Problem 22 - Calculus

Question

Find the indefinite (or definite) integral. $$ \int \cot 3 x d x $$

Step-by-Step Solution

Answer

\( \int \cot(3x) \, dx = \frac{1}{3} \ln |\sin(3x)| + C \)

Step 1: Rewrite Cotangent Function

To solve the integral of \( \cot(3x) \), we first rewrite the cotangent function in terms of sine and cosine. Remember that \( \cot(x) = \frac{\cos(x)}{\sin(x)} \). Thus, we can express the integral as: \[ \int \cot(3x) \, dx = \int \frac{\cos(3x)}{\sin(3x)} \, dx \]

Step 2: Use Substitution Method

To integrate \( \frac{\cos(3x)}{\sin(3x)} \), we use the substitution method. Let \( u = \sin(3x) \). Then, the derivative \( du = 3\cos(3x) \, dx \) or \( \frac{du}{3} = \cos(3x) \, dx \). Substitute \( u \) and \( du \) into the integral:\[ \int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{1}{u} \, du \]

Step 3: Integrate Using Logarithmic Rule

The integral \( \int \frac{1}{u} \, du \) is a standard logarithmic integral. This integrates to \( \ln |u| \). So, we have:\[ \frac{1}{3} \ln |u| + C \]

Step 4: Substitute Back the Original Variable

Now, replace \( u \) with the original trigonometric substitution \( \sin(3x) \) to revert back to the variable \( x \): \[ \frac{1}{3} \ln |\sin(3x)| + C \] This represents the indefinite integral of \( \cot(3x) \).

Step 5: Consolidate the Solution

The final answer which represents the indefinite integral of \( \int \cot(3x) \, dx \) is:\[ \frac{1}{3} \ln |\sin(3x)| + C \] where \( C \) is the constant of integration.

Key Concepts

Indefinite Integral

In calculus, an indefinite integral is the antiderivative of a function. It represents a collection of functions whose derivative is the original function. Unlike definite integrals, which compute a numerical value, indefinite integrals yield a function plus a constant. This constant, often denoted as "C," is significant because the process of differentiation loses constant terms. Hence, when finding an indefinite integral, always include this constant of integration.

• Denotes an entire family of functions.
• Always include the constant of integration, "C," in your answer.

For example, integrating a simple function like \( f(x) = x^2 \) yields \( F(x) = \frac{x^3}{3} + C \). No matter the complexity of the function, these properties remain unchanged.

Integration by Substitution

Integration by substitution is a powerful technique in integral calculus. It's akin to the reverse process of the chain rule in differentiation. This method involves changing variables to simplify the integration process. By making a proper substitution, one can evaluate more complex integrals with greater ease.

When you encounter a composite function, use substitution to simplify:

• Select a substitution (e.g., \( u = g(x) \)) that simplifies the integral.
• Determine the differential, \( du \), in terms of \( dx \).
• Rewrite the integral in terms of \( u \).
• Integrate with respect to \( u \).
• Replace \( u \) with the original expression to revert to the initial variable.

In the provided example, substitution transformed the integral of \( \cot(3x) \) into a simpler form by letting \( u = \sin(3x) \). This substitution streamlined the process and made integration feasible.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the involved variables. They are very useful when solving calculus problems, particularly when dealing with integrals of trigonometric functions.

In the problem, the identity for cotangent, \( \cot(x) = \frac{\cos(x)}{\sin(x)} \), was used:

• Helps to rewrite trigonometric integrals into a form amenable to integration.
• Enables recognition of patterns that can simplify the integral significantly.

Using trigonometric identities, such as transforming \( \cot(3x) \) to \( \frac{\cos(3x)}{\sin(3x)} \), is often the first step in managing complex integrals, allowing for further methods like substitution to be applied.

Logarithmic Integration

Logarithmic integration is a technique that arises when solving integrals involving functions of the form \( \frac{1}{x} \). The result of these integrals involves the natural logarithm, \( \ln \). This type of integration is straightforward once the function is set up correctly.
For example, the integral \( \int \frac{1}{u} \, du \) yields \( \ln|u| + C \). This arises commonly when using substitution to transform the original function.

• Recognize that \( \int \frac{1}{x} \, dx = \ln|x| + C \).
• Ensure that the absolute value is used because logarithms of negative numbers aren't defined in the real number system.
• Use transformations to apply this rule in broader integration contexts.

In our problem, the substitution led to an integral that fit the logarithmic form, allowing us to find the integral in terms of \( \ln|\sin(3x)| \). This approach highlights the versatility and importance of logarithmic integration in calculus.