Problem 48 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int \frac{1-\cos 6 t}{2} d t $$

Step-by-Step Solution

Answer

The antiderivative is \( \frac{t}{2} - \frac{1}{12}\sin(6t) + C \).

Step 1: Identify the Integral

We have been given the integral \( \int \frac{1 - \cos 6t}{2} \, dt \). We need to find its most general antiderivative.

Step 2: Split the Integral

Rewrite the integral to simplify it: \( \int \frac{1}{2} \, dt - \int \frac{\cos 6t}{2} \, dt \). This can be treated as two separate integrals for simplicity.

Step 3: Solve the First Integral

The first integral is \( \int \frac{1}{2} \, dt = \frac{t}{2} + C_1 \), where \( C_1 \) is a constant of integration.

Step 4: Solve the Second Integral

For the second integral, \( \int \frac{\cos 6t}{2} \, dt \), we apply the substitution rule. The antiderivative of \( \cos(6t) \) is \( \frac{1}{6}\sin(6t) \). Hence, the integral evaluates to \( \frac{1}{12}\sin(6t) + C_2 \), where \( C_2 \) is another constant of integration.

Step 5: Combine the Results

Combine the results from Steps 3 and 4 to obtain the most general antiderivative: \( \frac{t}{2} - \frac{1}{12}\sin(6t) + C \), where \( C \) is a constant that combines both \( C_1 \) and \( C_2 \).

Step 6: Check by Differentiation

Differentiate the result \( \frac{t}{2} - \frac{1}{12}\sin(6t) + C \). The derivative of \( \frac{t}{2} \) is \( \frac{1}{2} \), and the derivative of \( -\frac{1}{12}\sin(6t) \) is \( -\frac{1}{12} \cdot 6 \cos(6t) = -\frac{1}{2}\cos(6t) \). Combining these gives \( \frac{1}{2} - \frac{1}{2}\cos(6t) \), which simplifies to \( \frac{1 - \cos(6t)}{2} \), verifying the original integrand.

Key Concepts

Indefinite Integrals

Indefinite integrals, often associated with finding antiderivatives, are crucial in calculus. An indefinite integral of a function is a function whose derivative yields the original function. It incorporates the 'constant of integration,' represented by the letter "C," which accounts for any constant's presence in differentiating to the original function.
Indefinite integrals are represented without upper or lower limits of integration, hence the term 'indefinite.' When you see an integral expression like \( \int f(x) \, dx \), you are tasked with finding a function \( F(x) \) such that \( F'(x) = f(x) \). This requires reversing the process of differentiation, which can often be achieved by identifying known integral formulas or applying suitable integration techniques.

• The result of an indefinite integral is an expression plus a constant, \( C \).
• The integral symbol denoted by \( \int \) suggests an operation of "integration."
• The 'dx' indicates the variable of integration.

Substitution Method

The substitution method simplifies integrating functions, transforming them into easier forms. It's akin to reversing the chain rule in differentiation, where you substitute a part of the integral with a single variable to streamline the process.
In our exercise, especially with trigonometric integrals like \( \int \frac{\cos 6t}{2} \, dt \), substitution is beneficial. By setting an inner function, say \( u = 6t \), you change the variables seamlessly, making the integral simpler to handle.

• Start by identifying an inner function within the integral, often part of a composite function.
• Let \( u \) equal that function, e.g., \( u = 6t \), and compute \( du \).
• Substitute \( u \) into the integral, replacing all occurrences of the original variable.
• Work with the simpler integral, solve it, and then substitute back to the original variable.

Differentiation

Differentiation is the process of finding the derivative of a function. It is a fundamental operation in calculus that provides the rate at which a function is changing at any point. Differentiation is essential for checking solutions of indefinite integrals, ensuring their correctness.
In our exercise, after determining the antiderivative, differentiation helps verify the solution. We used differentiation on the result \( \frac{t}{2} - \frac{1}{12}\sin(6t) + C \) to ensure it matches the original integrand \( \frac{1 - \cos(6t)}{2} \).

• Find the derivative by applying rules such as the power rule, product rule, or chain rule as necessary.
• Check if the derived function equals the original function under the integral sign.
• If it matches, the antiderivative is confirmed.

Trigonometric Integrals

Trigonometric integrals involve integrating trigonometric functions such as sine, cosine, tangent, etc. These integrals require a keen understanding of trigonometric identities and formulas to simplify and solve them effectively.
In problems like our exercise, where \( \cos(6t) \) appears in the integral, you need to be familiar with how to integrate these to find antiderivatives correctly. Recognizing patterns and applying the substitution method often help simplify these integrals.

• Understand trigonometric identities such as \( \sin^2(x) + \cos^2(x) = 1 \) to aid in simplification.
• Identify when substitution will make integrating simpler.
• Know the basic antiderivatives, such as \( \int \cos(x) \, dx = \sin(x) + C \).