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The integrals in Exercises \(1-34\) converge. Evaluate the integrals without using tables. $$\int_{1}^{2} \frac{d s}{s \sqrt{s^{2}-1}}$$

Short Answer

Expert verified
The integral evaluates to \( \frac{\pi}{3} \).

Step by step solution

01

Identify the Integral Form

The integral \( \int \frac{ds}{s \sqrt{s^2 - 1}} \) matches the form of a common integral type: the inverse trigonometric substitution. Specifically, it suggests the use of the substitution related to \( \sec \theta \) since \( \sqrt{s^2-1} \) resembles \( \sqrt{\sec^2\theta - 1} = \tan \theta \).
02

Substitute the Variable

Let \( s = \sec \theta \). Then \( ds = \sec \theta \tan \theta \, d\theta \) and \( \sqrt{s^2 - 1} = \sqrt{\sec^2 \theta - 1} = \tan \theta \). Replace these into the integral, changing limits as well: when \( s = 1, \theta = 0 \) and when \( s = 2, \theta = \pi/3 \).
03

Rewrite the Integral

Substituting, the integral becomes \( \int \frac{\sec \theta \tan \theta \, d\theta}{\sec \theta \tan \theta} = \int d\theta \). This simplifies to the integration of \( \theta \) between the bounds \( 0 \) and \( \pi/3 \).
04

Evaluate the Integral

Integration of \( d\theta \) is \( \theta + C \). Evaluating from \( 0 \) to \( \pi/3 \), we find \( \left. \theta \right|_0^{\pi/3} = \frac{\pi}{3} - 0 = \frac{\pi}{3} \).

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Key Concepts

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

Inverse Trigonometric Substitution in Integration
In mathematics, inverse trigonometric substitution is a powerful technique often used to solve integrals involving expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{x^2 - a^2} \), or \( \sqrt{x^2 + a^2} \). These expressions can be transformed using the identities of inverse trigonometric functions. For instance, if you have
  • \( \sqrt{x^2 - 1} \), then \( x = \sec \theta \) or \( x = \csc \theta \)
  • \( \sqrt{1 - x^2} \), then \( x = \sin \theta \) or \( x = \cos \theta \)
  • \( \sqrt{x^2 + 1} \), then \( x = \tan \theta \) or \( x = \cot \theta \)
By substituting these identities, the complex radicals become simpler trigonometric expressions. This substitution not only simplifies the integral but also transforms it into one that is easier to evaluate.
In our original exercise, we encountered \( \int \frac{ds}{s \sqrt{s^2 - 1}} \), which prompted substitution with \( s = \sec \theta \), transforming the integral into a much simpler form. This strategy effectively changed a challenging expression into one that required integrating a basic function like \( \theta \).
Understanding Definite Integrals
Definite integrals represent the area under a curve between two points along the x-axis. They are denoted with an integral sign and limits of integration, such as \( \int_a^b f(x) \, dx \). The result gives the net area, accounting for areas above and below the x-axis.
The integral in the given exercise, \( \int_{1}^{2} \frac{d s}{s \sqrt{s^{2}-1}} \), is a definite integral, evaluated specifically between the bounds of 1 and 2. This means we compute the integral as a difference of the antiderivative evaluated at these two points:
  • Calculate the antiderivative of the function within the integral.
  • Evaluate the antiderivative at the upper and lower bounds.
  • Subtract the value at the lower bound from the value at the upper bound.
The transition of limits during substitution is crucial. Inverse trigonometric substitutions require us to adjust these limits accordingly. In our scenario, changing interval endpoints from \( s \)-values to \( \theta \)-values facilitates a more straightforward computation.
Exploring Various Integration Techniques
Integration techniques are diverse strategies used to find the antiderivative of a function. Some common methods include
  • substitution,
  • integration by parts,
  • partial fraction decomposition,
  • and trigonometric substitutions, such as those involving inverse trig functions.
Each technique serves particular forms of integrals more effectively. For integrals involving powers and radicals, inverse trigonometric substitution is a common choice because it simplifies the function.
Substitution, broadly speaking, involves replacing one expression to make integration more manageable. Specifically, inverse trigonometric substitution uses identities of sine, cosine, tangent, etc., converting complex expressions into simpler trigonometric forms.
In our specific case, the use of integration techniques like substitution with \( s = \sec \theta \) reduced the integral to basic terms, effectively simplifying what might appear complicated at first glance. Mastery of these techniques enriches your toolkit in solving various integral calculus problems efficiently.

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Most popular questions from this chapter

The infinite paint can or Gabriel's horn As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x$$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\) -axis, diverges also. By comparing the two integrals, we see that, for every finite value \(b>1\) . $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x$$ However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. \begin{equation} \begin{array}{l}{\text { a. Calculate it. }} \\ {\text { b. This solid of revolution is sometimes described as a can that }} \\ {\text { does not hold enough paint to cover its own interior. Think }} \\ {\text { about that for a moment. It is common sense that a finite }}\\\ {\text { amount of paint cannot cover an infinite surface. But if we fill }} \\\ {\text { the horn with paint (a finite amount), then we will have cov- }} \\\ {\text { ered an infinite surface. Explain the apparent contradiction. }}\end{array} \end{equation}

Find the values of \(p\) for which each integral converges. \begin{equation} \text { a. }\int_{1}^{2} \frac{d x}{x(\ln x)^{p}} \quad \text { b. } \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \end{equation}

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