Problem 8 - Calculus Early Transcendentals

Question

Evaluate \(\int \cos a x d x\) and \(\int \sin a x d x,\) where \(a\) is a constant.

Step-by-Step Solution

Answer

Question: Evaluate the integrals \(\int \cos a x d x\) and \(\int \sin a x d x\), where \(a\) is a constant. Answer: \(\int \cos a x d x = \frac{\sin(ax)}{a} + C\) and \(\int \sin a x d x = -\frac{\cos(ax)}{a} + C\), where \(C\) is the integration constant.

Step 1: 1. Identify the Function

In this case, we have \(f(x) = \cos a x\) and \(a\) is a constant.

Step 2: 2. Find the Antiderivative

To find the antiderivative of \(f(x) = \cos a x\), we can use integration by substitution. Let \(u = ax\), then \(du/dx = a\). Therefore, \(dx = du/a\). Now, substitute these variables in the integral: \(\int \cos a x d x = \int \cos u\left(\frac{du}{a}\right) = \frac{1}{a} \int \cos(u) du\). Now, we know the antiderivative of \(\cos(u)\) is \(\sin(u)\), thus, \(\frac{1}{a} \int \cos(u) du = \frac{\sin u}{a}\). Finally, substitute \(u\) back to its original form: \(\frac{\sin u}{a} = \frac{\sin(ax)}{a} + C\), where \(C\) is the integration constant. So, the result of the integral \(\int \cos a x d x\) is \(\frac{\sin(ax)}{a} + C\). For \(\int \sin a x d x\):

Step 3: 1. Identify the Function

In this case, we have \(f(x) = \sin a x\) and \(a\) is a constant.

Step 4: 2. Find the Antiderivative

To find the antiderivative of \(f(x) = \sin a x\), we can use integration by substitution. Let \(u = ax\), then \(du/dx = a\). Therefore, \(dx = du/a\). Now, substitute these variables in the integral: \(\int \sin a x d x = \int \sin u\left(\frac{du}{a}\right) = \frac{1}{a} \int \sin(u) du\). Now, we know the antiderivative of \(\sin(u)\) is \(-\cos(u)\), thus, \(\frac{1}{a} \int \sin(u) du = -\frac{\cos u}{a}\). Finally, substitute \(u\) back to its original form: \(- \frac{\cos u}{a} = - \frac{\cos(ax)}{a} + C\), where \(C\) is the integration constant. So, the result of the integral \(\int \sin a x d x\) is \(-\frac{\cos(ax)}{a} + C\).

Key Concepts

Antiderivative

In calculus, finding the antiderivative, also known as indefinite integration, is the reverse process of differentiation. Think of it as finding a function whose derivative is the given function. When dealing with integrals like \( \int \cos ax \, dx \) and \( \int \sin ax \, dx \), you're essentially finding a function whose derivative is \( \cos(ax) \) or \( \sin(ax) \), respectively.

An antiderivative of a function \( f(x) \) is a function \( F(x) \) such that the derivative of \( F(x) \) is \( f(x) \). It is often expressed with an integration constant \( C \), because integration introduces a constant of integration that accounts for all possible antiderivatives. For example:

• The antiderivative of \( \cos(ax) \) is \( \frac{1}{a} \sin(ax) + C \), because the derivative of \( \sin(ax) \) is \( a\cos(ax) \).
• Similarly, the antiderivative of \( \sin(ax) \) is \( -\frac{1}{a} \cos(ax) + C \), because the derivative of \( -\cos(ax) \) is \( a\sin(ax) \).

This technique is extremely useful in solving differential equations and understanding the area under curves.

Integration by Substitution

Integration by substitution is a crucial method in calculus used to simplify the process of finding integrals. It's similar to the reverse of the chain rule in differentiation. The main idea is to substitute a part of the integral with a new variable that makes the problem easier to handle.

For instance, when you have integrals like \( \int \cos ax \, dx \) or \( \int \sin ax \, dx \), substituting \( u = ax \) simplifies the integration:

• With \( du = a \, dx \), you rearrange to \( dx = \frac{du}{a} \). This substitution transforms the integral into one with respect to \( u \), making it easier to solve.
• For \( \int \cos(ax) \, dx \), substitute and get \( \frac{1}{a} \int \cos(u) \, du \).
• For \( \int \sin(ax) \, dx \), a similar substitution leads to \( \frac{1}{a} \int \sin(u) \, du \).

After making the substitution and finding the antiderivative in terms of \( u \), you convert back to the original variable. This method vastly simplifies complex integrals, especially those involving composite functions.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in mathematics and appear frequently in calculus problems. They describe the relationships between angles and sides of triangles as well as oscillatory phenomena, like waves.

The integral of trigonometric functions often involves recognizing identities and using substitution methods to simplify them:

• When integrating \( \cos(ax) \), the antiderivative is \( \frac{1}{a} \sin(ax) \), because differentiation of \( \sin(ax) \) yields \( a \cos(ax) \).
• When integrating \( \sin(ax) \), the antiderivative is \( -\frac{1}{a} \cos(ax) \), as differentiating \( -\cos(ax) \) gives \( a \sin(ax) \).

Understanding these functions and their properties helps solve a wide range of problems. They serve both pure and applied math, finding uses in physics, engineering, and beyond. Memorizing the integrals of these basic trigonometric functions can significantly speed up solving calculus problems.