Problem 35 - 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. $$\int(-2 \cos t) d t$$

Step-by-Step Solution

Answer

The most general antiderivative is \(-2 \sin t + C\).

Step 1: Understand the Given Function

The integral to solve is \( \int (-2 \cos t) \, dt \). This requires finding the antiderivative of \(-2 \cos t\).

Step 2: Find the Antiderivative Formula

The antiderivative of \( \cos t\) is \( \sin t\). Therefore, the antiderivative of \(-2 \cos t\) is \(-2 \cdot \sin t\), applying the constant multiple rule in integration.

Step 3: Add the Constant of Integration

Since we are looking for the most general antiderivative, we must add a constant \( C \) to our solution: \(-2 \sin t + C\).

Step 4: Differentiate to Check the Solution

Differentiate \(-2 \sin t + C\) with respect to \( t \) to verify it matches the original integrand. The derivative of \(-2 \sin t\) is \(-2 \cos t\), and the derivative of a constant \( C \) is zero. Therefore, the derivative is \(-2 \cos t\), which matches the original function.

Key Concepts

Integration Techniques

When tackling the problem of finding an antiderivative—also known as solving an indefinite integral—it's important to be familiar with various integration techniques. In this particular exercise, the function we need to integrate is \(-2 \cos t\). Integration techniques involve identifying patterns and employing specific strategies to find the antiderivative. For instance:

• **Basic Integration Rules**: For basic functions, memorize common integrals like \( \int \cos t \, dt = \sin t + C \).
• **Constant Multiple Rule**: If a function is multiplied by a constant, such as \( -2 \), you can take the constant out of the integral: \( \int a \cdot f(t) \, dt = a \int f(t) \, dt \).

Applying these techniques to our exercise, we recognize that the integral of \(-2 \cos t\) involves the antiderivative of \(\cos t\), multiplied by \(-2\). This yields the antiderivative as \(-2 \sin t + C\), where \(C\) is a constant of integration.

Constant of Integration

When finding the most general antiderivative, the constant of integration, denoted usually as \(C\), plays a crucial role. This constant accounts for the family of functions in the solution. Each different value of \(C\) represents a different antiderivative.

Why do we add this constant? Because integration is the inverse process of differentiation. Differentiating any constant results in zero, which means that during integration, we lose any specific constant that might have been part of the original function.

In our example \(-2 \cos t\), the antiderivative solution is \(-2 \sin t + C\). By including \(C\), we ensure that our result includes every possible function that, when differentiated, yields the original \(-2 \cos t\). Essentially, this provides us with the complete set of solutions.

Checking Solutions by Differentiation

Checking your solution through differentiation is an essential step in integrating a function. This not only verifies your work but also reinforces the connection between these fundamental concepts.

To check if \(-2 \sin t + C\) is the correct antiderivative of \(-2 \cos t\), differentiate it with respect to \(t\). The derivative of \(-2 \sin t\) is obtained as follows:

• The derivative of \(-2 \sin t\) is \(-2 \cos t\), due to the derivative rule \( \frac{d}{dt} \sin t = \cos t \).
• The derivative of the constant \(C\) is zero.

Therefore, the differentiation of \(-2 \sin t + C\) is exactly \(-2 \cos t\), proving that our original integration was executed correctly. This method of checking highlights the harmonious relationship between integration and differentiation, confirming the accuracy of your solution.