Question

In the following exercises, find each indefinite integral by using appropriate substitutions. $$ \int \ln (\cos x) \tan x d x $$

Short answer

The integral is \(-\frac{1}{2} (\ln(\cos x))^2 + C\).

Step-by-step solution

Identify the Substitution

To simplify the integral \( \int \ln (\cos x) \tan x \, dx \), we should identify a portion of the integrand as a potential substitution. Notice that the derivative of \( \ln(\cos x) \) is related to \( \tan x \). Given this, let us consider the substitution \( u = \ln(\cos x) \).

Differentiate and Substitute

Find the derivative of \( u = \ln(\cos x) \). Using the chain rule, \( \frac{du}{dx} = \frac{1}{\cos x}(-\sin x) = -\tan x \). Therefore, \( du = -\tan x \, dx \). Rearrange this to express \( dx \) in terms of \( du \): \( dx = -\frac{du}{\tan x} = -du \). Substitute \( u \) and \( dx \) into the integral.

Rewrite the Integral

Using the substitution \( u = \ln(\cos x) \) and \( dx = -du \), the integral becomes: \[ \int \ln(\cos x) \tan x \, dx = \int u(-du) \]. This simplifies to \( -\int u \, du \).

Integrate

Now integrate \( -\int u \, du \). The integral of \( u \, du \) is \( \frac{u^2}{2} \), so \( -\int u \, du = -\frac{u^2}{2} + C \), where \( C \) is the constant of integration.

Back Substitution

Replace \( u \) with its original expression \( \ln(\cos x) \) to go back to the variable \( x \): \[ -\frac{1}{2} (\ln(\cos x))^2 + C \]. This represents the indefinite integral in terms of \( x \).

Key concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify the process of integrating complex expressions. This method revolves around the idea of replacing part of the integrand with a new variable, typically denoted as \( u \).
This substitution should be chosen such that its derivative appears in the integrand, making the integration process more straightforward.In the exercise, the integral \( \int \ln(\cos x) \tan x \, dx \) was simplified through a substitution. We chose \( u = \ln(\cos x) \). This choice was strategic because the derivative of \( u \), using the chain rule, involves \( \tan x \), which is also a part of the original integrand. This makes the substitution effective and the integral easier to solve.Once the substitution is made, the integral can be transformed into a new integral in terms of \( u \), allowing you to handle it with methods for simpler expressions.

Chain Rule

The chain rule is a fundamental differentiation technique used to find the derivative of composite functions. It is essential when dealing with functions like \( f(g(x)) \). The chain rule states that the derivative of \( f(g(x)) \) with respect to \( x \) is \( f'(g(x)) \cdot g'(x) \).
In the context of the given exercise, when substituting \( u = \ln(\cos x) \), applying the chain rule allows us to find \( \frac{du}{dx} \). The derivative of \( \, \ln(\cos x) \) can be expressed using the chain rule as:

• \( \frac{d}{dx} \ln(\cos x) = \frac{1}{\cos x} \cdot -\sin x \)
• Which simplifies to \( -\tan x \)

Using the chain rule simplifies the process of finding derivatives of complex expressions, which in turn helps in the substitution process for integration. Without it, calculating \( \frac{du}{dx} \) would be far more complex.

Integration Techniques

There are several integration techniques used to solve indefinite integrals, each suitable for different kinds of problems. These techniques include substitution, integration by parts, partial fraction decomposition, and more. Choosing the right method is crucial to simplifying the integration process.In our problem, the substitution method provided a clear path to simplify the integral \( \int \ln(\cos x) \tan x \, dx \). This approach involves changing the variable and the limits accordingly if it were a definite integral. The reformulated integral in terms of \( u \) becomes \( -\int u \, du \), which is simpler to integrate.
Knowing these techniques and when to apply them is fundamental for effective problem-solving in calculus:

• The substitution method is particularly useful when you spot a function and its derivative in the integrand.
• Integration by parts is ideal for products of functions.
• Partial fraction decomposition is great for rational functions.

Having a toolkit of integration techniques empowers you to tackle diverse integration challenges with confidence.