Problem 11 - A First Course in Complex Analysis with Applications

Question

Evaluate the given trigonometric integral. $$ \int_{0}^{2 \pi} \frac{\cos ^{2} \theta}{2+\sin \theta} d \theta $$

Step-by-Step Solution

Answer

The integral over a zero interval evaluates to 0.

Step 1: Determine the Substitution

To simplify the integral, use the substitution \( u = 2 + \sin \( \theta \) \), making sure to find the corresponding differential \( du \). The derivative of \( u \) is \cos \( \theta \), hence \( du = \cos \( \theta \) d\theta \).

Step 2: Change Limits of Integration

Change the limits of integration according to the substitution. At \( \theta = 0 \), \( u = 2 + \sin(0) = 2 \). At \theta = 2\pi, \( u = 2 + \sin(2\pi) = 2 \). This means the limits of integration in terms of \( u \) are both 2.

Step 3: Transform the Integral

The integral in terms of \( u \) is \int_{2}^{2} \rac{(u-2)^{2}}{u} du \. Note that \(u = 2 + \sin\theta\), \cos(\theta)=du/du,\; and \theta = \cos(\theta) h du. However, the upper and lower limits are the same.

Step 4: Evaluate the Zero Interval

The immediate observation with the changed limits from step 2 indicates the integral will evaluate over an interval from 2 to 2, which inherently means no area is captured under the curve due to identical integration limits. Consequently, the interval of integration is actually a degenerate interval, over which definite integral evaluates to zero.

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus used to simplify integrals. By introducing a new variable, often denoted as \( u \), we transform the integral into a simpler form. This method is particularly useful in trigonometric integrals, like the one given in the exercise.Here’s how the process works:

• First, identify a part of the integrand that can be replaced with a new variable \( u \). In this problem, we chose \( u = 2 + \sin \theta \).
• Next, compute the derivative of \( u \) with respect to \( \theta \), which gives \( du = \cos \theta \, d\theta \). This differential substitution will replace \( d\theta \) in the integral.

The substitution helps transform the original integral into one involving \( u \). By this approach, we can often simplify complex integrals to make them more manageable.

Definite Integral

A definite integral calculates the net area under a curve between two specified limits. It has both upper and lower limits and results in a real number representing the area.In this exercise, we changed the integral limits according to our substitution from \( \theta \) to \( u \). Originally, \( \theta \) ranged from \( 0 \) to \( 2\pi \), corresponding to new limits where \( u = 2 \) at both \( \theta = 0 \) and \( \theta = 2\pi \).Key components of definite integrals include:

• Boundary values: These represent the interval over which integration occurs.
• Integration: The process of finding the cumulative sum (or area).
• Area under the curve: The result of integration is the signed area under a curve, which can be positive, negative, or zero depending on the limits and function behavior.

Zero Interval

In some situations, like in this exercise, the limits of the definite integral yield what is known as a zero interval.A zero interval occurs when the upper and lower limits of integration are the same. Here, both limits were transformed to the value of \( u = 2 \). Integrating over a zero interval means there is no width to the interval, and hence, no area to calculate. There are important implications:

• No captured area: Since the interval width is zero, the definite integral evaluates directly to zero.
• Simplicity of evaluation: Recognizing a zero interval immediately simplifies the problem as there is nothing to compute further.

This principle is crucial in evaluating integrals and recognizing when an integral does not require elaborate computation.