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Chapter 5: Problem 42

Evaluate the definite integrals. $$ \int_{2}^{4} 24 \cot (\pi / x) / x^{2} d x $$

Short Answer

Expert verified
The definite integral evaluates to \( \frac{12}{\pi} \ln(2) \).

Step by step solution

01

Understand the Integral

We are given the integral \( \int_{2}^{4} \frac{24 \cot \left( \frac{\pi}{x} \right)}{x^2} \, dx \). This is a definite integral, meaning we need to evaluate the integral within specific bounds, from \( x=2 \) to \( x=4 \). Our goal is to evaluate this expression to find the area under the curve that the function describes.
02

Perform Substitution

To simplify the integration process, let's use a substitution method. Let \( u = \frac{\pi}{x} \), which implies that \( x = \frac{\pi}{u} \). Then, \( dx = -\frac{\pi}{u^2}\, du \). Also, change the limits: when \( x = 2 \), \( u = \frac{\pi}{2} \), and when \( x = 4 \), \( u = \frac{\pi}{4} \). Substituting these into the integral gives us \( \int_{\pi/2}^{\pi/4} 24 \cot(u) \cdot \left( \frac{u^2}{\pi^2} \right) \cdot \left(-\frac{\pi}{u^2}\right) \, du \).
03

Simplify the Integral

Simplify the expression obtained after substitution: \( \int_{\pi/2}^{\pi/4} -24 \frac{\pi}{\pi^2} \cot(u) \, du \). This simplifies further to \( -\frac{24}{\pi} \int_{\pi/2}^{\pi/4} \cot(u) \, du \). The negative sign will flip the limits of integration.
04

Integrate the New Expression

We now have \( \frac{24}{\pi} \int_{\pi/4}^{\pi/2} \cot(u) \, du \). The integral of \( \cot(u) \) is \( \ln|\sin(u)| + C \). Therefore, evaluate: \[ \frac{24}{\pi} \left[ \ln|\sin(u)| \right]_{\pi/4}^{\pi/2} \].
05

Evaluate the Definite Integral

Substitute the limits into the antiderivative: \( \ln|\sin(\pi/2)| - \ln|\sin(\pi/4)| \). This results in \( \ln(1) - \ln\left(\frac{\sqrt{2}}{2}\right) \). Simplifying, we find \( 0 - (\ln(\sqrt{2}/2)) = -\ln\left(\frac{\sqrt{2}}{2}\right) \).
06

Calculate the Final Result

Simplify \( -\ln\left(\frac{\sqrt{2}}{2}\right) \) to \( \ln(2^{1/2}) = \frac{1}{2}\ln(2) \). Multiply this result by \( \frac{24}{\pi} \) to obtain \[ \frac{24}{\pi} \times \frac{1}{2}\ln(2) = \frac{24}{2\pi} \ln(2) = \frac{12}{\pi} \ln(2) \].
07

Conclusion

The definite integral evaluates to \( \frac{12}{\pi} \ln(2) \), which is the final result of the given integral.

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Key Concepts

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

Substitution Method in Integration
The substitution method in integration is a powerful technique used to simplify complex integrals. It's somewhat like reversing the chain rule you might know from derivatives. The idea is to change variables to make an integral easier to solve.

In this particular exercise, we're working with the integral \( \int\frac{24 \cot(\pi/x)}{x^2}\, dx \). This expression is daunting at first glance, but substitution provides a way to tackle it. Here's why and how we do it:
  • **Identifying the Substitution:** The substitution \( u = \frac{\pi}{x} \) was chosen here because it simplifies the trigonometric function and the fraction. This substitution allows us to express \( x \) and \( dx \) in terms of \( u \), leading to a new, often simpler integral.
  • **Change of Variable:** After defining \( u = \frac{\pi}{x} \), we derive that \( x = \frac{\pi}{u} \) and \( dx = -\frac{\pi}{u^2}\, du \). This transformation changes not only the integral's variable but also its limits, which are crucial since it's a definite integral.
  • **Simplifying the Integral:** By substituting the defined \( u \) into the integral, we simplify the original problem to a form that's more straightforward to integrate.

    Through substitution, integrals that seem challenging become manageable, demonstrating the method's value in calculus.
Trigonometric Integrals
Trigonometric integrals, like the one involving the function \( \cot(u) \), can initially seem intimidating. However, understanding basic trigonometric relationships simplifies the integration process.

Trigonometric functions often involve periodicity and symmetry, which we exploit. Let's take a closer look at how trigonometry aids in integrating these functions:
  • **Basic Identities:** Recognize core identities such as \( \cot(u) = \frac{\cos(u)}{\sin(u)} \), helping transform the integral into simpler logarithmic terms.
  • **Integration Rules:** For instance, integrating \( \cot(u) \) translates to \( \ln|\sin(u)| + C \). This direct result derives from the properties of derivatives of logarithmic functions.
  • **Evaluating with Limits:** In definite integrals, trigonometric identities must comply with the bounds of integration. Here, \( u \) changes from \( \pi/4 \) to \( \pi/2 \), allowing us to determine the values of trigonometric functions at these limits.
  • **Handling the Negative Sign:** During substitution, a negative sign from derivative calculations often flips the integration limits, reflecting the integral symmetrically over the variable range.

    Incorporating trigonometry, particularly for integrands like \( \cot(u) \), offers insight and paves the way for straightforward solutions.
Integration Techniques
Integration is one of calculus' fundamental operations, used to find areas, volumes, and solve differential equations. The technique you apply depends on the form of the integrand. Here's an overview of key techniques with examples:
  • **Substitution:** As illustrated by the given exercise, substitution rearranges variables to simplify the integration. Similar to variable substitution in algebra, it transforms the integral into an evaluateable form.
  • **Integration by Parts:** Utilizing the product rule's reversal, this technique is relevant when dealing with product of functions. Unfortunately, it doesn't apply directly here, but know when it's appropriate for future problems.
  • **Partial Fractions:** For rational functions, decomposing fractions into simpler components is beneficial. Again, not used here but a powerful tool in integrals involving rational expressions.
  • **Trigonometric Substitution:** Sometimes, substituting trigonometric identities simplifies integration, especially in cases involving quadratic expressions under radicals. Its application wasn't required in this problem because the transformation to a trigonometric function was immediate.

    Comprehending when and how to use these techniques equips you with a robust set of tools for tackling various integrals efficiently.

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    Most popular questions from this chapter

    Calculate \(A=\int_{-1}^{1}|x| d x\). Let \(N\) be an even positive integer that is not divisible by \(4 .\) Show that if a uniform partition of order \(N\) is used, then the Midpoint Rule approximation of \(A\) is exact, but the Simpson's Rule approximation is not.

    A function \(f\) is defined on a specified interval \(I=[a, b] .\) Calculate the area of the region that lies between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=\exp (x)-e \quad I=[-1,2] $$

    A function \(f,\) an interval \(I,\) and an even integer \(N\) are given. Approximate the integral of \(f\) over \(I\) by partitioning \(I\) into \(N\) equal length subintervals and using the Midpoint Rule, the Trapezoidal Rule, and then Simpson's Rule. $$ f(x)=x^{3} \quad I=[-1 / 2,7 / 2], N=4 $$

    Income data for three countries are given in the following tables. In each table, \(x\) represents a percentage, and \(L(x)\) is the corresponding value of the Lorenz function, as described in Example \(3 .\) In each of Exercises \(23-27,\) use the specified approximation method to estimate the coefficient of inequality for the indicated country. (The values \(L(0)=0\) and \(L(100)=100\) are not included in the tables, but they should be used.) $$ \begin{array}{|c|r|c|c|c|}\hline x & 20 & 40 & 60 & 80 \\\\\hline L(x) & 5 & 20 & 30 & 55\\\\\hline\end{array}$$ Income Data, Country A $$\begin{array}{|c|c|c|c|}\hline x & 25 & 50 & 75 \\\\\hline L(x) & 15 & 25 & 40 \\\\\hline\end{array}$$ Income Data, Country B $$\begin{array}{|c|r|r|r|r|r|r|r|r|r|}\hline \boldsymbol{x} & 10 & 20 & 30 & 40 & 50 & 60 & 70 & 80 & 90 \\\\\hline \boldsymbol{L}(\boldsymbol{x}) & 4 & 8 & 14 & 22 & 32 & 42 & 56 & 70 & 82 \\\\\hline\end{array}$$ Income Data, Country \(\mathbf{C}\) Country C Simpson's Rule

    Find the area of the region that is bounded by the graphs of \(y=f(x)\) and \(y=g(x)\) for \(x\) between the abscissas of the two points of intersection. $$ f(x)=x^{2}+5 x+9 \quad g(x)=-x^{2}+19 x-15 $$
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