Question

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

Short answer

The integral is \( \csc x \ln(\csc x) - \csc x + C \).

Step-by-step solution

Identify substitution

To solve the integral \( \int \ln (\csc x) \cot x \, dx \), we first look for a substitution that simplifies the integrand. Notice that the derivative of \( \csc x \) is \( -\csc x \cot x \). We can leverage this derivative to choose a substitution. Let \( u = \csc x \). Then, \( du = -\csc x \cot x \, dx \), which implies that \( -du = \csc x \cot x \, dx \).

Substitute into integral

Substitute \( u = \csc x \) into the integral. Our integrand \( \ln(\csc x) \cot x \ dx \) becomes \( \ln(u) (-du) \). Thus, the integral can be rewritten as:\[ \int \ln(u) \, (-du) = -\int \ln(u) \, du \]

Integrate by parts

We now have \( -\int \ln(u) \, du \). To integrate this, we use integration by parts, which states that \( \int u \, dv = uv - \int v \, du \). Let \( v = \ln(u) \) and \( dv = \frac{1}{u} \, du \). Then, let \( dw = du \) and \( w = u \). The integration becomes:\[ \int \ln(u) \, du = u \ln(u) - \int u \cdot \frac{1}{u} \, du = u \ln(u) - \int du = u \ln(u) - u + C \]

Substitute back to original variable

Replace \( u = \csc x \) back into the result from integration by parts. The solution is:\[ \csc x \ln(\csc x) - \csc x + C \]

Simplify final expression

Recognize that \( \csc x = \frac{1}{\sin x} \) if further simplification in terms of sine is needed. The integral evaluates to:\[ \csc x \ln(\csc x) - \csc x + C \] or equivalently in terms of sine:\[ \frac{1}{\sin x} \ln \left( \frac{1}{\sin x} \right) - \frac{1}{\sin x} + C \] This is the simplified expression for the original integral.

Key concepts

Substitution in Integration

Substitution is a powerful technique in integration and is typically used to simplify an integrand by introducing a new variable. In the case of the integral \( \int \ln (\csc x) \cot x \, dx \), we aim to find a substitution that turns a complex trigonometric expression into something more manageable.

By observing the derivative of \( \csc x \), which is \( -\csc x \cot x \), we can choose our substitution as \( u = \csc x \). The differential, \( du \), then becomes \( -\csc x \cot x \, dx \). This relationship allows us to express \( \cot x \, dx \) in terms of \( du \), specifically, \( -du = \csc x \cot x \, dx \). This substitution transforms the original integral into a simpler form: \( \int \ln(u) \, (-du) = -\int \ln(u) \, du \).

The substitution method is a key tool when dealing with integrals involving products of functions where a function and its derivative appear.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation and is used to integrate products of functions. The rule states: \( \int u \, dv = uv - \int v \, du \). This method is particularly useful for integrals involving logarithmic functions because direct integration can be challenging.

In our problem, we reach a stage where the integral becomes \( -\int \ln(u) \, du \). To apply integration by parts, we identify \( v = \ln(u) \) and \( dv = \frac{1}{u} \, du \), while choosing \( w = u \), \( dw = du \). This leads us to:

• \( uv = u \ln(u) \)
• \( \int v \, du = \int u \cdot \frac{1}{u} \, du = \int du \) which simplifies to \( u \)

Thus, the integral simplifies to \( u \ln(u) - u + C \). Integration by parts effectively reduces the problem by transferring part of the integrand's complexity from one function to another.

Trigonometric Substitution

Trigonometric substitution involves substituting trigonometric identities to simplify integrals. Although it is not the primary technique used here, understanding its principles helps in contexts where integrals involve \( \sqrt{a^2 - x^2} \), \( \sqrt{x^2 - a^2} \), or similar expressions.

In this exercise, recognizing that \( \csc x \) and \( \cot x \) can be rewritten using basic trigonometric identities prepares us for possible simplification steps in more complex integrals. For example, expressing \( \csc x = \frac{1}{\sin x} \) might lead to alternative substitutions or simplifications at final stages.
Keep in mind that trigonometric substitutions are more directly applied when the integral involves radical expressions or specific trigonometric forms.

Logarithmic Composition

Logarithmic composition in integration refers to integrals involving logarithmic functions. These usually lead to complex problems due to the non-algebraic nature of logarithms. In this integral, \( \int \ln(\csc x) \cot x \, dx \), the composition \( \ln(u) \) is effectively handled through integration by parts.

The integral of a logarithm, such as \( \int \ln(u) \, du \), requires us to express it in a form suitable for integration by parts. After applying substitution, calculating \( \int u \, dv = uv - \int v \, du \) facilitates the integration of the logarithmic element. The result \( u \ln(u) - u + C \) exemplifies how logarithmic expressions can be managed using integration by parts within a substitution framework. Remember, logarithms often necessitate special methods due to their inherent complexity.