Problem 37 - Thomas Calculus

Question

In Exercises \(17-54\) , find the most general antiderivative or indefinite integral. Check your answers by differentiation. $$ \int 7 \sin \frac{\theta}{3} d \theta $$

Step-by-Step Solution

Answer

The antiderivative is \(-21 \cos \frac{\theta}{3} + C\).

Step 1: Understand the Basic Form of Integration

The function to be integrated is \( 7 \sin \left( \frac{\theta}{3} \right) \). This involves a sine function, which typically integrates to a cosine function. The antiderivative of \( \sin x \) is \(-\cos x\). We must also consider any transformations involving the variable.

Step 2: Apply the Constant Multiple Rule

The integral \( \int 7 \sin \left( \frac{\theta}{3} \right) d\theta \) can be rewritten as \( 7 \int \sin \left( \frac{\theta}{3} \right) d\theta \). This simply pulls the constant multiplier out of the integral.

Step 3: Use Substitution Method for Integration

Let \( u = \frac{\theta}{3} \). Therefore, \( du = \frac{1}{3} d\theta \), or \( d\theta = 3 du \). Substitute these into the integral: \( 7 \int \sin(u) \cdot 3 \, du = 21 \int \sin(u) \, du \).

Step 4: Integrate the Simplified Expression

The integral \( \int \sin(u) \, du = -\cos(u) + C \), where \( C \) is the constant of integration. Therefore, the antiderivative is \( 21(-\cos(u)) + C = -21 \cos(u) + C \).

Step 5: Substitute Back in Terms of Original Variable

Replace \( u \) with \( \frac{\theta}{3} \) to express the antiderivative in terms of \( \theta \), resulting in \( -21 \cos \left( \frac{\theta}{3} \right) + C \).

Step 6: Check by Differentiation

Differentiate \( -21 \cos \left( \frac{\theta}{3} \right) + C \) to verify its correctness. The derivative of \(-21 \cos \left( \frac{\theta}{3} \right)\) is \( 21 \cdot \frac{1}{3} \sin \left( \frac{\theta}{3} \right) \) due to the chain rule, which simplifies back to \( 7 \sin \frac{\theta}{3} \), confirming our integration was done properly.

Key Concepts

Antiderivatives

Antiderivatives are fundamental in calculus as they reverse the process of differentiation. When you find an antiderivative of a function, it means you are determining what function, when differentiated, would return the original function you started with. This is closely related to the concept of indefinite integrals, which are essentially the family of all possible antiderivatives for a given function. Every antiderivative comes with a constant of integration, usually denoted as \( C \), due to the fact that differentiation of a constant is zero. In the problem given, we started with \( 7 \sin \left( \frac{\theta}{3} \right) \). Identifying the antiderivative involves looking for what function would differentiate to give \( \sin \left( \frac{\theta}{3} \right) \). Since the antiderivative of \( \sin x \) is known to be \(-\cos x\), we can apply similar reasoning to our function, taking care to address any transformations involving \( \theta \). This leads to the result \(-21 \cos \left( \frac{\theta}{3} \right) + C \).

Integration Techniques

Integration techniques are essential tools that allow us to evaluate integrals of complex functions. In this exercise, we demonstrate two critical techniques: the constant multiple rule and the substitution method.

• Constant Multiple Rule: This rule states that when integrating, you can factor out any constant from the integral. For example, in \( 7 \int \sin \left( \frac{\theta}{3} \right) d\theta \), the number 7 is a constant. We can pull it out of the integral, simplifying our work to \( 7 \int \sin \left( \frac{\theta}{3} \right) d\theta \). This rule helps in managing coefficients that multiply the function inside the integral.
• Substitution Method (u-substitution): This technique is vital when dealing with composition inside a function, such as \( \sin \left( \frac{\theta}{3} \right) \). Here, we set \( u = \frac{\theta}{3} \) and find \( du \), allowing us to rewrite the integral in terms of \( u \). This simplification helps to tackle integrals involving transformations or chains, making them easier to evaluate.

Integrating \( 21 \int \sin(u) \, du \) gives us \(-21 \cos(u) + C\), which we finally substitute back to \( \theta \) terms, reaching our solution.

Trigonometric Integrals

Trigonometric integrals involve integrating functions derived from trigonometric functions, such as sine, cosine, and tangent. These integrals appear frequently in calculus, especially within physics and engineering, due to their wave-like nature modeling periodic phenomena.For our exercise, the function \( 7 \sin \left( \frac{\theta}{3} \right) \) is based on the trigonometric sine function. Trigonometric integrals often have straightforward antiderivatives, provided you remember the basic rules:

• The integral of \( \sin(x) \) is \(-\cos(x)\).
• The integral of \( \cos(x) \) is \( \sin(x)\).

Handling trigonometric integrals often requires understanding their properties and transformations due to amplitudes, frequencies (represented by coefficients inside the angle), and phase shifts. In the problem, after applying the substitution \( u = \frac{\theta}{3} \), we integrated \( \sin(u) \), utilizing the known antiderivative \(-\cos(u)\). This demonstrates how changes in the function's domain affect the integral, subsequently showing why substituting back to the original variable is crucial for the solution.