Problem 38 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(33-38\). $$ \int_{-\pi / 2}^{\pi / 2} \cos x \cos 7 x d x $$

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Identify the Integral and Use Trigonometric Identity

We need to evaluate the integral \( \int_{-\pi/2}^{\pi/2} \cos x \cos 7x \, dx \). We can use the product-to-sum identities for trigonometric functions, which state that \( \cos a \cos b = \frac{1}{2}(\cos(a-b) + \cos(a+b)) \). Applying this identity, we replace \( a = x \) and \( b = 7x \).

Step 2: Apply Trigonometric Identity to Rewrite the Integrand

Using the identity, we have:\[ \cos x \cos 7x = \frac{1}{2}(\cos(7x - x) + \cos(7x + x)) = \frac{1}{2}(\cos(6x) + \cos(8x)) \]So, the integral simplifies to: \[ \int_{-\pi/2}^{\pi/2} \left( \frac{1}{2} \cos(6x) + \frac{1}{2} \cos(8x) \right) dx \]

Step 3: Split the Integral into Two Parts

The integral can be separated into two parts: \[ \frac{1}{2} \int_{-\pi/2}^{\pi/2} \cos(6x) \, dx + \frac{1}{2} \int_{-\pi/2}^{\pi/2} \cos(8x) \, dx \] We will evaluate each integral separately.

Step 4: Evaluate \( \int_{-\pi/2}^{\pi/2} \cos(6x) \, dx \)

The integral \( \int \cos(6x) \, dx \) is equal to \( \frac{1}{6} \sin(6x) + C \). Evaluating the definite integral, we find:\[ \int_{-\pi/2}^{\pi/2} \cos(6x) \, dx = \left[ \frac{1}{6} \sin(6x) \right]_{-\pi/2}^{\pi/2} \]Substituting the limits of integration, we get:\[ \frac{1}{6} \sin(6(\pi/2)) - \frac{1}{6} \sin(6(-\pi/2)) = \frac{1}{6} \sin(3\pi) - \frac{1}{6} \sin(-3\pi) = 0 \]

Step 5: Evaluate \( \int_{-\pi/2}^{\pi/2} \cos(8x) \, dx \)

Similarly, we evaluate \( \int \cos(8x) \, dx \), which equals \( \frac{1}{8} \sin(8x) + C \), and compute:\[ \int_{-\pi/2}^{\pi/2} \cos(8x) \, dx = \left[ \frac{1}{8} \sin(8x) \right]_{-\pi/2}^{\pi/2} \]Using the limits of integration, we substitute:\[ \frac{1}{8} \sin(4\pi) - \frac{1}{8} \sin(-4\pi) = 0 - 0 = 0 \]

Step 6: Combine the Results of Both Integrals

Both integrals evaluate to 0. Therefore, the original integral is:\[ \frac{1}{2} \times 0 + \frac{1}{2} \times 0 = 0 \] Thus, the value of the integral \( \int_{-\pi/2}^{\pi/2} \cos x \cos 7x \, dx \) is 0.

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical equations that involve trigonometric functions like sine, cosine, and tangent. These identities hold true for all values of the involved variables. In calculus, trigonometric identities are often used to simplify integrals or solve equations.

For instance, the identity:

• \( \cos(a) \cos(b) = \frac{1}{2}(\cos(a-b) + \cos(a+b)) \)

can be extremely useful for integrating products of cosine functions. In our exercise, this identity was used to transform the product of two cosine functions into a sum of two cosine terms, which are easier to integrate. Understanding these identities helps simplify the involved mathematical expressions and make integration possible.

Product-to-Sum Formulas

Product-to-sum formulas are an invaluable tool in trigonometry, especially for solving integrals that involve trigonometric products. These formulas convert products of sines and cosines into sums or differences, making the integration process more straightforward. The general formula looks like:

• \( \cos(a) \cos(b) = \frac{1}{2}(\cos(a-b) + \cos(a+b)) \)

In the original problem, the product-to-sum formula was used to convert \( \cos(x) \cos(7x) \) into \( \frac{1}{2}(\cos(6x) + \cos(8x)) \).

By transforming the product of cosines into a sum, the integrals became easier to handle and separate into simpler parts. This substitution simplifies the integration process, as integrating sums of trigonometric functions is usually direct.

Integration Techniques

Integrating trigonometric functions often requires specific techniques to handle the expressions adequately. Once we have rewritten a complicated trigonometric expression using identities or formulas, the integration step itself involves applying integration rules.

In the given example, each resulting integral after applying the product-to-sum formulas was straightforward. We used the formula:

• \( \int \cos(kx) \, dx = \frac{1}{k} \sin(kx) + C \)

This formula helps integrate each cosine term easily. For definite integrals, after finding the antiderivative, substitute the bounds to find the value.

For both parts of the integral \( \int_{-\pi/2}^{\pi/2} \cos(6x) \, dx \) and \( \int_{-\pi/2}^{\pi/2} \cos(8x) \, dx \), their results are zero because sine of any multiple of \(\pi\) is zero. Hence, understanding these basic integration techniques is crucial for solving integrals efficiently.