Problem 70 - 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. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}(\cos ((m-n) x)-\cos ((m+n) x))\\\&\sin m x \cos n x=\frac{1}{2}(\sin ((m-n) x)+\sin ((m+n) x))\\\&\cos m x \cos n x=\frac{1}{2}(\cos ((m-n) x)+\cos ((m+n) x))\end{aligned}$$ $$\int \cos x \cos 2 x d x$$

Step-by-Step Solution

Answer

Based on the solution above, explain in your own words how the integral of $$\int \cos x \cos 2x dx$$ can be solved. First, we use the identity for \(\cos mx \cos nx\), where \(m = 1\) and \(n = 2\), to simplify the integral. This allows us to rewrite the integral as $$\int \frac{1}{2} (\cos (-x) + \cos (3x)) dx$$. Then, we separate the integral into two parts and factor out the constant \(\frac{1}{2}\). Next, we integrate each part individually. We know that the antiderivative of \(\cos x\) is \(\sin x\), so we can easily find the antiderivative of both parts. Our result is $$\frac{1}{2} (\sin (-x) + \frac{1}{3} \sin (3x)) + C$$, which is the final answer.

Step 1: Use the given identity to simplify the integral

Using the identity for \(\cos mx \cos nx\), where \(m = 1\) and \(n = 2\), we have: $$\int \cos x \cos 2x dx = \int \frac{1}{2} (\cos (x - 2x) + \cos (x + 2x)) dx = \int \frac{1}{2} (\cos (-x) + \cos (3x)) dx$$

Step 2: Separate the integral into two parts

We can separate the integral into two parts and factor out the constant \(\frac{1}{2}\): $$\int \frac{1}{2} (\cos (-x) + \cos (3x)) dx = \frac{1}{2} \int (\cos (-x) + \cos (3x)) dx$$

Step 3: Integrate each part

Now, we can integrate the two parts: $$\frac{1}{2} \int (\cos (-x) + \cos (3x)) dx = \frac{1}{2} \left(\int \cos (-x) dx + \int \cos (3x) dx \right)$$ Using the fact that \(\cos (-x) = \cos(x)\), and knowing the antiderivative of \(\cos x\) is \(\sin x\): $$= \frac{1}{2} (\sin (-x) + \frac{1}{3} \sin (3x)) + C$$

Step 4: Simplify the result

Finally, we simplify the result and arrive at the final answer: $$\frac{1}{2} \int \cos x \cos 2x dx = \frac{1}{2} (\sin (-x) + \frac{1}{3} \sin (3x)) + C$$

Key Concepts

Trigonometric Identities

Trigonometric identities are mathematical equations that relate various trigonometric functions to one another. They are essential tools in calculus, particularly when dealing with integrals and derivatives of trigonometric functions. Trigonometric identities can simplify complex expressions, making them easier to integrate or differentiate. For example, when integrating a product of sine and cosine functions, such as in the integral \(\int \cos x \cos 2x dx\), we can utilize the identity \(\cos m x \cos n x = \frac{1}{2}(\cos((m-n)x) + \cos((m+n)x))\) to express the integral in terms of a sum of cosines. This leads to a much simpler integral that is easier to solve. Learning these identities is a powerful step toward mastering integral calculus, as it expands the toolbox you can use to tackle a wide range of problems.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the antiderivatives of functions. In essence, it deals with the accumulation of quantities, such as areas under curves. When we calculate the integral of a function, we're finding the total amount that function accumulates over a certain interval. This requires understanding how to reverse the process of differentiation, leading us to the antiderivative or indefinite integral of a function.

For instance, in the context of trigonometric functions, integral calculus allows us to find the area under a curve described by a function like \(\cos x\) over an interval. By applying techniques like the one illustrated in the exercise where we integrate \(\int \cos x \cos 2x dx\), we can determine the antiderivative and, if needed, calculate the definite integral over specific limits.

Integration by Substitution

Integration by substitution, also known as u-substitution, is a method used in integral calculus to simplify integrals by changing the variable of integration. This technique is analog to the chain rule in differentiation and can transform a complicated integral into a simpler form that is easier to solve.

Although it was not utilized directly in the original exercise, understanding this approach is crucial when dealing with more complex integrals. For example, if you encounter a problematic portion within an integral, you can substitute it with a new variable u, find du by differentiating your substitution, and then replace the original complicated part with a simpler u-expression that can be integrated more conveniently. This strategy is especially handy when integrating functions involving compositions of trigonometric and polynomial expressions.

Antiderivatives of Trigonometric Functions

Antiderivatives are the inverse operation to differentiation. When we take a trigonometric function and find its antiderivative, we're looking for another function that, when differentiated, gives us the original trigonometric function. For example, the antiderivative of \(\cos x\) is \(\sin x\), because when you differentiate \(\sin x\), you get \(\cos x\).

In our exercise, we needed to find the antiderivatives for \(\cos x\) and \(\cos 3x\). Knowing that these are \(\sin x\) and \(\sin 3x\), respectively, allowed us to solve the integral. It's important for students to be familiar with the basic antiderivatives of trigonometric functions, as they frequently appear in various integrals. By memorizing these fundamental antiderivatives, students can streamline the integration process and solve problems more efficiently.