Problem 58 - University Calculus: Early Transcendentals

Question

Find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation. $$\int(2 \cos 2 x-3 \sin 3 x) d x$$

Step-by-Step Solution

Answer

The antiderivative is \(\sin 2x - \cos 3x + C\).

Step 1: Identify the Antiderivative Components

We start by identifying the components of the integral: we have two separate functions, \(2 \cos 2x\) and \(-3 \sin 3x\). The task is to find the antiderivative of each component separately.

Step 2: Find the Antiderivative of Each Component

The antiderivative of \(\cos ax\) is \(\frac{1}{a}\sin ax\). Thus, the antiderivative of \(2 \cos 2x\) is \(\frac{2}{2} \sin 2x = \sin 2x\). Similarly, the antiderivative of \(-3 \sin 3x\) is \(-\frac{3}{3} \cos 3x = -\cos 3x\).

Step 3: Combine the Antiderivatives

Combine the antiderivatives of the individual components: \( \sin 2x - \cos 3x\). Remember to add the constant of integration, \(C\), as we are finding the indefinite integral. So the most general antiderivative is \(\sin 2x - \cos 3x + C\).

Step 4: Check Your Work by Differentiating

Differentiate the result to ensure it matches the original integrand: the derivative of \(\sin 2x\) is \(2 \cos 2x\) and the derivative of \(-\cos 3x\) is \(3 \sin 3x\), so the derivative of \(\sin 2x - \cos 3x\) is \(2 \cos 2x - 3 \sin 3x\), the original integrand. This confirms that our antiderivative is correct.

Key Concepts

Antiderivative

An antiderivative, often referred to as an indefinite integral, is a function that reverses the process of differentiation. When you integrate a function, you find the original function whose derivative is the given function. This process is opposite to what we do when differentiating. For each function you integrate, you generally obtain a family of functions, which is why the result includes a constant of integration, denoted as \(C\). This reflects the idea that there are multiple functions (differing by a constant) that can be differentiated to return to the given function.

• To integrate, determine the form of the function you're dealing with, and apply the general rules or known formulas, such as those for sinusoidal functions.
• Always remember to add the constant of integration \(C\).

In our example, we identified two separate terms to integrate individually, leading to \(\sin 2x - \cos 3x + C\). This demonstrates how recognizing patterns helps in finding antiderivatives swiftly.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures the rate at which a function changes. In simple terms, a derivative provides the slope of a function at a given point.To check the accuracy of an antiderivative, we can differentiate it and see if we return to the original function (the integrand). In the example exercise, after finding the antiderivative \(\sin 2x - \cos 3x + C\), differentiating this function should yield the original function \(2 \cos 2x - 3 \sin 3x\).

• Differentiating \(\sin 2x\) gives us the function \(2 \cos 2x\).
• Differentiating \(-\cos 3x\) yields \(3 \sin 3x\).

Thus, combining these derivatives confirms our solution’s correctness. Differentiation acts as a useful tool to verify integration work, ensuring no steps in constructing the antiderivative were overlooked.

Trigonometric Functions

Trigonometric functions include sine, cosine, and other related functions. They are fundamental in calculus due to their periodic properties and easily recognizable differentiation and integration formulas.In our particular case, we dealt with integrals of \(\cos ax\) and \(-\sin ax\). These functions have straightforward integration rules:

• The integral of \(\cos ax\) is \(\frac{1}{a}\sin ax\).
• The integral of \(-\sin ax\) becomes \(-\frac{1}{a}\cos ax\).

Recognizing these forms quickly is crucial. It allows you to conserve effort and reduce mistakes, especially when preparing for exams or tackling tricky calculus problems. Remembering these integral rules and their differentiation counterparts helps in efficiently solving calculus exercises and in verifying solutions through differentiation.