Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-28, perform the indicated integrations. \(\int \cot ^{3} 2 t d t\)

Step-by-Step Solution

Answer

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

Step 1: Recognize the Trigonometric Identity

To solve the integral \( \int \cot^3 2t \ dt \), we begin by recognizing a useful identity. Recall that \( \cot x = \frac{\cos x}{\sin x} \). Thus, \( \cot^3 2t = \left( \frac{\cos 2t}{\sin 2t} \right)^3 = \frac{\cos^3 2t}{\sin^3 2t} \).

Step 2: Split the Integrand

Next, we split the integrand. We can express \( \cot^3 2t \) as \( \cot 2t \cdot \cot^2 2t \). This leads to \( \cot^3 2t = \cot 2t \cdot (\csc^2 2t - 1) \), using the identity \( \cot^2 x + 1 = \csc^2 x \).

Step 3: Simplify and Integrate

Substitute \( \cot^3 2t = \cot 2t \cdot (\csc^2 2t - 1) \) into the integral, resulting in two separate integrals: \( \int \cot 2t \cdot \csc^2 2t \, dt \) and \( -\int \cot 2t \, dt \).

Step 4: Solve the First Integral

For \( \int \cot 2t \cdot \csc^2 2t \, dt \), let \( u = \csc 2t \), then \( du = -2\cot 2t \csc 2t \, dt \). Therefore, the integral becomes \(-\frac{1}{2} \int du = -\frac{1}{2} u + C = -\frac{1}{2} \csc 2t + C \).

Step 5: Solve the Second Integral

For the integral \( -\int \cot 2t \, dt \), use the substitution \( v = \sin 2t \), then \( dv = 2\cos 2t \, dt \), which gives \( \int \frac{\cos 2t}{\sin 2t} \cdot 1 \, dt = \frac{1}{2} \ln |\sin 2t| + C \).

Step 6: Combine the Results

Combine the results from steps 4 and 5: \[-\frac{1}{2} \csc 2t - \frac{1}{2} \ln |\sin 2t| + C' \], where \( C' \) represents the combined constant of integration.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integration to simplify complex expressions. It's especially useful when dealing with integrals involving quadratic expressions or trigonometric functions. In the given exercise, the integral of the form \[\int \cot^3 2t \, dt\]is encountered.
To break down this problem, we employed trigonometric identities, starting with rewriting \(\cot 2t\) using its fundamental identity \(\cot x = \frac{\cos x}{\sin x}\). Thus, \(\cot^3 2t\) becomes\[\frac{\cos^3 2t}{\sin^3 2t}.\]
Using trigonometric substitution helps by converting the expression into a form that's more manageable and can be expressed in terms of only sine or cosine, as in this case. By transforming a trigonometric power or ratio, this approach often leads to simpler integrals.

Integrals of Trigonometric Functions

When integrating trigonometric functions, one often utilizes identities and transformations to make the process feasible. In our specific problem, the goal was to simplify the integral of \(\cot^3 2t\). Breaking it into smaller parts is key to tackling this problem.

• Recognize identities: Here, the conversion \(\cot^3 2t = \cot 2t \cdot (\csc^2 2t - 1)\) was crucial.
• Split integrals: This step allows the integration to happen over standard forms, leading to \[ \int \cot 2t \cdot \csc^2 2t \, dt - \int \cot 2t \, dt. \]
• Integrate using standard forms: The integral of \(\cot 2t \cdot \csc^2 2t\) involves recognizing a derivative, while \(-\int \cot 2t \, dt\) is simplified by a direct conversion to a logarithmic form.

Overall, handling integrals of trigonometric functions demands a blend of identities and systematic substitutions.

Change of Variables

The technique of change of variables or substitution is pivotal for simplifying integrals. In this exercise, it is effectively utilized twice to transform and solve the integrals:

• For \(\int \cot 2t \cdot \csc^2 2t \, dt\):
  • Let \(u = \csc 2t\), leading to \(du = -2\cot 2t \csc 2t \, dt\).
  • Thus, the integral reduces to a basic form \(-\frac{1}{2} \int du\), giving \(-\frac{1}{2} \csc 2t\).
• For \(-\int \cot 2t \, dt\):
  • Let \(v = \sin 2t\), then \(dv = 2\cos 2t \, dt\).
  • The integral forms into \(\frac{1}{2} \ln |\sin 2t|\) by rearranging derivatives.

Substitution methods like these simplify complex integrals by turning them into more manageable versions that are easier to evaluate, ultimately leading to a clearer and quicker integration process.