Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-28, perform the indicated integrations. \(\int \cot ^{5} 2 t d t\)

Step-by-Step Solution

Answer

The integral is \(-\frac{1}{6 \sin^3 2t} + C\).

Step 1: Express Cotangent in Terms of Trigonometric Functions

The cotangent function can be expressed as the ratio of cosine to sine: \( \cot 2t = \frac{\cos 2t}{\sin 2t} \). Therefore, \( \cot^5 2t = \left(\frac{\cos 2t}{\sin 2t}\right)^5 = \frac{\cos^5 2t}{\sin^5 2t} \).

Step 2: Simplify Using Powers of Cotangent

Separate one \( \cot 2t \) to form a derivative component: \( \cot^5 2t = \cot^4 2t \cdot \cot 2t = \frac{\cos^4 2t}{\sin^4 2t} \cdot \cot 2t \).

Step 3: Express Cotangent in a Different Form

Use \( \cot^2 2t = \csc^2 2t - 1 \) to express \( \cot^4 2t = (\cot^2 2t)^2 = (\csc^2 2t - 1)^2 \), then integrate with respect to \( t \).

Step 4: Substitute to Simplify Integration

Let \( u = \sin 2t \), then \( du = 2 \cos 2t \, dt \) or \( \frac{1}{2} du = \cos 2t \, dt \). This transforms the integral to \( \int (u^{-4}) \frac{1}{2} du \).

Step 5: Perform the Integration

Integrate the expression \( \frac{1}{2} \int u^{-4} \, du \). The antiderivative of \( u^{-4} \) is \( -\frac{1}{3} u^{-3} \). Thus, \( \frac{1}{2} \times -\frac{1}{3} \times u^{-3} = -\frac{1}{6} u^{-3} \).

Step 6: Substitute Back to Original Variable

Replace \( u \) with \( \sin 2t \) to get \[ -\frac{1}{6} (\sin 2t)^{-3} = -\frac{1}{6 \sin^3 2t} \].

Step 7: Apply Integration Constant

Don't forget to add the constant of integration \( C \) at the end to indicate the family of antiderivatives, resulting in \(-\frac{1}{6 \sin^3 2t} + C \).

Key Concepts

Integration Techniques

When we tackle integration, understanding various techniques is crucial. Integration is essentially the reverse of differentiation, helping us find antiderivatives. Several techniques can be employed to solve complex integrals more effectively:

• Substitution Method: This is handy when we can substitute a part of the problem with a new variable, simplifying the integral into something more manageable. In our example, letting \( u = \sin 2t \) simplifies the integral significantly.
• Integration by Parts: Occasionally useful when the integral is a product of functions. Similar to the product rule in differentiation.
• Trigonometric Identities: Utilizing these can transform trigonometric integrals into familiar forms. Changing the cotangent into sine and cosine ratios is an example here.

Each technique helps to break down a complex integral into a format that's easier to integrate. Practice and familiarity will help in selecting and applying these methods with confidence.

Trigonometric Functions

Trigonometric functions are core building blocks in calculus and are vital in understanding both derivatives and integrals of oscillatory behavior. By defining the cotangent function in terms of sine and cosine for instance, we can simplify integrals significantly. Here’s a quick refresher:

• Sine and Cosine: Foundational functions, where \( \sin \theta \) relates to the y-coordinate and \( \cos \theta \) relates to the x-coordinate on the unit circle.
• Cotangent: Defined as \( \cot \theta = \frac{\cos \theta}{\sin \theta} \); it can be rewritten using these basic trigonometric functions for easier integration.

Using identities such as \( \cot^2 \theta = \csc^2 \theta - 1 \), we can further manipulate and simplify complex trigonometric expressions. This strategic transformation is crucial in integral calculus to simplify functions and solve them effectively.

Antiderivatives

An antiderivative, often simply called an integral, is a function whose derivative is the original function we started with. Finding the antiderivative is crucial in integral calculus and can be particularly challenging. However, by understanding a few key points, it becomes more approachable:

• Concept of Antiderivative: If \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \). It provides the area under the curve described by \( f(x) \).
• Steps to Find Antiderivatives: Convert complex expressions into simpler forms using substitutions, or other transformations involving trigonometric identities, then integrate using basic power rule or other techniques.
• Constant of Integration: Always add a constant \( C \) to indicate an indefinite integral. The constant accounts for all possible shifts vertically of the antiderivative function.

In our example, transforming the integral using substitution made it easier to apply the power rule, ultimately simplifying our solution to an antiderivative form and ensuring we include the constant of integration, which is essential for covering all possibilities of the function's integral solutions.