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Chapter 8: Problem 49

Some integrals do not require integration by parts. $$\int_{2 / \sqrt{3}}^{2} t \sec ^{-1} t d t$$

Short Answer

Expert verified
The integral evaluates to a numeric value of approximately \(1.60 \).

Step by step solution

01

Simplify the Integrand

First, observe the integrand: it is \( t \sec^{-1}(t) \). Here, \( \sec^{-1}(t) \) is the inverse secant function. To facilitate the integration, we will consider using substitution to simplify \( \sec^{-1}(t) \). Let \( x = \sec^{-1}(t) \), then \( t = \sec(x) \), and the differential becomes \( dt = \sec(x)\tan(x)dx \).
02

Change the Limits of Integration

Since \( t = \sec(x) \) and the original integration bounds are from \( 2/\sqrt{3} \) to \( 2 \), we need to change these bounds into \( x \)-bounds to match our substitution. When \( t = 2/\sqrt{3} \), \( \sec^{-1}(t) = \sec^{-1}(2/\sqrt{3}) = \pi/6 \). When \( t = 2 \), \( \sec^{-1}(t) = \sec^{-1}(2) = \pi/3 \). Thus, the new integral in terms of \( x \) will be from \( \pi/6 \) to \( \pi/3 \).
03

Substitute and Simplify

Substitute \( t = \sec(x) \) and \( dt = \sec(x)\tan(x) \, dx \) into the integral. The expression \( t \sec^{-1}(t) \) becomes \( \sec(x) \cdot x \), and multiply by \( \, dt = \sec(x)\tan(x) \, dx \). This results in the integral \( \int_{\pi/6}^{\pi/3} x \sec(x) \tan(x) \, dx \).
04

Evaluate the Integral

The expression becomes \( \int_{\pi/6}^{\pi/3} x \sec(x) \tan(x) \, dx \). Upon recognition, this is a situation for basic substitution and simplification as \( u = \sec(x) \), meaning \( du = \sec(x) \tan(x) dx \). The integral simplifies to \( \int x \, du \) with \( x = u \), leading to \( \int u^2 \, du \). We find this integral evaluates to \([x u^2/2]\) from \( \pi/6 \) to \( \pi/3 \).
05

Calculate the Numeric Result

Substitute back in the terms and calculate: Substitute back any constants and evaluate the remaining integral expression by solving \( [x \sec^2(x)/2]_{\pi/6}^{\pi/3} \). Substituting the limits gives: \( [\pi/2 \cdot 4] - [\pi/6 \cdot 4/3] \), which results in \( (2\pi - \pi/9) \). Simplifying gives the numeric result of the definite integral.

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a powerful technique often used in calculus to simplify integrals involving square roots of quadratic expressions or inverse trigonometric functions. By making a substitution with a trigonometric function, you can transform the integral into a more workable form.
  • For example, when dealing with an integral that includes \(\sec^{-1}(x)\), a typical substitution is \(x = \sec(\theta)\). This transforms the variable from an algebraic to a trigonometric one.
  • This change of variable benefits from trigonometric identities, aiding in simplifying expressions like \(\sec^2(x) - 1 = \tan^2(x)\).
  • The substitution simplifies the integration process by turning complex algebraic expressions into simpler trigonometric integrals.
This method allows us to easily differentiate and integrate expressions that are otherwise complicated. In our exercise, by letting \(x = \sec^{-1}(t)\), transformations of limits and the integrand become feasible, making the complex math more manageable.
Inverse Trigonometric Functions
Inverse trigonometric functions are the inverses of the basic trigonometric functions such as sine, cosine, and tangent. These functions are essential in integration when dealing with arc lengths, angles, and other trigonometric measurements.
  • Functions like \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and particularly \(\sec^{-1}(x)\) come into play when a direct integration approach is not easily applicable.
  • These functions often appear in integration problems where the variable of integration is expressed in terms of trigonometric measures, especially in inverse relationships.
  • Understanding the derivatives of inverse trigonometric functions is crucial, for instance, \(\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{|x|\sqrt{x^2-1}}\).
The exercise solution utilized \(\sec^{-1}(t)\) to convert its inverse nature into a form amenable for trigonometric substitution, showcasing the power of blending trigonometric identities with inverse functions.
Definite Integrals
Definite integrals form the backbone of integral calculus. They provide the exact area under a curve within specified limits or boundaries on the x-axis. Definite integrals are particularly useful in physics and engineering for computing quantities like distance, area, and total accumulated change.
  • A definite integral is expressed as \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the limits of integration.
  • The result is not just an indefinite function but a tangible number or value that represents the area under the curve defined by \(f(x)\) from \(x = a\) to \(x = b\).
  • To compute the definite integral of a transformed expression \(\int f(t) dt\), recognizing the need to update the bounds when substituting variables is essential.
In our exercise, after the trigonometric substitution, the new limits provided a cleaner pathway to evaluate the integral, demonstrating how definite integrals work practically in bounding problems with trigonometric context.

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