Problem 86 - Thomas Calculus

Question

a. Express \(\int \cot ^{3} \theta d \theta\) in terms of \(\int \cot \theta d \theta .\) Then evaluate \(\int \cot ^{3} \theta d \theta .\left(\text {Hint} : \cot ^{2} \theta=\csc ^{2} \theta-1 .\right)\) b. Express \(\int \cot ^{5} \theta d \theta\) in terms of \(\int \cot ^{3} \theta d \theta\) c. Express \(\int \cot ^{7} \theta d \theta\) in terms of \(\int \cot ^{5} \theta d \theta\) d. Express \(\int \cot ^{2 k+1} \theta d \theta,\) where \(k\) is a positive integer, in terms of \(\int \cot ^{2 k-1} \theta d \theta\)

Step-by-Step Solution

Answer

Express each higher odd integral in terms of lower odd integrals using \( \cot^2\theta = \csc^2\theta - 1 \) and reduce recursively.

Step 1: Simplify using Trigonometric Identity

Given the hint, we know that \( \cot^2\theta = \csc^2\theta - 1 \). Express \( \int \cot^3 \theta \, d\theta \) as \( \int \cot\theta \cdot (\cot^2\theta) \, d\theta \). Substitute \( \cot^2\theta = \csc^2\theta - 1 \) to get \( \int \cot\theta(\csc^2\theta - 1) \, d\theta = \int \cot\theta \csc^2\theta \, d\theta - \int \cot\theta \, d\theta \).

Step 2: Integrate Terms Separately

The expression in Step 1 has two integrals: \( \int \cot\theta \csc^2\theta \, d\theta \) and \( -\int \cot\theta \, d\theta \). The integral \( \int \cot\theta \csc^2\theta \, d\theta \) can be evaluated using substitution. Let \( u = \cot\theta \), then \( du = -\csc^2\theta \, d\theta \), simplifying to \( -\int u \, du = -\frac{u^2}{2} + C_1 = -\frac{\cot^2\theta}{2} + C_1 \). Integrate \( \int \cot\theta \, d\theta \) to get \( \ln|\sin\theta| + C_2 \).

Step 3: Combine Results

Combine the results from Step 2: \( \int \cot^3\theta \, d\theta = -\frac{\cot^2\theta}{2} - \ln|\sin\theta| + C \), simplifying to \( -\frac{1}{2} \cot^2\theta - \ln|\sin\theta| + C \).

Step 4: Generalize for Higher Powers

For higher powers, express \( \int \cot^{2k+1} \theta \, d\theta \) in terms of \( \int \cot^{2k-1} \theta \, d\theta \). Using the identity \( \cot^2\theta = \csc^2\theta - 1 \), rewrite \( \int \cot^{2k+1} \theta \, d\theta \) as \( \int \cot\theta \cdot \cot^{2k} \theta \, d\theta = \int \cot\theta \cdot (\csc^2\theta - 1)^{k} \, d\theta \). Apply integration by parts or substitution to reduce the power, eventually leading to a relation involving \( \int \cot^{2k-1} \theta \, d\theta \).

Step 5: Express Subsequent Integrals

From Step 4, express \( \int \cot^5\theta \, d\theta \) using \( \int \cot^3\theta \, d\theta \) and similarly express \( \int \cot^7 \theta \, d\theta \) using \( \int \cot^5 \theta \, d\theta \). For each, simplify by letting \( \int \cot^3\theta \, d\theta = I_3 \), so \( \int \cot^5\theta \, d\theta = \int \cot\theta \cdot (\csc^2\theta - 1) \cdot I_3 \, d\theta \).

Step 6: Recursive Formula for Odd Powers

The recursive relation derived is \( \int \cot^{2k+1} \theta \, d\theta = \frac{-1}{2k} \cot^{2k}\theta\cot\theta \) plus a term involving \( \int \cot^{2k-1} \theta \, d\theta \), which shows how each integral can be expressed in terms of its lower power counterpart.

Key Concepts

Integration Techniques

Trigonometric integrals often require special strategies to evaluate, especially when dealing with odd powers of trigonometric functions like cotangent. In these cases, integration by substitution and familiarity with trigonometric identities are key. For instance, in the exercise, to evaluate \( \int \cot^3 \theta \, d\theta \), we use the identity \( \cot^2 \theta = \csc^2 \theta - 1 \) to simplify the integral into parts that are manageable through substitution. This typically involves recognizing a part of the integrand as a derivative of another part, which permits substitution. In general, breaking down complex trigonometric powers into simpler forms reduces the problem into a series of simpler integrals that are easier to solve one at a time. Additionally, knowing when to apply substitution or partial fractions is beneficial across solving calculus problems.

Trigonometric Identities

Understanding trigonometric identities like \( \cot^2 \theta = \csc^2 \theta - 1 \) is invaluable. These identities allow us to transform integrals into forms that are easier to evaluate. The given problem leverages this identity to express \( \int \cot^3 \theta \, d\theta \) in terms of \( \int \cot \theta \, d\theta \). Identifying these relationships enables us to break down complex integrals into basic functions that we already know how to integrate. Such identities are not mere academic exercises but practical tools in solving real integrals encountered in calculus problems. By using trigonometric identities, you simplify the process of integration, making them a powerful component in your calculus toolkit.

Calculus Problem Solving

Approaching calculus problems systematically is crucial. This involves being methodical in using integration techniques and applying trigonometric identities. With the given exercise, a step-by-step approach is emphasized. We start by using identities to simplify the integral, then separately tackle each part using known integration methods like substitution and direct integration. For higher powers of cotangent functions, we use recursive methods—expressing higher powers in terms of lower ones. This reduces complex calculus problems to manageable tasks. In essence, problem-solving in calculus often means breaking down a problem into components and tackling each piece methodically. Practicing this way trains you to recognize patterns and solutions with ease, making even the most daunting problems approachable.