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Chapter 5: Problem 335

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

Short Answer

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

Step by step solution

01

Recognize the Function to be Substituted

The integral given is \(\int \ln(\cos x) \tan x \, dx\). Notice that the derivative of \(\cos x\) involves \(\tan x\), which hints at a possible substitution using \(u = \ln(\cos x)\). This choice simplifies the integral.
02

Find the Derivative for Substitution

To confirm the choice of substitution \(u = \ln(\cos x)\), compute its derivative. We have \(\frac{du}{dx} = \frac{1}{\cos x}(-\sin x) = -\tan x\). This derivative simplifies to \(du = -\tan x \, dx\). Thus, \(-du = \tan x \, dx\).
03

Substitute and Simplify the Integral

Replace \(\ln(\cos x)\) by \(u\) and \(\tan x \, dx\) by \(-du\) in the integral: \(\int \ln(\cos x) \tan x \, dx = \int u (-du)\). Simplify this to \(-\int u \, du\).
04

Integrate the Simplified Function

To solve \(-\int u \; du\), use the power rule for integration. This results in \(-\left(\frac{1}{2}u^2 + C\right)\) where \(C\) is the constant of integration.
05

Back Substitute to Original Variable

Replace \(u\) with \(\ln(\cos x)\) to revert to the original variable. Thus, the indefinite integral becomes \(-\frac{1}{2}(\ln(\cos x))^2 + C\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indefinite Integrals
Indefinite integrals are a fundamental concept in calculus, often referred to as antiderivatives. They represent a family of functions whose derivative yields the original function. When you find an indefinite integral, you are essentially looking for a function that, when differentiated, will return the integrand. Unlike definite integrals, indefinite integrals do not have limits of integration, so their result includes a constant of integration, often denoted as "C."

Key points about indefinite integrals include:
  • Resulting in a general solution that represents a family of functions.
  • Adding a constant of integration "C" because the differentiation of a constant is zero.
  • Played a major role in solving problems related to areas, volumes, and many other physical and theoretical applications.
Understanding indefinite integrals is crucial because they form the backbone of various techniques used in integral calculus.
Integration by Substitution
Integration by substitution is a powerful technique often used to simplify integrals. It involves changing variables to make integrals easier to solve. This technique is akin to the reverse process of the chain rule for differentiation.

Here's how integration by substitution works:
  • Identify a part of the integrand that can be replaced with a new variable to simplify the integral.
  • Compute the derivative of the chosen part to express "dx" in terms of the new variable.
  • Substitute all occurrences in the integrand and "dx" with these expressions.
  • Simplify and integrate the resulting function with respect to the new variable.
  • Finally, substitute back the original variable.
In the context of the problem given, the substitution of \( u = \ln(\cos x) \) makes the integral more manageable by transforming it into a basic polynomial integral.
Integral of Logarithmic Functions
Integrating logarithmic functions can sometimes pose a conceptual challenge, but with approaches like substitution, the task can be simplified. Logarithmic functions like \( \ln(x) \) frequently appear in calculus and have well-defined integral properties.

For example, when integrating \( \ln(x) \, \) common techniques like integration by parts or substitution can be applied. With substitution, identifying another function that simplifies the integral is key.
  • For the integration of \( \ln(\cos x) \), as in the example problem, substitution helps tackle the logarithm coupled with trigonometric expressions.
  • Using substitution, the problem becomes a much simpler task of finding the antiderivative of a polynomial.
  • This ultimately renders solving such integrals manageable and straightforward.
Mastering these techniques for integrating logarithmic functions allows tackling a broad range of more complex calculus problems.

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Most popular questions from this chapter

In the following exercises, use averages of values at the left (L) and right (R) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals. \(\\{(0,2),(1,0),(3,5),(5,5),(6,2),(8,0)\\} \quad\) over \([0,8]\)

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