx)\text{cos}(nx) = \frac{1}{2}[\text{sin}((m-n)x) + \text{sin}((m+n)x)] \)
• \( \text{cos}(mx)\text{cos}(nx) = \frac{1}{2}[\text{cos}((m-n)x) + \text{cos}((m+n)x)] \)

These identities are essential because they convert products into sum or differences, which are much easier to integrate. In the given problem, the identity \( \text{sin}(mx)\text{sin}(nx) \rightarrow \frac{1}{2}[\text{cos}((m-n)x) - \text{cos}((m+n)x)] \) was used to transform the integrand into a more manageable form for integration.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the antiderivatives, or integrals, of functions. Basically, it is the process of adding up infinitely small quantities to determine the whole.

The core concept involves finding the function that, when differentiated, yields the given integrand. There are various techniques used, such as substitution, integration by parts, and trigonometric substitution. One common strategy for integrating trigonometric functions is to utilize the aforementioned trigonometric identities.

In our exercise, after applying the identity, we are left with straightforward integrals of cosine functions: \( \frac{1}{2}\text{cos}(x) \text{ and } \frac{1}{2}\text{cos}(5x) \text{,} \) which are much simpler to integrate than the original expression. The integral calculus step is central to solving the problem at hand, demonstrating how different techniques in calculus can streamline complex problems.

Indefinite Integrals

Indefinite integrals represent a family of functions that are antiderivatives of a given function. They are called indefinite because there is an arbitrary constant of integration, usually denoted by C, that can take any value. Essentially, an indefinite integral gives all possible antiderivatives of a function.

In the problem, each term is integrated separately to find the indefinite integrals of \( \text{cos}(x) \text{ and } \text{cos}(5x) \text{.} \) The resulting antiderivatives, \( \text{sin}(x) \text{ and } \text{sin}(5x) \text{,} \) each come with their own constant of integration, which are later combined into a single constant C for simplicity.

When you encounter integrals of trigonometric functions, it is essential to remember that although the integrals are indefinite, any constants of integration will ultimately be combined. Correctly managing these constants is crucial to finding the correct family of antiderivative functions.