Problem 353 - Calculus Volume 1

Question

In the following exercises, use appropriate substitutions to express the trigonometric integrals in terms of compositions with logarithms. $$\int \ln (\csc x) \cot x d x$$

Step-by-Step Solution

Answer

Integral is \(-\frac{(\ln(\sin x))^2}{2} + C\).

Step 1: Identify Trigonometric Identity

Recognize that \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \). This helps us rewrite \( \ln(\csc x) \) using the identity \( \ln(\csc x) = \ln\left(\frac{1}{\sin x}\right) = -\ln(\sin x) \). Substitute this into the integral.

Step 2: Substitution and Simplification

Substitute \( -\ln(\sin x) \) in place of \( \ln(\csc x) \) into the integral. The integral becomes: \[ \int -\ln(\sin x) \cdot \cot x \, dx = - \int \ln(\sin x) \cdot \frac{\cos x}{\sin x} \, dx \].

Step 3: Set Up Substitution

Let \( u = \ln(\sin x) \). Then, \( \frac{du}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x \). This means \( du = \cot x \, dx \). Substitute \( u \) and \( du \) in the integral.

Step 4: Change Variables

Substitute into the integral: \( -\int u \, du \).

Step 5: Integrate

Integrate with respect to \( u \): \[ -\int u \, du = -\left( \frac{u^2}{2} \right) + C \].

Step 6: Substitute Back u

Substitute back \( u = \ln(\sin x) \). We have: \(-\frac{(\ln(\sin x))^2}{2} + C\).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in simplifying expressions involving trigonometric functions. They help in transforming complex trigonometric expressions into more manageable forms by utilizing known relationships between different trigonometric functions. For instance, in solving the integral \( \int \ln (\csc x) \cot x \, dx \), we can use the identities \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \). These identities enable us to rewrite \( \ln(\csc x) \) as \( \ln\left(\frac{1}{\sin x}\right) = -\ln(\sin x) \). This transformation is crucial because it simplifies the original integral, making it easier to work with using integration techniques.

Integration by Substitution

Integration by substitution is a method used to simplify the process of integration by changing variables. It involves substituting a part of the integrand (the function to be integrated) with a single variable. This change makes the integral simpler to solve. In our problem, we use substitution to tackle the integral \( \int \ln(\sin x) \cdot \frac{\cos x}{\sin x} \, dx \). By recognizing that \( u = \ln(\sin x) \), we can derive \( du = \cot x \, dx \) from the derivative of \( u \) relative to \( x \). By substituting \( u \) and \( du \) into the integral, we transform it into a straightforward integral \( -\int u \, du \), which is much easier to evaluate.

Logarithmic Functions

Logarithmic functions are an essential component of calculus, particularly in integration. The integral involving a logarithmic function can often be simplified by expressing the logarithm in terms of other functions. In our example, the function \( \ln(\csc x) \) is initially present. This logarithmic function gets simplified through the trigonometric identity \( \ln(\csc x) = -\ln(\sin x) \). Thus, transforming complex integrals involving logarithms into more straightforward forms that can be easily integrated using techniques like substitution. Appreciating how logarithmic functions interact with trigonometric identities is crucial for solving integrals efficiently.

Cotangent Function

The cotangent function \( \cot x \) is one of the six primary trigonometric functions. It is the reciprocal of the tangent function and can be expressed as \( \cot x = \frac{\cos x}{\sin x} \). Cotangent function often appears in various calculus problems, including trigonometric integrals. In our exercise, \( \cot x \) is an integral part of the expression \( \ln(\csc x) \cot x \). Recognizing \( \cot x \) allows us to substitute successfully when finding \( du \) as \( du = \cot x \, dx \). Understanding the cotangent function's behavior and its relationships with other trigonometric functions is beneficial for integrals involving trigonometric expressions.