Problem 48 - Thomas Calculus in SI Units

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. $$\int \frac{1-\cos 6 t}{2} d t$$

Step-by-Step Solution

Answer

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

Step 1: Simplify the Integral Expression

The integrand is \( \frac{1 - \cos 6t}{2} \). Recognize it as a half-angle identity and rewrite it as the integral of \( \frac{1}{2} - \frac{1}{2} \cos 6t \). This simplifies the expression and makes it easier to integrate.

Step 2: Distribute the Integral

Distribute the integral over the difference: \[ \int \left( \frac{1}{2} - \frac{1}{2} \cos 6t \right) \, dt = \frac{1}{2} \int dt - \frac{1}{2} \int \cos 6t \, dt. \] This separates the original problem into two simpler integrals.

Step 3: Find the First Antiderivative

Focus on \( \frac{1}{2} \int dt \). The antiderivative of a constant \( \frac{1}{2} \) with respect to \( t \) is \( \frac{1}{2}t \).

Step 4: Find the Second Antiderivative

For \( \frac{1}{2} \int \cos 6t \, dt \), use the substitution method. Let \( u = 6t \), then \( du = 6 \, dt \) so that \( dt = \frac{1}{6} du \). The integral becomes \(-\frac{1}{12} \int \cos u \, du = -\frac{1}{12} \sin u + C_1 \). Substitute back with \( u = 6t \) to get \(-\frac{1}{12} \sin 6t \).

Step 5: Combine the Antiderivatives

Combine the results of steps 3 and 4 for the original integral: \[ \frac{1}{2}t - \frac{1}{12} \sin 6t + C. \] The constant \( C \) represents the constant of integration.

Step 6: Verify Through Differentiation

Differentiate the result \( \frac{1}{2}t - \frac{1}{12} \sin 6t + C \) to verify correctness. The derivative is \( \frac{1}{2} - \frac{1}{2} \cos 6t \), which matches the original integrand, confirming the solution is correct.

Key Concepts

Antiderivatives

In integral calculus, an antiderivative of a function is a function whose derivative is the original function. Think of it as the opposite of differentiation. When you find an antiderivative, you are essentially reversing the process of differentiation.
For the exercise, the integral given is \( \int \frac{1-\cos 6t}{2} \, dt \). Our goal is to find its most general antiderivative, which involves integrating the expression to get back the original function before it was differentiated. This process includes recognizing patterns and identities, such as the half-angle identity used to rewrite the integral.
It's important to remember that integrals without limits, like this one, result in an indefinite integral, which always includes a constant of integration \( C \). This constant helps represent all possible antiderivatives of the function.

Integration Techniques

Integration techniques are various strategies used to find antiderivatives. In the exercise, techniques like distribution of the integral and substitution were used to find the antiderivative of the given expression.
- **Distribution of the Integral**: By recognizing that \( \frac{1-\cos 6t}{2} \) can be rewritten as \( \frac{1}{2} - \frac{1}{2} \cos 6t \), you simplify the problem. This allows the integral to be split into two more manageable parts: \[ \int \left( \frac{1}{2} - \frac{1}{2} \cos 6t \right) \, dt = \frac{1}{2} \int dt - \frac{1}{2} \int \cos 6t \, dt. \]- **Substitution**: This method involves replacing a part of the integrand with a new variable to simplify integration. In the second integral, \( u = 6t \) was used so that the integration became simpler. After integrating with respect to \( u \), the result was changed back to the original variable.
These techniques make complex integrals more approachable and are essential tools for solving them efficiently.

Verification through Differentiation

Verification through differentiation means checking your result by deriving the antiderivative you found to see if it gives you back the original integrand. This is a crucial step to ensure correctness of your solution.
In the exercise, after finding the antiderivative expression \( \frac{1}{2}t - \frac{1}{12} \sin 6t + C \), you differentiate this expression with respect to \( t \):- The derivative of \( \frac{1}{2}t \) is \( \frac{1}{2} \).- The derivative of \(-\frac{1}{12} \sin 6t \) involves using the chain rule, giving \( -\frac{1}{2} \cos 6t \).- The constant \( C \) disappears after differentiation.Adding these derivatives gives \( \frac{1}{2} - \frac{1}{2} \cos 6t \), which matches the original integrand \( \frac{1-\cos 6t}{2} \). By performing this check, you confirm that the antiderivative found is indeed correct.