Problem 37 - 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. $$\int 7 \sin \frac{\theta}{3} d \theta$$

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Integral

The given integral is \( \int 7 \sin \frac{\theta}{3} \, d\theta \). To simplify, notice that the integral is a constant factor times a basic trigonometric function. Begin by factoring out the constant, 7: \[ 7 \int \sin \left( \frac{\theta}{3} \right) \, d\theta \].

Step 2: Use Substitution Method

Perform the substitution \( u = \frac{\theta}{3} \). This means \( du = \frac{1}{3} \, d\theta \) or \( d\theta = 3 \, du \). Replace \( \frac{\theta}{3} \) in the integrand and \( d\theta \) with \( 3 \, du \): \[ 7 \int \sin(u) \cdot 3 \, du = 21 \int \sin(u) \, du \].

Step 3: Integrate the Simplified Integral

Find the integral of \( 21 \sin(u) \, du \). The integral of \( \sin(u) \) is \( -\cos(u) \). So:\[ 21 \int \sin(u) \, du = -21 \cos(u) + C \], where \( C \) is the constant of integration.

Step 4: Substitute Back the Original Variable

Substitute back for \( u \), using \( u = \frac{\theta}{3} \), into the expression. \[-21 \cos \left( \frac{\theta}{3} \right) + C \].

Step 5: Verify by Differentiation

Differentiate \(-21 \cos \left( \frac{\theta}{3} \right) + C \) with respect to \( \theta \):\( \frac{d}{d\theta} [-21 \cos(\theta/3)] = -21 \left(-\sin(\theta/3)\right) \cdot \frac{1}{3} = 7 \sin(\theta/3) \), which matches the original integrand, confirming the solution.

Key Concepts

Antiderivative

In calculus, an antiderivative is a function whose derivative is the original function. It's the reverse process of differentiation. Instead of finding the rate of change, you're looking to find the original function.
The antiderivative, or indefinite integral, of a function returns all functions whose derivative is the integrand. It is written as an integral sign with a function and a differential. The result includes a constant of integration (usually denoted by \( C \)) because differentiation of a constant is zero, and any constant would disappear when taking the derivative.

• For example, if you have \( f'(x) = 2x \), the antiderivative would be \( f(x) = x^2 + C \).
• The presence of \( C \) accounts for the fact that multiple functions could have the same derivative.

Finding an antiderivative requires familiarity with basic derivative rules, and sometimes some trial and error. With practice, recognizing patterns in derivatives can make finding antiderivatives quicker and easier.

Substitution Method

The substitution method is a powerful technique for evaluating indefinite integrals. It simplifies the integration process by changing variables to make integration more straightforward.
Here's how it works:

• Select a substitution: Choose a part of the integrand to substitute with a single variable. For instance, in this exercise, \( u = \frac{\theta}{3} \) was chosen.
• Differentiate: Find the differential of your substitution (\( du \)). Here, \( du = \frac{1}{3} \, d\theta \), so \( d\theta = 3 \, du \).
• Rewrite the integral: Substitute all parts of \( \theta \) in the integral. Replace \( \sin(\theta/3) \) and \( d\theta \) with \( \sin(u) \) and \( 3 \, du \) respectively. The integral then becomes \( 21 \int \sin(u) \, du \).
• Integrate: Perform the integration with respect to \( u \).
• Substitute back: After integrating, replace \( u \) with the original expression to revert to the original variable.

This technique is especially effective when the integrand includes a composite function, as it is easier to integrate the substituted expression.

Trigonometric Function

Trigonometric functions, like sine and cosine, are fundamental in calculus, especially in integration and differentiation.
In this exercise, the function \( \sin(\theta/3) \) is an example of a trigonometric function within an integral. Such functions often require specific techniques for integration.
The properties of trigonometric functions that are largely used in integration include:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The integral of \( \sin(x) \) is \( -\cos(x) + C \).
• Trig identities may be applied to simplify integrands.

Being comfortable with these basic trigonometric properties, and knowing when to apply identities, aids vastly in solving integrals involving these functions. Trigonometric functions appear frequently in problems involving periodic phenomena or angular measurements, making them vital for calculus.