Integration using Inverse Trigonometric Functions

10 Key excerpts on "Integration using Inverse Trigonometric Functions"

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  Precalculus

  A Prelude to Calculus

  • Sheldon Axler(Author)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  This chapter concludes with an investigation into transformations of trigono- metric functions, which are used to model periodic events. Redoing function transformations in the context of trigonometric functions will also help us review the results from Chapter 1 on how function transformations change graphs. 351 352 Chapter 5 Trigonometric Algebra and Geometry 5.1 Inverse Trigonometric Functions Learning Objectives By the end of this section you should be able to • compute values of cos -1 , sin -1 , and tan -1 ; • sketch the radius of the unit circle corresponding to the arccosine, arcsine, and arctangent of a number; • use the inverse trigonometric functions to find angles in a right triangle, given the lengths of two sides; • find the angles in an isosceles triangle, given the lengths of the sides; • use tan -1 to find the angle a line with given slope makes with the horizontal axis. Several of the most important functions in mathematics are defined as the inverse functions of familiar functions. For example, the cube root is defined as the inverse function of x 3 , and the logarithm base 3 is defined as the inverse function of 3 x . In this section, we will define the inverses of the cosine, sine, and tangent The inverse trigonometric functions provide remarkably useful tools for solving many problems. functions. These inverse functions are called the arccosine, the arcsine, and the arctangent. Neither cosine nor sine nor tangent is one-to-one when defined on its usual domain. Thus we will need to restrict the domains of these functions to obtain one-to-one functions that have inverses. The Arccosine Function Recall that a function is called one-to-one if it assigns distinct values to distinct numbers in its domain. The cosine function, whose domain is the entire real line, is As usual, we will assume throughout this section that all angles are measured in radians unless explicitly stated otherwise. not one-to-one because, for example, cos 0 = cos 2π.

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  Calculus

  Late Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  Each definition has advantages and disadvantages, but we will use the current definition to conform with the conventions used by the CAS programs Mathematica, Maple, and Sage. 390 Chapter 6 / Exponential, Logarithmic, and Inverse Trigonometric Functions continuous at all points where the denominator is nonzero and either x ≤ −1 or x ≥ 1. Note that π + 4 tan −1 x = 0 if x = tan (−π / 4) = −1. Thus, f is continuous on the intervals (−∞, −1) and [1, +∞). EVALUATING INVERSE TRIGONOMETRIC FUNCTIONS A common problem in trigonometry is to find an angle whose sine is known. For example, you might want to find an angle x in radian measure such that sin x = 1 2 (4) and, more generally, for a given value of y in the interval −1 ≤ y ≤ 1 you might want to solve the equation sin x = y (5) Because sin x repeats periodically, this equation has infinitely many solutions for x; however, if we solve this equation as x = sin −1 y then we isolate the specific solution that lies in the interval [−π / 2, π / 2], since this is the range of the inverse sine. For example, Figure 6.7.2 shows four solutions of Equation (4), namely, −11π / 6, −7π / 6, π / 6, and 5π / 6. Of these, π / 6 is the solution in the interval [−π / 2, π / 2], so sin −1 ( 1 2 ) = π / 6 (6) Figure 6.7.2 In general, if we view x = sin −1 y as an angle in radian measure whose sine is y, then the restriction −π / 2 ≤ x ≤ π / 2 imposes the geometric requirement that the angle x in standard position terminate in either the first or fourth quadrant or on an axis adjacent to those quadrants. TECHNOLOGY MASTERY Refer to the documentation for your calculating utility to determine how to calculate inverse sines, inverse cosines, and inverse tangents; and then con- firm Equation (6) numerically by show- ing that sin −1 (0.5) ≈ 0.523598775598 . . . ≈ π/6 Example 3 Find exact values of (a) sin −1 (1 / √ 2 ) (b) sin −1 (−1) by inspection, and confirm your results numerically using a calculating utility.

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  Elementary Special Functions in Mathematics

  • (Author)
  • 2014(Publication Date)
  • Learning Press

    (Publisher)

  ________________________ WORLD TECHNOLOGIES ________________________ Chapter 5 Inverse Trigonometric Functions and Trigonometric Functions Inverse trigonometric functions In mathematics, the inverse trigonometric functions or cyclometric functions are the inverse functions of the trigonometric functions, though they do not meet the official definition for inverse functions as their ranges are subsets of the domains of the original functions. Since none of the six trigonometric functions are one-to-one (by failing the horizontal line test), they must be restricted in order to have inverse functions. For example, just as the square root function is defined such that y 2 = x , the function y = arcsin( x ) is defined so that sin( y ) = x . There are multiple numbers y such that sin( y ) = x ; for example, sin(0) = 0, but also sin(π) = 0, sin(2π) = 0, etc. It follows that the arcsine function is multivalued: arcsin(0) = 0, but also a rcsin(0) = π, arcsin(0) = 2π, etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each x in the domain the expression arcsin( x ) will evaluate only to a single value, called its principal value. These properties apply to all the inverse trigonometric functions. The principal inverses are listed in the following table. Name Usual notation Definition Domain of x for real result Range of usual principal value (radians) Range of usual principal value (degrees) arcsine y = arcsin x x = sin y −1 ≤ x ≤ 1 −π/2 ≤ y ≤ π/2 −90° ≤ y ≤ 90° arccosine y = arccos x x = cos y −1 ≤ x ≤ 1 0 ≤ y ≤ π 0° ≤ y ≤ 180° arctangent y = arctan x x = tan y all real numbers −π/2 < y < π/2 −90° < y < 90° arccotangent y = arccot x x = cot y all real numbers 0 < y < π 0° < y < 180°

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  Trigonometry

  • Cynthia Y. Young(Author)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  As with all equations, the goal is to find the value(s) of the variable that make the equation true. You will find that trigonometric equations (like algebraic equations) can have no solution, one or more solutions, or even an infinite number of solutions. SOLVING TRIGONOMETRIC EQUATIONS 6.1 INVERSE TRIGONOMETRIC FUNCTIONS 6.2 SOLVING TRIGONOMETRIC EQUATIONS THAT INVOLVE ONLY ONE TRIGONOMETRIC FUNCTION 6.3 SOLVING TRIGONOMETRIC EQUATIONS THAT INVOLVE MULTIPLE TRIGONOMETRIC FUNCTIONS • Inverse Sine Function • Inverse Cosine Function • Inverse Tangent Function • Remaining Inverse Trigonometric Functions • Finding Exact Values for Expressions Involving Inverse Trigonometric Functions • Solving Trigonometric Equations by Inspection • Solving Trigonometric Equations Using Algebraic Techniques • Solving Trigonometric Equations That Require the Use of Inverse Functions • Solving Trigonometric Equations 292 CHAPTER 6 Solving Trigonometric Equations SKILLS OBJECTIVES ■ ■ Find the exact values of an inverse sine function. ■ ■ Find the exact values of an inverse cosine function. ■ ■ Find the exact values of an inverse tangent function. ■ ■ Find the exact values of the cotangent, cosecant, and secant inverse functions. ■ ■ Use identities to find exact values of trigonometric expressions involving inverse trigonometric functions. CONCEPTUAL OBJECTIVES ■ ■ Understand that the domain of the sine function is restricted C 2 p 2 , p 2 D in order for the inverse sine function to exist. ■ ■ Understand that the domain of the cosine function is restricted 30, p4 in order for the inverse cosine function to exist. ■ ■ Understand that the domain of the tangent function is restricted A 2 p 2 , p 2 B in order for the inverse tangent function to exist.

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  Calculus

  Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  382 CHAPTER 6 Exponential, Logarithmic, and Inverse Trigonometric Functions 53. Writing A student objects that it is circular reasoning to make the definition ln x =  x 1 1 t dt since to evaluate the integral we need to know the value of ln x. Write a short paragraph that answers this student’s objection. 54. Writing Write a short paragraph that compares Definition 6.6.1 with the definition of the natural logarithm function given in Section 6.1. Be sure to discuss the issues surrounding continu- ity and differentiability. 6.6 | Quick Check Answers 1. −1 2. a. 5 6 b. 7 12 3. e 4. y = 2 +  x 0 cos t 3 dt 5. − e −x 1 + e −4x 6.7 Derivatives and Integrals Involving Inverse Trigonometric Functions A common problem in trigonometry is to find an angle x using a known value of sin x, cos x, or some other trigonometric function. Problems of this type involve the computation of inverse trigonometric functions. In this section we will study these functions from the viewpoint of general inverse functions, with the goal of developing derivative formulas for the inverse trigonometric functions. We will also derive some related integration formulas that involve inverse trigonometric functions. Inverse Trigonometric Functions The six basic trigonometric functions do not have inverses because their graphs repeat periodi- cally and hence do not pass the horizontal line test. To circumvent this problem we will restrict the domains of the trigonometric functions to produce one-to-one functions and then define the “inverse trigonometric functions” to be the inverses of these restricted functions. The top part of Figure 6.7.1 shows geometrically how these restrictions are made for sin x, cos x, tan x, and sec x, and the bottom part of the figure shows the graphs of the corresponding inverse functions sin −1 x, cos −1 x, tan −1 x, sec −1 x (also denoted by arcsin x, arccos x, arctan x, and arcsec x).

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  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  208 • Inverse Functions and Inverse Trig Functions 10.2 Inverse Trig Functions Now it’s time to investigate the inverse trig functions. We’ll see how to define them, what their graphs look like, and how to differentiate them. Let’s look at them one at a time, beginning with inverse sine. 10.2.1 Inverse sine Let’s start by looking at the graph of y = sin( x ) once again: 1 0 -1 -3 π -5 π 2 -2 π -3 π 2 -π -π 2 3 π 5 π 2 2 π 3 π 2 π π 2 y = sin( x ) Does the sine function have an inverse? You can see from the above graph that the horizontal line test fails pretty miserably. In fact, every horizontal line of height between -1 and 1 intersects the graph infinitely many times, which is a lot more than the zero or one time we can tolerate. So, using the tactic described in Section 1.2.3 in Chapter 1, we throw away as little of the domain as possible in order to pass the horizontal line test. There are many options, but the sensible one is to restrict the domain to the interval [ -π/ 2 , π/ 2]. Here’s the effect of this: 1 0 -1 -3 π -5 π 2 -2 π -3 π 2 -π -π 2 3 π 5 π 2 2 π 3 π 2 π π 2 y = sin( x ) , -π 2 ≤ x ≤ π 2 The solid portion of the curve is all we have left after we restrict the domain. Clearly we can’t go to the right of π/ 2 or else we’ll start repeating the values immediately to the left of π/ 2 as the curve dips back down. A similar thing happens at -π/ 2. So, we’re stuck with our interval. OK, if f ( x ) = sin( x ) with domain [ -π/ 2 , π/ 2], then it satisfies the hor-izontal line test, so it has an inverse f -1 . We’ll write f -1 ( x ) as sin -1 ( x ) or arcsin( x ). (Beware: the first of these notations is a little confusing at first, since sin -1 ( x ) does not mean the same thing as (sin( x )) -1 , even though sin 2 ( x ) = (sin( x )) 2 and sin 3 ( x ) = (sin( x )) 3 .) So, what is the domain of the inverse sine function? Well, since the range of f ( x ) = sin( x ) is [ -1 , 1], the domain of the inverse function is [ -1 , 1].

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  Student Solutions Manual Analytic Trigonometry with Applications

  • Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen(Authors)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  Chapter 5 Inverse Trigonometric Functions; Trigonometric Equations 199 Chapter 5 Inverse Trigonometric Functions; Trigonometric Equations EXERCISE 5.1 Inverse Sine, Cosine, and Tangent Functions 1. For a function to be one-to-one, each element of the range must correspond to a single element of the domain. Since there are, for example, an infinite number of domain elements (x = n, n any integer) corresponding to range element 0, y = sin x is not a one-to-one function. 3. No. For example, sin  6 = 1 2 and sin 5 6 = 1 2 , so more than one element of the domain [0, ] corresponds to the element 1 2 of the range. 5. Make a careful drawing. Measure   53°. 7. Make a careful drawing. Measure   66°. 9. Make a careful drawing. Measure   31°. Exercise 5.1 Inverse Sine, Cosine, and Tangent Functions 200 11. y = sin –1 0 is equivalent to sin y = 0. No reference triangle can be drawn, but the only y between –  2 and  2 which has sine equal to 0 is y = 0. Thus, sin –1 0 = 0. (1, 0) 13. y = arccos 3 2 is equivalent to cos y = 3 2 . What y between 0 and  has cosine equal to 3 2 ? y must be associated with a reference triangle in the first quadrant. Reference triangle is a special 30°–60° triangle. y =  6 , arccos 3 2 =  6 y 2 1 3 15. y = tan –1 1 is equivalent to tan y = 1. What y between –  2 and  2 has tangent equal to 1? y must be associated with a reference triangle in the first quadrant. Reference triangle is a special 45° triangle. y =  4 , tan –1 1 =  4 y 1 2 1 17. y = cos –1 1 2 is equivalent to cos y = 1 2 . What y between 0 and  has cosine equal to 1 2 ? y must be associated with a reference triangle in the first quadrant. Reference triangle is a special 30°–60° triangle. y =  3 , cos –1 1 2 =  3 2 y 1 3 19. Calculator in radian mode: cos –1 (–0.9999) = 3.127 21. Calculator in radian mode: tan –1 4.056 = 1.329 23. 3.142 is not in the domain of the inverse sine function. –1  3.142  1 is false. arcsin 3.142 is not defined.

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  Precalculus

  Functions and Graphs

  • Earl Swokowski, Jeffery Cole(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  The following general relationships involving f and f 2 1 were discussed in Section 4.1. The Inverse Trigonometric Functions 6.6 460 CHAPTER 6 Analytic Trigonometry Copyright 2019 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. We shall use relationship 1 to define each of the inverse trigonometric functions. The sine function is not one-to-one, since different numbers, such as p y 6, 5 p y 6 , and 2 7 p y 6 , yield the same function value s 1 2 d . If we restrict the domain to f 2 p y 2, p y 2 g , then, as illustrated by the blue portion of the graph of y 5 sin x in Figure 1, we obtain a one-to-one (increasing) function that takes on every value of the sine function once and only once. We use this new function with domain f 2 p y 2, p y 2 g and range f 2 1, 1 g to define the inverse sine function. The domain of the inverse sine function is f 2 1, 1 g , and the range is f 2 p y 2, p y 2 g . The notation y 5 sin 2 1 x is sometimes read “ y is the inverse sine of x .” The equation x 5 sin y in the definition allows us to regard y as an angle, so y 5 sin 2 1 x may also be read “ y is the angle whose sine is x ” (with 2 p y 2 # y # p y 2 ). The inverse sine function is also called the arcsine function, and arcsin x may be used in place of sin 2 1 x . If t 5 arcsin x , then sin t 5 x , and t may be interpreted as an arc length on the unit circle U with center at the origin. We will use both notations— sin 2 1 and arcsin—throughout our work. Several values of the inverse sine function are listed in the next chart.

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  Anton's Calculus

  Early Transcendentals

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  Solution. By definition, sin −1 x is the inverse of the restricted sine function whose domain is the interval [− ∕ 2,  ∕ 2] and whose range is the interval [−1, 1]. Since sin x is continuous on the interval [− ∕ 2,  ∕ 2], Theorem 1.5.7 implies sin −1 x is continuous on the interval [−1, 1]. Arguments similar to the solution of Example 1 show that each of the inverse trigono- metric functions is continuous at each point of its domain. Example 2 Where is the function f (x) = sec −1 x  + 4 tan −1 x continuous? Solution. The fraction will be continuous at all points where the numerator and denom- inator are both continuous and the denominator is nonzero. Since sec −1 x is continuous on its domain, the numerator is continuous if x ≤ −1 or x ≥ 1. Since tan −1 x is continuous everywhere, the denominator is continuous everywhere, so the fraction will be continu- ous at all points where the denominator is nonzero and either x ≤ −1 or x ≥ 1. Note that  + 4 tan −1 x = 0 if x = tan(− ∕ 4) = −1. Thus, f is continuous on the intervals (−∞, −1) and [1, +∞). EVALUATING INVERSE TRIGONOMETRIC FUNCTIONS A common problem in trigonometry is to find an angle whose sine is known. For example, you might want to find an angle x in radian measure such that sin x = 1 2 (4) 1.7 Inverse Trigonometric Functions 59 and, more generally, for a given value of y in the interval −1 ≤ y ≤ 1 you might want to solve the equation sin x = y (5) Because sin x repeats periodically, this equation has infinitely many solutions for x; however, if we solve this equation as x = sin −1 y then we isolate the specific solution that lies in the interval [− ∕ 2,  ∕ 2], since this is the range of the inverse sine. For example, Figure 1.7.2 shows four solutions of Equation (4), namely, −11 ∕ 6, −7 ∕ 6,  ∕ 6, and 5 ∕ 6.

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  Calculus

  Early Transcendental Single Variable

  • Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  lim x→−∞ tan −1 x = − π 2 lim x→+∞ tan −1 x = π 2 (1–2) lim x→−∞ sec −1 x = lim x→+∞ sec −1 x = π 2 (3) CONTINUITY OF INVERSE TRIGONOMETRIC FUNCTIONS Example 1 Use Theorem 1.5.7 to prove that sin −1 x is continuous on the interval [−1, 1]. Solution. By definition, sin −1 x is the inverse of the restricted sine function whose do- main is the interval [−π / 2, π / 2] and whose range is the interval [−1, 1]. Since sin x is continuous on the interval [−π / 2, π / 2], Theorem 1.5.7 implies sin −1 x is continuous on the interval [−1, 1]. Arguments similar to the solution of Example 1 show that each of the inverse trigono- metric functions is continuous at each point of its domain. Example 2 Where is the function f (x) = sec −1 x π + 4 tan −1 x continuous? Solution. The fraction will be continuous at all points where the numerator and denom- inator are both continuous and the denominator is nonzero. Since sec −1 x is continuous on its domain, the numerator is continuous if x ≤ −1 or x ≥ 1. Since tan −1 x is con- tinuous everywhere, the denominator is continuous everywhere, so the fraction will be continuous at all points where the denominator is nonzero and either x ≤ −1 or x ≥ 1. Note that π + 4 tan −1 x = 0 if x = tan (−π / 4) = −1. Thus, f is continuous on the intervals (−∞, −1) and [1, +∞). EVALUATING INVERSE TRIGONOMETRIC FUNCTIONS A common problem in trigonometry is to find an angle whose sine is known. For example, you might want to find an angle x in radian measure such that sin x = 1 2 (4) 1.7 Inverse Trigonometric Functions 59 and, more generally, for a given value of y in the interval −1 ≤ y ≤ 1 you might want to solve the equation sin x = y (5) Because sin x repeats periodically, this equation has infinitely many solutions for x; however, if we solve this equation as x = sin −1 y then we isolate the specific solution that lies in the interval [−π / 2, π / 2], since this is the range of the inverse sine.

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