Problem 46 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int(2 \cos 2 x-3 \sin 3 x) d x $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Integral

We need to find the indefinite integral of the function \( \int (2 \cos 2x - 3 \sin 3x) \, dx \). The goal is to determine an antiderivative for this expression, which means finding a function whose derivative gives us the original integrand.

Step 2: Integrate Each Term Separately

We integrate each term in \( \int (2 \cos 2x - 3 \sin 3x) \, dx \) separately. The integral of \( 2 \cos 2x \, dx \) is \( \frac{2}{2} \sin 2x = \sin 2x \), and the integral of \( -3 \sin 3x \, dx \) is \( \frac{-3}{3} \cos 3x = -\cos 3x \).

Step 3: Combine the Integrals

Combine the results from step 2 to obtain the general antiderivative: \( \sin 2x - \cos 3x \), and add the constant of integration \( C \). Thus, the antiderivative is \( \sin 2x - \cos 3x + C \).

Step 4: Verify the Result by Differentiation

Differentiate the antiderivative \( \sin 2x - \cos 3x + C \) with respect to \( x \). The derivative is \( 2 \cos 2x - 3 \sin 3x + 0 \), which matches the original integrand \( 2 \cos 2x - 3 \sin 3x \). This confirms our solution is correct.

Key Concepts

Antiderivatives

Antiderivatives are functions that reverse the process of differentiation. When you find an antiderivative of a function, you're essentially determining an original function before it was differentiated. This is also known as finding an indefinite integral. In simpler terms, if you have a function, the antiderivative tells you the original curve that could have been differentiated to result in the given function.In the problem we are discussing, the antiderivative of the function \( \int (2 \cos 2x - 3 \sin 3x) \, dx \) becomes \( \sin 2x - \cos 3x + C \). The constant \( C \) is important because when differentiating an antiderivative, any constant goes to zero, thus multiple original functions could have led to the same derivative.To find an antiderivative:

• Identify each term separately if the function consists of multiple terms.
• Use known integration formulas, such as integrating \( \cos \) or \( \sin \) functions.
• Don’t forget to add the constant of integration \( C \).

Differentiation

Differentiation is the process of finding the derivative of a function. It provides the rate at which a function changes at any given point. In the realm of calculus, differentiation and integration are inverse operations.After finding the antiderivative in our exercise, we verify its correctness by using differentiation. The derivative of the antiderivative \( \sin 2x - \cos 3x + C \) is calculated as follows:

• Find the derivative of \( \sin 2x \), which involves applying the chain rule, resulting in \( 2 \cos 2x \).
• The derivative of \(-\cos 3x\) similarly uses the chain rule, yielding \(-3 \sin 3x\).
• The constant \( C \) differentiates to zero.

Thus, the derivative \( 2 \cos 2x - 3 \sin 3x \) matches the original integrand, ensuring our solution's accuracy.

Trigonometric Functions

Trigonometric functions include sine, cosine, and tangent, among others. These functions are crucial in calculus due to their repetitive (periodic) nature, and they frequently appear in problems involving integrals and derivatives.In the given exercise using trig functions:

• We dealt with \( \cos 2x \) and \( \sin 3x \). These are basic trigonometric functions modified by coefficients within their arguments.
• It's crucial to remember how integrals and derivatives behave with these functions. For example, integrating \( \cos(ax) \) results in \( \frac{1}{a} \sin(ax) \), while differentiating \( \sin(ax) \) gives \( a \cos(ax) \).

These rules are essential as they let you accurately perform integration and differentiation, affirming that you reach the correct solutions.