Problem 36 - Thomas Calculus

Question

In Exercises \(35-64\) , use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$ \int_{0}^{\pi / 2} \cot \theta d \theta $$

Step-by-Step Solution

Answer

The integral diverges due to a singularity at \( \theta = 0 \).

Step 1: Rewrite the Integral

Start by rewriting the given integral in terms of sine and cosine since \( \cot \theta = \frac{\cos \theta}{\sin \theta} \). The integral becomes: \[ \int_{0}^{\pi / 2} \frac{\cos \theta}{\sin \theta} \, d \theta \] This can be further simplified by recognizing it as: \[ \int_{0}^{\pi / 2} \frac{1}{\sin \theta} \cos \theta \, d \theta \]

Step 2: Use Substitution

Let \( u = \sin \theta \), then \( \frac{du}{d\theta} = \cos \theta \) or \( du = \cos \theta \, d\theta \). Substitute into the integral:\[ \int \frac{1}{u} \, du \]The limits change as \( \theta \) changes from 0 to \( \pi/2 \), so \( u \) changes from 0 to 1.

Step 3: Evaluate the New Integral

Now evaluate the integral:\[ \int \frac{1}{u} \, du = \ln |u| + C \]So, replacing \( u \) with \( \sin \theta \), the integral is:\[ \left[ \ln |\sin \theta| \right]_{0}^{\pi/2} \]

Step 4: Evaluate the Limits

Evaluate the definite integral by applying the limits:- At \( \theta = \pi/2 \), \( \sin(\pi/2)=1 \) so \( \ln |1| = 0 \).- At \( \theta = 0 \), \( \sin(0)=0 \) which involves \( \ln |0| \), leading to a singularity.Thus the integral diverges.

Key Concepts

Convergence Tests

Convergence tests help us determine whether an integral converges or diverges. When working with improper integrals, like the one in our exercise, these tests are invaluable tools.
To determine the behavior of the integral \[ \int_{0}^{\pi / 2} \cot \theta \, d \theta \] we consider both the Direct Comparison Test and the Limit Comparison Test. These tests require comparing our integral with a simpler function whose convergence is known.

• In the Direct Comparison Test, if the function we are comparing to converges and is greater than our function, then our function also converges. Conversely, if the simpler function diverges and is less than our function, then ours diverges as well.
• The Limit Comparison Test, on the other hand, involves calculating the limit of the ratio of the two functions as they approach the bounds of integration. If this limit is a positive finite number, the behavior of the series matches the comparison function.

Because the function \( \cot \theta \) behaves like \( \frac{1}{\theta} \) near zero, and the integral of \( \frac{1}{\theta} \) diverges, we can infer that the original integral diverges.

Substitution Method

The substitution method is a powerful technique for simplifying integrals by making them easier to evaluate.
In this specific problem, we start by noting that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \). This allows us to rewrite the integral:
\[ \int_{0}^{\pi / 2} \frac{1}{\sin \theta} \cos \theta \, d \theta \] Substitution involves choosing a new variable that simplifies the integral. In this case, we let \( u = \sin \theta \).

• Then, the derivative \( \frac{du}{d\theta} = \cos \theta \) gives us \( du = \cos \theta \, d\theta \).
• With this substitution, the integral changes to \( \int \frac{1}{u} \, du \).

The substitution simplifies the integral significantly, allowing us to recognize it as the natural logarithm: \( \ln |u| + C \), which is much easier to evaluate.

Trigonometric Integrals

Trigonometric integrals often present themselves in calculus problems, especially when dealing with integrals involving trigonometric functions.
These integrals require familiarity with trigonometric identities and substitution techniques. In this problem, we encountered an integral of \( \cot \theta \).
The substitution method transformed it into an integral concerning \( \frac{1}{\sin \theta} \). Trigonometric integrals like these often involve direct use of identities:

• Knowing that \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) allows us to simplify the expression.
• This conversion is essential to integrate the function using more straightforward forms and substitution methods.

By mastering these trigonometric relationships, solving their integrals becomes much more manageable.