Problem 64 - Calculus Early Transcendentals

Question

Evaluate the following definite integrals. Use Theorem 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 order to evaluate the integral \(\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x\), we used hyperbolic trigonometric identities to rewrite the integrand and applied Theorem 10. We made a variable substitution and evaluated the definite integral, ultimately finding the final answer: \(\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x = \ln \left|\frac{(1+\tanh \frac{\ln 5}{2})(1-\tanh \frac{\ln 9}{2})}{(1-\tanh \frac{\ln 5}{2})(1+\tanh \frac{\ln 9}{2})}\right|\).

Step 1: Rewrite the integrand using hyperbolic trigonometric identities

In order to apply Theorem 10, we first need to rewrite the integrand using hyperbolic identities. Let's start by recalling that \(\cosh^2 x - \sinh^2 x = 1\). Therefore, we can rewrite the denominator as follows: \(4 - \sinh^2 x = (2\cosh x)(2-\cosh x)\) Now, our integral becomes: \(\int_{\ln 5}^{\ln 9} \frac{\cosh x}{(2\cosh x)(2-\cosh x)} d x\)

Step 2: Simplify the integrand

Next, we can simplify the integrand by cancelling out the common factors in the numerator and denominator: \(\int_{\ln 5}^{\ln 9} \frac{\cosh x}{(2\cosh x)(2-\cosh x)} d x= \int_{\ln 5}^{\ln 9} \frac{1}{(2)(2-\cosh x)} d x\)

Step 3: Apply Theorem 10

According to Theorem 10, \(\int \frac{1}{1-\cos 2 u} d u = -\frac{1}{2} \ln \left|\tan(u-\frac{\pi}{4})\right| + C \) In order to apply this theorem to our integral, we need make a substitution: Let \(2u = x\), so \(u = \frac{x}{2}\). Thus, we have \(d u = \frac{1}{2} d x\) Now, we can rewrite our integral as: \(\int_{\ln 5}^{\ln 9} \frac{1}{(2)(2-\cosh x)} d x = 2 \int_{\frac{\ln 5}{2}}^{\frac{\ln 9}{2}} \frac{1}{(2)(2-\cosh 2u)} d u\)

Step 4: Calculate the integral

Applying Theorem 10, we get: \(2 \int_{\frac{\ln 5}{2}}^{\frac{\ln 9}{2}} \frac{1}{(2)(2-\cosh 2u)} d u = -2\left(\frac{1}{2} \ln \left|\frac{1+\tanh u}{1-\tanh u}\right|\right) \Bigg | _{\frac{\ln 5}{2}}^{\frac{\ln 9}{2}}\) Now, evaluate the definite integral: \(-\ln \left|\frac{1+\tanh \frac{\ln 9}{2}}{1-\tanh \frac{\ln 9}{2}}\right| + \ln \left|\frac{1+\tanh \frac{\ln 5}{2}}{1-\tanh \frac{\ln 5}{2}}\right|\)

Step 5: Simplify the answer

Finally, simplify the answer using logarithmic properties: \(-\ln \left|\frac{1+\tanh \frac{\ln 9}{2}}{1-\tanh \frac{\ln 9}{2}}\right| + \ln \left|\frac{1+\tanh \frac{\ln 5}{2}}{1-\tanh \frac{\ln 5}{2}}\right| = \ln \left|\frac{(1+\tanh \frac{\ln 5}{2})(1-\tanh \frac{\ln 9}{2})}{(1-\tanh \frac{\ln 5}{2})(1+\tanh \frac{\ln 9}{2})}\right|\) This is the final answer for the given integral: \( \int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x = \ln \left|\frac{(1+\tanh \frac{\ln 5}{2})(1-\tanh \frac{\ln 9}{2})}{(1-\tanh \frac{\ln 5}{2})(1+\tanh \frac{\ln 9}{2})}\right|\)

Key Concepts

Hyperbolic Trigonometric Identities

When working with integrals involving hyperbolic functions like cosh (for hyperbolic cosine) and sinh (for hyperbolic sine), it's essential to understand their identities. One of the most important identities is \[ \cosh^2 x - \sinh^2 x = 1. \]This identity is similar to the Pythagorean identity in traditional trigonometry. It helps to convert or simplify expressions that include hyperbolic functions. In this exercise, the identity was used to rewrite the denominator of the integrand, transforming \(4 - \sinh^2 x\) into \((2\cosh x)(2-\cosh x)\). Thus, simplifying the integration process. It's useful to familiarize yourself with these identities to identify when they can be applied effectively.

Theorem Application

In calculus, theorems are powerful tools. They provide ready-made solutions for complex integration problems. Here, Theorem 10 was applied to solve the integral. The theorem states:\[ \int \frac{1}{1-\cos 2u} \, du = -\frac{1}{2} \ln \left|\tan(u-\frac{\pi}{4})\right| + C \]In this problem, the theorem's application involved substituting the given integral into a form that fits the theorem. Substitution simplifies the problem by transforming it into a recognizable pattern that a theorem can solve. It’s crucial to correctly identify when and how a theorem can apply, as this can lead to a solution with less computational effort.

Integration Techniques

Integration is a fundamental technique in calculus that helps to find the total accumulation of a function. In this exercise, we used substitution as the main integration technique. Substitution is a method used when an integral takes the form of a known pattern, but requires a change of variable to fit this pattern.When looking at an integral, identify parts that resemble a standard form or theorem, then redefine variables and limits accordingly. In this solution, by setting \(2u = x\) and consequently \(du = \frac{1}{2}dx\), the integral was adjusted to match the conditions of Theorem 10. This transformed the integral into something manageable and allowed solving directly using the theorem.

Logarithmic Simplification

Logarithmic simplification is a key step when solving integrals that result in logarithmic expressions. In the final stages of this problem, the answer required simplifying a logarithmic expression. We started with an expression:\[-\ln \left|\frac{1+\tanh \frac{\ln 9}{2}}{1-\tanh \frac{\ln 9}{2}}\right| + \ln \left|\frac{1+\tanh \frac{\ln 5}{2}}{1-\tanh \frac{\ln 5}{2}}\right| \]By using logarithmic properties, specifically \[ \ln a - \ln b = \ln \frac{a}{b}, \]we simplified the expression to: \[ \ln \left|\frac{(1+\tanh \frac{\ln 5}{2})(1-\tanh \frac{\ln 9}{2})}{(1-\tanh \frac{\ln 5}{2})(1+\tanh \frac{\ln 9}{2})}\right|.\]This step requires a thorough understanding of logarithm rules such as product, quotient, and power rules. These simplifications help in achieving a clearer, more elegant answer and are often needed to format the result consistent with standard mathematical practices.