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Chapter 12: Problem 31

Evaluate the indefinite integral.\(\int \tanh x \ln (\cosh x) d x\)

Short Answer

Expert verified
The indefinite integral evaluates to \[ \frac{1}{2} (\ln(\cosh x ) )^2 + C \]

Step by step solution

01

Identify the Function Components

Here, the integral involves the hyperbolic tangent function \( \tanh x \) and the natural logarithm of the hyperbolic cosine function \( \ln (\cosh x ) \).
02

Simplify the Integrand

Recall that \( \tanh x = \frac{\sinh x}{\cosh x} \). Substituting \( \tanh x \) into the integral, we get: \[ \int \frac{\sinh x}{\cosh x} \ln (\cosh x ) dx \]
03

Use Substitution Method

Let \( u = \cosh x \), then \( du = \sinh x dx \). Rewriting the integral in terms of \(u\), we obtain: \[ \int \frac{du}{u} \ln ( u ) \]
04

Solve the Integral

This is a standard integral where the form is \[\int \frac{\ln(u)}{u} du \]. This integral is known to equal: \[ \frac{1}{2} (\ln(u))^2 + C \]
05

Substitute Back the Original Variable

Replace \( u \) with \( \cosh x \): \[ \frac{1}{2} ( \ln(\cosh x ) )^2 + C \]

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Key Concepts

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

Hyperbolic Functions
Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas rather than circles. The core hyperbolic functions include:
  • Sinh (sinh x): Hyperbolic sine
  • Cosh (cosh x): Hyperbolic cosine
  • Tanh (tanh x): Hyperbolic tangent
The hyperbolic tangent function, \(\tanh x\), is defined as:
\[ \tanh x = \frac{\sinh x}{\cosh x} \] This is analogous to the tangent in trigonometry but with hyperbolic functions. Understanding these functions can greatly help in integrating expressions involving hyperbolic terms.
Substitution Method
The substitution method is a powerful technique for solving integrals. It involves substituting a part of the integrand with a new variable, simplifying the integral into an easily recognizable form. In the given exercise, we used the substitution:
\[ u = \cosh x \]
This means that the derivative of \(\cosh x\) with respect to \(x\) is \(\sinh x dx \), so:
\[ du = \sinh x dx \]
Substituting these into the integral, we transformed:
\[ \int \frac{\sinh x}{\cosh x}\ln (\cosh x) dx \] into:
\[ \int \frac{du}{u}\ln (u) \] This simplification helps us solve the integral more easily.
Logarithmic Integration
Logarithmic integration occurs when dealing with integrals that have logarithms in the integrand. For the integral of the form:
\[ \int \frac{\ln(u)}{u} du \] we use a known result:
\[ \int \frac{\ln(u)}{u} du = \frac{1}{2} [\ln(u)]^2 + C \] In the solution, we applied this result directly after substitution. Then, we substituted back the original variable \(u = \cosh x\) to get:
\[ \frac{1}{2} [\ln(\cosh x)]^2 + C \] This shows how the integral simplifies by proper use of substitution and known integral results.

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Most popular questions from this chapter

Express the indefinite integral in terms of an inverse hyperbolic function and as a natural logarithm.\(\int \frac{3 x d x}{\sqrt{x^{4}+6 x^{2}+5}}\)

Express the indefinite integral in terms of an inverse hyperbolic function and as a natural logarithm.\(\int \frac{d x}{4 e^{x}-e^{-x}}\)

Prove that the hyperbolic sine function is an odd function and the hyperbolic cosine function is an even function.

Prove the identities.\(\cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y\)

Prove the identities.(a) \(\sinh 2 x=2 \sinh x \cosh x ;\) (b) \(\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x=2 \sinh ^{2} x+1=2 \cosh ^{2} x-1\)
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