Problem 44 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(39-56\) $$ \int_{0}^{\pi / 3} \frac{4 \sin \theta}{1-4 \cos \theta} d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \(- \ln(3)\).

Step 1: Identify the Integral

We are given the integral \( \int_{0}^{\pi / 3} \frac{4 \sin \theta}{1-4 \cos \theta} d \theta \). The task is to evaluate this definite integral.

Step 2: Choose a Suitable Method of Integration

Since the integrand has a form that suggests a potential substitution due to the presence of \( \sin \theta \) and \( \cos \theta \), let's try a trigonometric substitution. We will use the substitution \( u = 1 - 4 \cos \theta \).

Step 3: Compute the Derivative

Differentiate the substitution to find \( du \): \( du = 4 \sin \theta \, d\theta \). This implies \( \sin \theta \, d\theta = \frac{1}{4} du \).

Step 4: Change the Limits of Integration

Substitute the limits of \( \theta \) into \( u \): When \( \theta = 0 \), \( u = 1 - 4 \cos(0) = 1 - 4 = -3 \). When \( \theta = \frac{\pi}{3} \), \( u = 1 - 4(\frac{1}{2}) = 1 - 2 = -1 \).

Step 5: Substitute and Simplify the Integral

Substituting into the integral gives: \[ \int_{-3}^{-1} \frac{4 \sin \theta}{1-4 \cos \theta} d\theta = \int_{-3}^{-1} \frac{4}{u} \cdot \frac{1}{4} du = \int_{-3}^{-1} \frac{1}{u} du. \]

Step 6: Evaluate the New Integral

The integral \( \int \frac{1}{u} du \) evaluates to \( \ln |u| + C \). Thus, \[ \int_{-3}^{-1} \frac{1}{u} \, du = \left[ \ln |u| \right]_{-3}^{-1}. \]

Step 7: Substitute the Limits Back

Substitute the limits into the result: \[ \left[ \ln |-1| \right] - \left[ \ln |-3| \right] = \ln(1) - \ln(3) = 0 - \ln(3). \] Thus, the value of the integral is \(-\ln(3)\), which can be simplified to \(\ln(\frac{1}{3})\).

Step 8: Compile the Result

The evaluated integral is \(- \ln(3)\). This means the integral of \( \int_{0}^{\pi / 3} \frac{4 \sin \theta}{1-4 \cos \theta} d\theta \) is equal to the logarithmic expression \(- \ln(3)\).

Key Concepts

Trigonometric Substitution

In calculus, trigonometric substitution is a powerful technique for evaluating integrals involving trigonometric functions like sine and cosine. By substituting one expression with another, it often simplifies the equation. For example, in the integral \( \int \frac{4 \sin \theta}{1-4 \cos \theta} d \theta \), we used the substitution \( u = 1 - 4 \cos \theta \).

• This substitution transforms complex trig integrands into simpler forms.
• The derivative \( du = 4 \sin \theta \, d\theta \) then enables us to replace \( \sin \theta \, d\theta \) with \( \frac{1}{4} du \).

Thus, the substitution transforms the problem into an integral in terms of \( u \), making it easier to solve.

Limits of Integration

Limits of integration are the values that define the range over which we are integrating a function. When a substitution is made, these limits must also be changed to match the new variable. In our exercise, the original limits of integration were from \( \theta = 0 \) to \( \theta = \frac{\pi}{3} \).

• Substitute these values into the equation \( u = 1 - 4 \cos \theta \) to determine the new bounds.
• For \( \theta = 0 \), we find \( u = -3 \).
• For \( \theta = \frac{\pi}{3} \), \( u = -1 \).

This ensures that the integral reflects the correct region in terms of the new variable \( u \). Adjusting limits is essential for accurate evaluation of definite integrals.

Logarithmic Integration

Logarithmic integration involves integrating to find an expression that contains a logarithm. Typically, it comes into play when the integral involves a function of the form \( \frac{1}{u} \). In the solution to our problem, after substitution, the integral \( \int \frac{1}{u} \, du \) arises.

• The integral evaluates to \( \ln |u| + C \), where \( C \) is a constant of integration.
• This log transformation is key when dealing with variables that appear in a denominator.

Therefore, knowledge of logarithmic integration is crucial for successfully handling such manipulations in calculus.

Definite Integral Evaluation

Definite integral evaluation is about finding the numerical value of an integral over a specified interval. In this exercise, after substituting and finding the indefinite integral, we plugged back in the limits to evaluate it.

• Use the result from the indefinite integral \( \ln |u| \) and substitute back the limits from \( u = -3 \) to \( u = -1 \).
• This produces \( \left[ \ln |-1| \right] - \left[ \ln |-3| \right] = 0 - \ln(3) \).
• The computed value is \(-\ln(3)\), simplifying to \( \ln(\frac{1}{3}) \).

Thus, definite integral evaluation allows us to derive a precise answer from the integral expression over a given interval.