Question

Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms. $$\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x$$

Short answer

Based on the given information and step by step solution, provide a short answer to the problem: The definite integral, $$\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x$$, is equal to $$-\frac{1}{2} \ln\left|\frac{5-\cosh^2{(\ln 9)}}{5-\cosh^2{(\ln 5)}}\right|$$ when evaluated using Theorem 7.7, also known as the "hyperbolic substitution" technique.

Step-by-step solution

Perform the substitution u = cosh x

Let us consider the substitution $$u = \cosh x$$. We need to find its derivative with respect to $$x$$: $$\frac{d u}{d x} = \frac{d (\cosh x)}{d x} = \sinh x$$ Now, we change the integral with the substitution: $$\int_{\ln 5}^{\ln 9} \frac{u}{4-\sinh ^{2} x} d x$$

Solve for \(\sinh^2{x}\) in terms of u

Using the identity $$\cosh^2 x - \sinh^2 x = 1$$, we can solve for \(\sinh^2 x\) in terms of $$u$$: $$\sinh^2 x = u^2 - 1$$ Now we can substitute this expression for \(\sinh^2{x}\) back into the integral: $$\int_{\ln 5}^{\ln 9} \frac{u}{4-(u^2 - 1)} d x$$

Simplify the integrand and change the limits of integration

Let's simplify the integrand first: $$\int_{\ln 5}^{\ln 9} \frac{u}{4-(u^2 - 1)} d x = \int_{\ln 5}^{\ln 9} \frac{u}{5-u^2} d x$$ Now we need to change the limits of integration. Since the substitution is $$u = \cosh x$$, then the new limits will be $$\cosh{(\ln 5)}$$ and $$\cosh{(\ln 9)}$$: $$\int_{\cosh{(\ln 5)}}^{\cosh{(\ln 9)}} \frac{u}{5-u^2} d x$$

Perform the integration

Now, we can integrate the function in terms of $$u$$: $$\int \frac{u}{5-u^2} d x = -\frac{1}{2} \ln |5-u^2| + C$$

Compute the definite integral

Now, we can compute the definite integral by substituting the limits of integration: $$\left[-\frac{1}{2} \ln |5-u^2| \right]_{\cosh{(\ln 5)}}^{\cosh{(\ln 9)}} = -\frac{1}{2} \ln\left|\frac{5-\cosh^2{(\ln 9)}}{5-\cosh^2{(\ln 5)}}\right|$$ Thus, $$\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x = -\frac{1}{2} \ln\left|\frac{5-\cosh^2{(\ln 9)}}{5-\cosh^2{(\ln 5)}}\right|$$

Key concepts

Hyperbolic Functions

Hyperbolic functions are analogues of the ordinary trigonometric, or circular, functions. They are widely used in many areas of mathematics, including calculus. The most commonly used hyperbolic functions are the hyperbolic sine and cosine, denoted as \( \sinh x \) and \( \cosh x \), respectively. These functions are defined using the exponential function \( e^x \):

\( \sinh x = \frac{e^x - e^{-x}}{2} \) and \( \cosh x = \frac{e^x + e^{-x}}{2} \).

One crucial property that often comes in handy when solving integrals involving hyperbolic functions is the identity \( \cosh^2 x - \sinh^2 x = 1 \), which is analogous to the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \) for trigonometric functions. This property was used in the step-by-step solution to express \( \sinh^2 x \) in terms of \( u \) after the substitution \( u = \cosh x \) was made.

Integral Substitution

Integral substitution, or u-substitution, is a powerful technique in calculus for solving integrals. It's essentially the reverse process of the chain rule for differentiation. The main idea is to choose a new variable \( u \) that simplifies the integral, often by making it recognizable or removing complexity.

When applying u-substitution, the following steps are generally involved:

• Identify a portion of the integrand to substitute with \( u \).
• Compute the derivative of \( u \) with respect to \( x \) to find \( \frac{du}{dx} \), and then solve for \( dx \).
• Substitute \( u \) and \( dx \) into the integral and adjust the limits of integration if necessary.
• Solve the new integral in terms of \( u \).
• If the original problem was a definite integral, evaluate at the new limits, else if it was an indefinite integral, convert back to the original variable \( x \) after integration.

In the provided solution, \( u = \cosh x \) was selected, which led to a simpler integrand and ultimately to a form that was easier to integrate.

Logarithmic Expression

A logarithmic expression is one containing a logarithm, which is the inverse operation to exponentiation. In calculus, logarithms frequently appear as a result of integrating functions that involve division. The logarithm of a product, quotient, power, or root can often be simplified using the properties of logarithms.

The base-\( e \) logarithm, denoted as \( \ln(x) \) or natural log, is particularly prevalent. When evaluating definite integrals resulting in logarithmic expressions, as in the step-by-step example, we apply the fundamental theorem of calculus and compute the difference of the antiderivative evaluated at the upper and lower limits of integration.

In the context of the original problem, the integral resolved into the log of a difference of squares, which is directly related to the properties of hyperbolic functions and their connection to exponential expressions. It’s essential to handle absolute values properly since logarithms are only defined for positive arguments in real numbers. The exercise illustrates the importance of understanding properties of logarithms in simplifying and interpreting final results from definite integrals.