Problem 40 - Calculus Early Transcendentals

Question

Evaluate each definite integral. \(\int_{\ln 2}^{\ln 3} \operatorname{csch} x d x\)

Step-by-Step Solution

Answer

To summarize, the definite integral of the hyperbolic cosecant function (csch) between the natural logarithm of 2 and the natural logarithm of 3 can be calculated through several steps, including understanding the hyperbolic functions, applying a substitution method, simplifying using an identity, evaluating the integral, and finally simplifying the expression. The final answer for the given definite integral is approximately 0.639.

Step 1: Understand the hyperbolic cosecant function

The hyperbolic cosecant function (csch) is defined as the reciprocal of the hyperbolic sine function (sinh). That is: \(\operatorname{csch}(x)=\frac{1}{\operatorname{sinh}(x)}\) This means that our definite integral becomes: \(\int_{\ln 2}^{\ln 3} \frac{1}{\operatorname{sinh}(x)} dx\)

Step 2: Substitution

To make this integral easier to evaluate, we can use a substitution method. Let \(u=\operatorname{sinh}(x)\). Then, we have to find the derivative of u with respect to x, which is: \(\frac{du}{dx} = \frac{d}{dx}\operatorname{sinh}(x) = \operatorname{cosh}(x)\) Now, we will solve for \(dx\): \(dx = \frac{du}{\operatorname{cosh}(x)}\) We also need to change our limits of integration. When \(x = \ln 2\), \(u = \operatorname{sinh}(\ln 2)\). When \(x = \ln 3\), \(u = \operatorname{sinh}(\ln 3)\). Now, we substitute everything into our integral: \(\int_{\operatorname{sinh}(\ln 2)}^{\operatorname{sinh}(\ln 3)} \frac{1}{u} \frac{du}{\operatorname{cosh}(x)}\)

Step 3: Simplification using identity

Recall the identity: \(\operatorname{cosh}^{2}(x) - \operatorname{sinh}^{2}(x) = 1\) In terms of our variable \(u\), we can rewrite the identity as: \(\operatorname{cosh}^{2}(x) - u^2 = 1\) \(\operatorname{cosh}(x) = \sqrt{u^2 + 1}\) Now we can substitute this into our integral: \(\int_{\operatorname{sinh}(\ln 2)}^{\operatorname{sinh}(\ln 3)} \frac{1}{u} \frac{du}{\sqrt{u^2 + 1}}\)

Step 4: Evaluate the integral

Now we have the integral in a standard form: \(\int_{\operatorname{sinh}(\ln 2)}^{\operatorname{sinh}(\ln 3)} \frac{1}{u\sqrt{u^2 + 1}} du\) The integral of this function is the natural logarithm function with an absolute value argument: \(\int \frac{1}{u\sqrt{u^2 + 1}} du = \ln |u+\sqrt{u^2 + 1}| + C\) Now we apply the limits of integration: \(\ln |u+\sqrt{u^2 + 1}| \Big|_{\operatorname{sinh}(\ln 2)}^{\operatorname{sinh}(\ln 3)} = \left[ \ln(\operatorname{sinh}(\ln 3) +\sqrt{(\operatorname{sinh}(\ln 3))^2 + 1}) - \ln(\operatorname{sinh}(\ln 2) + \sqrt{(\operatorname{sinh}(\ln 2))^2 + 1}) \right]\)

Step 5: Simplify and find the final answer

Now we just need to simplify the expression and find the numerical value: \(\ln(\frac{\operatorname{sinh}(\ln 3) +\sqrt{(\operatorname{sinh}(\ln 3))^2 + 1}}{\operatorname{sinh}(\ln 2) + \sqrt{(\operatorname{sinh}(\ln 2))^2 + 1}})\) After calculating the numerical values, we get: \(\ln(\frac{3+2\sqrt{2}}{2+\sqrt{3}})\approx 0.639\) So, the final solution for the given definite integral is about 0.639.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogous to trigonometric functions but for hyperbolas rather than circles. They are fundamental in calculus for solving various types of integrals, especially those involving exponential growth.
The two main hyperbolic functions are

• sinh (hyperbolic sine): defined as \( ext{sinh}(x) = \frac{e^x - e^{-x}}{2}\)
• cosh (hyperbolic cosine): defined as \( ext{cosh}(x) = \frac{e^x + e^{-x}}{2}\).

In this particular exercise, we are working with the hyperbolic cosecant function, \( ext{csch}(x)\), which is the reciprocal of \( ext{sinh}(x)\).

• That means \( ext{csch}(x) = \frac{1}{ ext{sinh}(x)}\).

These functions behave differently than their trigonometric counterparts, but they follow similar identities, such as the relationship between sinh and cosh, where \( ext{cosh}^2(x) - ext{sinh}^2(x) = 1\). Understanding these properties is critical when evaluating definite integrals that include hyperbolic functions.

Substitution Method

The substitution method is a powerful technique used to simplify integrals by transforming the variables into something easier to integrate.
In our exercise, we used substitution to handle the tricky \(\text{csch}(x)\). By setting \(u = \text{sinh}(x)\), we transformed the integral into a more manageable form.
Here's why substitution helps:

• The original problem had a reciprocal hyperbolic function, \( \frac{1}{\text{sinh}(x)} \), which could be challenging to integrate directly.
• Substituting transforms the integral into \( \frac{1}{u} \) with respect to \( u \), reducing complexity.
• It changes the variable from \( x \) to \( u \), with the differential \( dx = \frac{du}{\text{cosh}(x)} \), further simplifying the process.

Overall, substitution is like changing lanes on a highway when traffic is heavy—to speed up your calculus journey without getting stuck.

Integration Limits

Integration limits are vital in defining a definite integral, telling us from where to where we're calculating the area under the curve. They change during substitution to match the new variable.
In the substitution step, we initially had limits from \( \ln 2 \) to \( \ln 3 \).
Due to the variable substitution as \( u = \text{sinh}(x) \), the limits had to be recalculated:

• For the lower limit: substitute \( x = \ln 2 \) making \( u = \text{sinh}(\ln 2) \).
• For the upper limit: substitute \( x = \ln 3 \) making \( u = \text{sinh}(\ln 3) \).

Replacing these correctly ensures the integral remains consistent and accurately calculated. Not transforming limits after substitution is a common pitfall that can lead to incorrect results.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) (approximately 2.718) is a fundamental constant. It's indispensable in solving integrals involving exponential growth or decay and appears frequently in calculus.

• In our problem, \( \ln \) helps manage the overall growth scale when computing limits.
• The integral simplifies into a form that can be solved using the expression \( \ln |u+\sqrt{u^2 + 1}|\), emphasizing the importance of understanding the dual nature of \(x\) and \(e^x\) relations.
• This transformation not only simplifies the integral but directly connects to calculating hyperbolic functions since they heavily rely on exponential functions.

Natural logarithms thus serve as both a tool for simplification in calculus and a bridge to deeper mathematical concepts like hyperbolic functions.