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Chapter 5: Problem 30

Evaluate the integrals in Exercises \(17-50\) $$ \int \csc \left(\frac{v-\pi}{2}\right) \cot \left(\frac{v-\pi}{2}\right) d v $$

Short Answer

Expert verified
Integral evaluates to \(-2 \csc\left(\frac{v-\pi}{2}\right) + C\).

Step by step solution

01

Identify the integral type

The integral involves the product of the cosecant and cotangent functions of a linear transformation of the variable. The integrand is \( \csc\left(\frac{v-\pi}{2}\right) \cot\left(\frac{v-\pi}{2}\right) \). This is a standard integral that can be solved using substitution.
02

Make a substitution

Let \( u = \frac{v-\pi}{2} \). Then, \( \frac{du}{dv} = \frac{1}{2} \) or equivalently, \( dv = 2 \, du \). This substitution simplifies the integral and changes the limits accordingly.
03

Rewrite the integral

Substitute \( u \) and \( dv = 2 \, du \) into the integral. The integral now becomes \( \int \csc(u) \cot(u) \cdot 2 \, du = 2 \int \csc(u) \cot(u) \, du \).
04

Solve the standard integral

The integral \( \int \csc(u) \cot(u) \, du \) is a standard integral. It evaluates to \(-\csc(u) + C\), where \(C\) is the constant of integration. Thus, the integral becomes \( 2(-\csc(u) + C) = -2 \csc(u) + 2C \).
05

Substitute back the original variable

Replace \( u \) back with \( \frac{v-\pi}{2} \), according to our substitution. The solution in terms of \( v \) is \(-2 \csc\left(\frac{v-\pi}{2}\right) + C'\), where \( C' \) absorbs the factor of 2 in \( 2C \).

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Key Concepts

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

Integration Techniques
Understanding integration techniques is crucial in calculus, especially when dealing with complex functions. Sometimes, integration requires manipulating the function into a more recognizable form. This can often involve:
  • Standard rules and formulas, such as power rule, exponential and trigonometric integrals.
  • Transformation of the integrand by substitution or integration by parts.
In our original problem, recognizing that the integrand \( \csc\left(\frac{v-\pi}{2}\right) \cot\left(\frac{v-\pi}{2}\right) \) matches a standard integral form helps simplify the problem. This allows the use of substitution to transform the function into a simpler equivalent.
Trigonometric Identities
Trigonometric identities simplify complex trigonometric expressions, especially when integrating. Common identities like Pythagorean identities, double angle, and product-to-sum formulas play a significant role.In the context of our integral, understanding how \( \csc \) and \( \cot \) relate to sine and cosine helps. Recall that:
  • \( \csc(x) = \frac{1}{\sin(x)} \)
  • \( \cot(x) = \frac{\cos(x)}{\sin(x)} \)
These identities illustrate how the integrand can be composed as a fraction of sine functions. Recognizing these patterns can make substitution or further simplification feasible in the solution process.
Substitution Method
The substitution method is a powerful tool in integral calculus. It simplifies integration by transforming the integrand into an easier form. Here's how it works effectively:
  • Select a substitution that simplifies the expressions. Often, this involves setting \( u \) equal to a part of the integrand.
  • Differentiate \( u \) to express \( du \) in terms of the original variable. This involves finding derivatives and understanding how variables transform.
In our exercise, setting \( u = \frac{v-\pi}{2} \) transforms the integral into one involving \( \csc(u) \cot(u) \), a standard form that simplifies directly to \( -\csc(u) \). By reversing the substitution at the end, we express the solution in terms of the original variable, obtaining \( -2 \csc\left(\frac{v-\pi}{2}\right) + C' \). This showcases how substitution can demystify seemingly complex integrals and streamline the integration process.

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Most popular questions from this chapter

In Exercises \(95-98,\) use a CAS to perform the following steps: $$ \begin{array}{l}{\text { a. Plot the functions over the given interval. }} \\\ {\text { b. Partition the interval into } n=100,200, \text { and } 1000 \text { subintervals }} \\ {\text { of equal length, and evaluate the function at the midpoint }} \\ {\text { of each subinterval. }}\\\ {\text { c. compute the average value of the function values generated in }} \\\ {\text { part (b). }} \\ {\text { d. Solve the equation } f(x)=(\text { average value) for } x \text { using the average}} \\ {\text { value calculated in part (c) for the } n=1000 \text { partitioning. }}\end{array} $$ $$ f(x)=\sin x \quad \text { on } \quad[0, \pi] $$

Use the Substitution Formula in Theorem 7 to evaluate the integrals in Exercises \(1-24 .\) $$ \int_{-1}^{1} \frac{5 r}{\left(4+r^{2}\right)^{2}} d r \quad \text { b. } \int_{0}^{1} \frac{5 r}{\left(4+r^{2}\right)^{2}} d r $$

Revenue from marginal revenue Suppose that a company's marginal revenue from the manufacture and sale of eggbeaters is $$\frac{d r}{d x}=2-2 /(x+1)^{2},$$ where \(r\) is measured in thousands of dollars and \(x\) in thousands of units. How much money should the company expect from a production run of \(x=3\) thousand eggbeaters? To find out, integrate the marginal revenue from \(x=0\) to \(x=3\) .

In Exercises \(75-78\) , let \(F(x)=\int_{a}^{x} f(t) d t\) for the specified function \(f\) and interval \([a, b] .\) Use a CAS to perform the following steps and answer the questions posed. \begin{equation} \begin{array}{l}{\text { a. Plot the functions } f \text { and } F \text { together over }[a, b] \text { . }} \\ {\text { b. Solve the equation } F^{\prime}(x)=0 . \text { What can you see to be true about }} \\ {\text { the graphs of } f \text { and } F \text { at points where } F^{\prime}(x)=0 \text { . Is your observation }} \\ {\text { borne out by Part } 1 \text { of the Fundamental Theorem coupled }} \\ {\text { with information provided by the first derivative? Explain your }} \\ {\text { answer. }}\\\\{\text { c. Over what intervals (approximately) is the function } F \text { increasing }} \\\ {\text { and decreasing? What is true about } f \text { over those intervals? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { d. Calculate the derivative } f^{\prime} \text { and plot it together with } F . \text { What can }} \\ {\text { you see to be true about the graph of } F \text { at points where } f^{\prime}(x)=0 ?} \\\ {\text { Is your observation borne out by Part } 1 \text { of the Fundamental Theorem? }} \\ {\text { Explain your answer. }}\end{array} \end{equation} $$f(x)=2 x^{4}-17 x^{3}+46 x^{2}-43 x+12,\left[0, \frac{9}{2}\right]$$

Each of the following functions solves one of the initial value problems in Exercises \(55-58 .\) Which function solves which problem? Give brief reasons for your answers. \begin{equation} \begin{array}{ll}{\text { a. } y} & {=\int_{1}^{x} \frac{1}{t} d t-3 \quad\quad \text { b. } y=\int_{0}^{x} \sec t d t+4} \\ {\text { c. } y} & {=\int_{-1}^{x} \sec t d t+4 \quad \text { d. } y=\int_{\pi}^{x} \frac{1}{t} d t-3}\end{array} \end{equation} $$y^{\prime}=\sec x, \quad y(-1)=4$$
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