Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \cot ^{3} 2 t d t $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Strategy

To solve the integral \( \int \cot ^{3} 2t \ dt \), notice that it's a trigonometric integral involving an odd power of cotangent. A useful technique is to express the integral in terms of sine and cosine to simplify the expression. We could use a trigonometric identity and a substitution to solve it.

Step 2: Express Cotangent in Sine and Cosine

Recall the identity \( \cot x = \frac{\cos x}{\sin x} \). Therefore, \( \cot^3 2t = \left( \frac{\cos 2t}{\sin 2t} \right)^3 = \frac{\cos^3 2t}{\sin^3 2t} \). This transforms our integral to:\[\int \frac{\cos^3 2t}{\sin^3 2t} \, dt.\] Decompose the expression into:\[\int \cot 2t \cdot \csc^2 2t \, dt.\]

Step 3: Use Trigonometric Identity and Substitution

Notice that \( \csc^2 x = 1 + \cot^2 x \). Let's use the substitution \( u = \csc^2 2t \), hence \( du = -2 \cot 2t \csc^2 2t \, dt \). Therefore, we can write: \[ \int \cot 2t \cdot \csc^2 2t \, dt = -\frac{1}{2} \int du. \]

Step 4: Integrate and Back-Substitute

Now integrate with respect to \( u \): \[ -\frac{1}{2} \int du = -\frac{1}{2}u + C. \]Re-substitute \( u = \csc^2 2t \) into the equation gives:\[ -\frac{1}{2} \csc^2 2t + C. \]

Key Concepts

Trigonometric Identities

Trigonometric identities are essential tools in integral calculus. They enable us to simplify complex trigonometric expressions, making it easier to integrate them. In the context of trigonometric integrals, identities express relationships between various functions, such as sine, cosine, and their reciprocals.In this exercise, the identity used is for cotangent, where \( \cot x = \frac{\cos x}{\sin x} \). This relation helps rewrite expressions involving powers of cotangent in terms of more manageable sine and cosine functions. Additionally, there's also the identity \( \csc^2 x = 1 + \cot^2 x \) used in the substitution process.Understanding these identities:

• Cotangent (\( \cot x \)): The reciprocal of tangent, expressing it in terms of cosine and sine simplifies manipulations.
• Cosecant squared (\( \csc^2 x \)): Useful in recognizing transformations or substitutions related to cotangent.

Knowing and applying these identities effectively is key to solving trigonometric integrals like the one given in this exercise.

Substitution Method

The substitution method, also known as u-substitution, is a powerful technique to solve integrals that seem complex at first glance. The method involves changing variables to simplify the integral's structure.To apply substitution:

• Choose a Substitution: Identify part of the integral that can be substituted with a single variable \( u \) to make integration easier. In our problem, \( u = \csc^2 2t \) is chosen.
• Differentiate: Calculate the derivative of \( u \), giving \( du = -2\cot 2t \csc^2 2t \, dt \).
• Rewrite the Integral: Substitute \( u \) and \( du \) into the integral. Here, this leads to \( -\frac{1}{2} \int du \).
• Simplify and Integrate: The substitution transforms the integral into a basic form that's straightforward to integrate, resulting in \( -\frac{1}{2} u + C \).
• Back-Substitute: Finally, replace \( u \) with the original trigonometric expression \( \csc^2 2t \).

Substitution transforms challenging integrals into manageable ones, simplifying trigonometric integral problems effectively.

Integral Calculus

Integral calculus is a fundamental area of mathematics that deals with the process of integration. It allows us to find areas under curves, compute accumulated quantities, and solve differential equations, among other applications.In the context of our exercise, the goal was to integrate the function \( \int \cot^{3} 2t \, dt \). This involves breaking down the problem using trigonometric identities and substitution to find a solution.Key insights about integral calculus include:

• Definite vs. Indefinite Integrals: The problem at hand is an indefinite integral, which implies integration results in a general form with an arbitrary constant \( C \).
• Techniques of Integration: Understanding when to apply certain techniques like substitution or integration by parts can greatly simplify the process.
• Application of Calculus: The use of calculus goes beyond just solving integrals; it finds applications in physics, engineering, and even economics.

Through practice, integral calculus becomes a versatile tool, providing solutions to seemingly complex mathematical problems.