Problem 64 - Calculus Early Transcendentals

Question

Definite integrals Evaluate the following definite integrals. Use Theorem \(6.10\) to express your answer in terms of logarithms. $$\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x$$

Step-by-Step Solution

Answer

In this problem, we were asked to evaluate the definite integral $$\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4 - \sinh^2 x} dx$$ using Theorem 6.10. We substituted for the hyperbolic functions, changed variables, and performed partial fraction decomposition and integration to ultimately find that the value of the definite integral is $$-\frac{2}{3}(\ln \frac{9}{5}).$$

Step 1: Theorem 6.10

According to Theorem 6.10, we can express hyperbolic functions in terms of logarithms: $$\cosh x = \frac{e^{x} + e^{-x}}{2}$$ $$\sinh x = \frac{e^{x} - e^{-x}}{2}$$ Now, we need to substitute these expressions into the integrand and simplify.

Step 2: Substituting and Simplifying

After substituting for \(\cosh x\) and \(\sinh x\) and simplifying, we get: $$\int_{\ln 5}^{\ln 9} \frac{\frac{1}{2}(e^{x} + e^{-x})}{4 - \frac{1}{4}(e^{x}-e^{-x})^{2}} dx$$ Simplify further: $$\int_{\ln 5}^{\ln 9} \frac{2(e^{x} + e^{-x})}{16 - (e^{x} - e^{-x})^{2}} dx$$

Step 3: Substitution Trick

Let's make a substitution to simplify the integral. Let \(u = e^{x}\). Then, \(du = e^{x} dx\). Also, since \(x = \ln u\), we have \(e^{-x} = \frac{1}{u}\). Now, we re-write the integral using this substitution: $$2\int_{5}^{9} \frac{u + \frac{1}{u}}{16u^{2} - (u^{2} - 2 + \frac{1}{u^{2}})}\ du,\quad\textrm{where}\quad dx = \frac{du}{u}$$ Now, the integral becomes: $$2\int_{5}^{9} \frac{u^{3} + u}{16u^{4} - u^{4} + 2u^{2} - 1} \ du$$

Step 4: Partial Fraction Decomposition

Now, let's perform partial fraction decomposition to simplify this expression further. Using partial fraction decomposition, we can express the given rational function as the sum of two simpler functions: $$ \frac{u^{3} + u}{15u^{4} + 2u^{2} - 1} = \frac{A}{u} + \frac{Bu + C}{u^{2} + 1} + \frac{Du^{3} + Eu^{2} + Fu + G}{(u^{2}+1)^{2}}$$ By solving for the coefficients \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), and \(G\), we find that: $$A = -\frac{1}{3},\quad B = \frac{1}{3},\quad C = 1,\quad D = 0,\quad E = 0,\quad F = 0,\quad G = -\frac{1}{3}$$ The partial fraction decomposition expression becomes: $$ -\frac{1}{3u} + \frac{u + 1}{3(u^{2} + 1)} - \frac{1}{3(u^{2}+1)^{2}}$$ Now, integrate each of these terms separately:

Step 5: Integrating Each Term

$$\int_{5}^{9} 2\left[-\frac{1}{3u} + \frac{u + 1}{3(u^{2} + 1)} - \frac{1}{3(u^{2}+1)^{2}}\right] du$$ Integrate: $$2\left[-\frac{1}{3}(\ln u)\Big|_{5}^{9} + \frac{1}{3}\int_{5}^{9}\frac{u + 1}{u^{2} + 1}du - \frac{1}{3}\int_{5}^{9}\frac{1}{(u^{2} + 1)^{2}} du\right]$$

Step 6: Evaluating Integrals

Integrate each term and evaluate the definite integral: $$2\left[-\frac{1}{3}(\ln 9 - \ln 5) + \frac{1}{3}(\arctan(u))\Big|_{5}^{9} - \frac{1}{3}\int_{5}^{9}\frac{1}{(u^{2} + 1)^{2}} du\right]$$ The remaining integral can be evaluated with substitution \(v = \arctan(u)\): $$-\frac{2}{3}(\ln \frac{9}{5}) + \frac{2}{3}(\arctan(9) - \arctan(5)) - \frac{2}{3}\int_{\arctan(5)}^{\arctan(9)}\frac{1}{(1 + \tan^2(v))} dv$$ Now, the integral of the last term is easily evaluated, as it is a known formula: $$-\frac{2}{3}(\ln \frac{9}{5}) + \frac{2}{3}(\arctan(9) - \arctan(5)) - \frac{2}{3}(v)\Big|_{\arctan(5)}^{\arctan(9)}$$ Finally, we get the value of the definite integral: $$-\frac{2}{3}(\ln \frac{9}{5}) + \frac{2}{3}(\arctan(9) - \arctan(5)) - \frac{2}{3}(\arctan(9) - \arctan(5))$$ So, the definite integral equals: $$-\frac{2}{3}(\ln \frac{9}{5})$$

Key Concepts

Logarithms

Logarithms are fundamentally useful in simplifying expressions and solving integrals, especially those involving exponential and hyperbolic functions. A logarithm is the inverse of an exponential function.
This means if you have an exponential equation like \( a^b = c \), the logarithm helps you find \( b \) through \( b = \log_a(c) \). In definite integrals, especially those evaluated using specific limits, logarithmic expressions often appear due to the properties of exponents.

• Natural Logarithms (denoted as \( \ln \)): Base \( e \) logarithms are the most common in calculus and are crucial in expressing hyperbolic functions and integrals over exponential limits.
• Logarithmic Properties: They simplify multiplications into additions and powers into multiplications which are helpful to perform integration.
• Inverse Functions: Understanding the inverse relationship between logarithms and exponentials is key to evaluating integrals with exponential functions.

Therefore, when integrating functions defined over limits like \( \ln 5 \) to \( \ln 9 \), converting exponential expressions into logarithmic forms can simplify the process significantly.

Theorem 6.10

Theorem 6.10 plays a pivotal role in transforming hyperbolic functions into expressions involving logarithms. This theorem essentially connects hyperbolic functions with exponentials, which can then be manipulated into logarithmic forms. It provides the foundation to express

• Hyperbolic Cosine (\(\cosh x\)): Given by \(\frac{e^x + e^{-x}}{2}\), let's us switch to exponentials.
• Hyperbolic Sine (\(\sinh x\)): Expressed as \(\frac{e^x - e^{-x}}{2}\), which also allows us to use exponentials effectively.

The importance of Theorem 6.10 lies in its ability to rewrite complex hyperbolic expressions into a form that can be easily integrated using known properties of logarithms. This becomes particularly useful in integrals, where the intricacies of hyperbolic functions can often complicate the evaluation process. By transforming hyperbolic terms into exponential terms, and further simplifying with logarithms, we can neatly solve otherwise challenging definite integrals.

Partial Fraction Decomposition

Partial Fraction Decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to integrate. It is especially effective when dealing with polynomial denominators.
Here are the main steps:

• Identify the Proper Form: Ensure that the numerator is of lower degree than the denominator. If not, perform polynomial division first.
• Decompose: Split the rational expression into simpler fractions, identifying each partial fraction’s structure, such as linear terms (e.g., \(A/u\)) or quadratic terms (e.g., \((Bu+C)/(u^2+1)\)).
• Solve for Coefficients: Use algebraic manipulation to find the specific values of coefficients (\(A\), \(B\), \(C\), etc.) by expanding and equating coefficients.

In the given integral, by transforming the expression using partial fraction decomposition, we are able to integrate each simple component separately. This decomposition process often reveals logarithmic and trigonometric integral components that can be solved using basic integral formulas or known results.

Hyperbolic Functions

Hyperbolic functions, like their trigonometric counterparts, describe a certain type of curve known as a hyperbola. These functions include the hyperbolic sine \(\sinh x\) and hyperbolic cosine \(\cosh x\).
Hyperbolic functions are analogous to trigonometric functions, but instead, they are expressed with exponential functions:

• \(\sinh x\): Defined as \((e^x - e^{-x})/2\), resembling sine in its properties but using exponential definitions.
• \(\cosh x\): Given by \((e^x + e^{-x})/2\), similar in form to cosine but leading to different graph shapes and properties.
• Identity: The identity \(\cosh^2 x - \sinh^2 x = 1\) parallels the Pythagorean identity in trigonometry.

In calculus, integrating hyperbolic functions often requires expressing them in exponential terms using the above formulas. This expression simplifies their integration, making use of familiar exponential and logarithmic properties, as demonstrated in the exercise problem. Understanding the behavior and transformations of these functions is crucial when encountering them in definite integrals.