Trigonometric Substitution

8 Key excerpts on "Trigonometric Substitution"

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  Calculus

  Resequenced for Students in STEM

  • David Dwyer, Mark Gruenwald(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  1 2 3 4 1 2 3 4 y = f (x) Figure 6.2 6.3 Trigonometric Substitution Integrals involving radical expressions like √ x 2 + c 2 , √ c 2 - x 2 , or √ x 2 - c 2 can often be integrated by letting θ be an inverse trigonometric function of x. After rewriting in terms of θ, the integral typically involves trigonometric functions of θ. For this reason, the technique is called Trigonometric Substitution. One way of selecting an appropriate Trigonometric Substitution is to use the Pythagorean Theorem to drive the construction of a right triangle having the radical expression as one of its side lengths and then to let θ be one of the acute angles in the triangle. We illustrate this technique in the following example. Example 1 Integration by Trigonometric Substitution Evaluate Z 1 √ x 2 + 9 dx, where x > 0. Solution We begin by constructing a right triangle with legs of lengths 3 and x and a hypotenuse of length √ x 2 + 9, as shown in Figure 6.3. From the triangle, it’s 362 CHAPTER 6. TECHNIQUES OF INTEGRATION clear that cos θ = 3 √ x 2 + 9 so that 1 √ x 2 + 9 = 1 3 cos θ To find dx, we look for the simplest ratio with x in the numerator and differentiate. In this case, we have tan θ = x 3 sec 2 θ dθ = dx 3 Solving for dx gives us θ 3 x √ x 2 + 9 Figure 6.3 dx = 3 sec 2 θ dθ Substituting and integrating gives us Z 1 √ x 2 + 9 dx = Z  1 3 cos θ  ( 3 sec 2 θ dθ ) = Z sec θ dθ = ln |sec θ + tan θ| + C Using Figure 6.3 and the right triangle definitions of the trigonometric functions, we obtain ln |sec θ + tan θ| = ln     hyp adj + opp adj     = ln      √ x 2 + 9 3 + x 3      Z 1 √ x 2 + 9 dx = ln      √ x 2 + 9 3 + x 3      + C In Example 1, we used the fact that x is positive to construct a triangle with x as a side length. Interestingly, the resulting expression is actually valid for all x, whether positive or not—a result which can be confirmed directly by differentiation.

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  Introduction to Integral Calculus

  Systematic Studies with Engineering Applications for Beginners

  • Ulrich L. Rohde, G. C. Jain, Ajay K. Poddar, A. K. Ghosh(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  (These integrals should not be confused with those discussed in Section 2.3). We shall introduce a very simple substitution, which can be uniformly used in evaluating all such integrals. But, as a prerequisite, it is necessary to first establish the following standard integrals, since the above integrals are reduced to quadratic algebraic functions, due to substitution: (1) ð 1 x 2 þ a 2 dx ¼ 1 a tan 1 x a   þ c (2) ð 1 x 2  a 2 dx ¼ 1 2a log x  a x þ a   þ c; x > a (3) ð 1 a 2  x 2 dx ¼ 1 2a log a þ x a  x   þ c; x < a Using these standard integrals, we can also evaluate integrals of the form Ð ðdx=ðax 2 þ bx þ cÞÞ. Details are discussed in Chapter 3b. 42 INTEGRATION USING TRIGONOMETRIC IDENTITIES 3a Integration by Substitution: Change of Variable of Integration 3a.1 INTRODUCTION So far we have evaluated integrals of functions, which are of standard forms and those, which can be reduced to standard forms by simple algebraic operations or trigonometric simplifi- cation methods including the use of trigonometric identities. Many integrals cannot be reduced to standard forms by these methods. We must, therefore, learn other techniques of integration. In this chapter, we shall discuss the method of substitution, which is applicable in reducing to standard forms, the integrals involving composite functions. It will be observed that this method involves change of variable of integration as against the earlier methods, wherein the variable of integration remains unchanged. Before introducing the theorem which governs the rule of integration by substitution, let us recall the chain rule for differentiation, as applied to a power of a function. If u ¼ f(x) is a differentiable function and r is a rational number, then d dx u rþ1 r þ 1  ! ¼ ðr þ 1Þu r ðr þ 1Þ  du dx ¼ u r : d dx ðuÞ; ðr is rational; r 6¼ 1Þ or d dx ½ f ðxÞ rþ1 r þ 1  ! ¼ ½ f ðxÞ r  f 0 ðxÞ ð1Þ From the above result, we obtain the following important rule for indefinite integrals.

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  Calculus, Metric Edition

  • James Stewart, Daniel K. Clegg, Saleem Watson, , James Stewart, James Stewart, Daniel K. Clegg, Saleem Watson(Authors)
  • 2020(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  For instance, in the integral y x x 2 2 1 dx we notice that if u - x 2 2 1, then du - 2x dx. Therefore we use the substitution u - x 2 2 1 instead of the method of partial fractions. 3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the solution, then we take a look at the form of the integrand f s xd. (a) Trigonometric functions. If f s xd is a product of powers of sin x and cos x, of tan x and sec x, or of cot x and csc x, then we use the substitutions recom- mended in Section 7.2. (b) Rational functions. If f is a rational function, we use the procedure of Section 7.4 involving partial fractions. (c) Integration by parts. If f s xd is a product of a power of x (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv according to the advice given in Section 7.1. If you look at the functions in Exercises 7.1, you will see that most of them are the type just described. (d) Radicals. Particular kinds of substitutions are recommended when certain radicals appear. (i) If sx 2 1 a 2 , sx 2 2 a 2 , or sa 2 2 x 2 occurs, we use a Trigonometric Substitution according to the table in Section 7.3. (ii) If s n ax 1 b occurs, we use the rationalizing substitution u - s n ax 1 b . More generally, this sometimes works for s n ts xd . Copyright 2021 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.

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  Calculus

  Single Variable

  • Carl V. Lutzer, H. T. Goodwill(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  Section 6.4 Trigonometric Substitution 440 The final step of our process is returning to the original variable, which we do by drawing a right triangle in which sec(θ) = t/1. This has been done in Figure 4.3. Figure 4.3: A right triangle in which sec(θ) = t/1. Based on this triangle, our answer is 1 2 t p t 2 - 1 - 1 2 ln |t + p t 2 - 1| + C Using the secant in a Trigonometric Substitution Example 4.5. Determine R 1 √ 3t 2 -2 dt when t > p 2/3. Solution: When we determine the structure of this integrand by temporarily setting the coefficients to 1, we see 1/ √ t 2 - 1, which is similar in nature to the integral that we did in Example 4.4. That leads us to try the same basic technique, but we’ll need the constant term to be a 1 to do it (as it is in the Pythagorean identity), so we begin by factoring the 2 out of the radical: 1 √ 3t 2 - 2 = 1 q 2 ( 3 2 t 2 - 1 ) = 1 √ 2 1 q 3 2 t 2 - 1 . In order for the term under the radical to have the form sec 2 (θ) - 1, we need sec 2 (θ) = 3 2 t 2 .

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  Precalculus with Limits

  • Ron Larson(Author)
  • 2021(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  ©iStockphoto/MirasWonderland 362 Chapter 5 Analytic Trigonometry 1.4 Functions GO DIGITAL 5.3 Solving Trigonometric Equations Use standard algebraic techniques to solve trigonometric equations. Solve trigonometric equations of quadratic type. Solve trigonometric equations involving multiple angles. Use inverse trigonometric functions to solve trigonometric equations. Introduction To solve a trigonometric equation, use standard algebraic techniques (when possible) such as collecting like terms, extracting square roots, and factoring. Your preliminary goal in solving a trigonometric equation is to isolate the trigonometric function on one side of the equation. For example, to solve the equation 2 sin x = 1, divide each side by 2 to obtain sin x = 1 2 .

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  Calculus: Single and Multivariable

  • Deborah Hughes-Hallett, William G. McCallum, Andrew M. Gleason, Eric Connally, Daniel E. Flath, Selin Kalaycioglu, Brigitte Lahme, Patti Frazer Lock, David O. Lomen, David Lovelock, Guadalupe I. Lozano, Jerry Morris, David Mumford, Brad G. Osgood, Cody L. Patterson, Douglas Quinney, Karen R. Rhea, Ayse Arzu Sahin, Ad(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  Warning We saw in the preceding example that we can apply the substitution method when a constant factor is missing from the derivative of the inside function. However, we may not be able to use substitution if anything other than a constant factor is missing. For example, setting  =  4 + 5 to find   2 √  4 + 5  does us no good because  2  is not a constant multiple of  = 4 3 . Substitution works if the integrand contains the derivative of the inside function, to within a constant factor. 7.1 INTEGRATION BY SUBSTITUTION 345 Some people prefer the substitution method over guess-and-check since it is more systematic, but both methods achieve the same result. For simple problems, guess-and-check can be faster. Example 5 Find   cos  sin  . Solution We let  = cos  since its derivative is − sin  and there is a factor of sin  in the integrand. This gives  =   ()  = − sin  , so − = sin  . Thus   cos  sin   =    (−) = (−1)     = −  +  = − cos  + . Example 6 Find    1 +   . Solution Observing that the derivative of 1 +   is   , we see  = 1 +   is a good choice. Then  =   , so that    1 +    =  1 1 +      =  1   = ln  +  = ln 1 +    +  = ln(1 +   ) + . (Since (1 +   ) is always positive.) Since the numerator is   , we might also have tried  =   . This substitution leads to the integral ∫ (1∕(1+)), which is better than the original integral but requires another substitution,  = 1+, to finish. There are often several different ways of doing an integral by substitution. Notice the pattern in the previous example: having a function in the denominator and its deriva- tive in the numerator leads to a natural logarithm. The next example follows the same pattern. Example 7 Find  tan  . Solution Recall that tan  = (sin )∕(cos ). If  = cos , then  = − sin  , so  tan   =  sin  cos   =  −  = − ln  +  = − ln  cos  + .

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  Mathematics N5 Student's Book

  TVET FIRST

  • JV John(Author)
  • 2022(Publication Date)
  • Macmillan

    (Publisher)

  In this module, we will learn some of the elementary methods of integration, including inspection, algebraic and Trigonometric Substitutions, using partial fractions and integration by parts. Note The full learning outcomes for each module are listed in the table at the back of the book. Integration techniques Types: o ∫[ f (x) ]nf '(x) dx = [ f (x) ] n + 1 _ n + 1 + c o ∫ f '(x) _ f (x) dx = ln f (x) + c o ∫ a f (x) f '(x) dx o ∫ e f (x) f '(x) dx Other substitutions 4.3: Integration by algebraic substitution Introduction to the indefinite integral Methods of integration Standard integrals Trigonometric identities 4.1: Revision of basic integration Readily determining the integral directly by examining the integrand 4.2: Integration by inspection Squares of sin, cos, tan and cot Products of powers of sin and cos: ∫ sin m x cos n x dx, m or n odd or both odd Powers of tan or cot Products of sin and/or cos with different coefficients 4.4: Integration of trigonometric functions Finding integrals using Trigonometric Substitution Using Trigonometric Substitution to derive the formulae: o ∫ 1 ___________ √ _ a 2 − b 2 x 2 dx = 1 _ b sin −1 bx _ a + c o ∫ √ _ a 2 − b 2 x 2 dx = a 2 _ 2b sin −1 bx _ a + x _ 2 √ _ a 2 − b 2 x 2 + c o ∫ 1 _ a 2 + b 2 x 2 dx = 1 _ ab tan −1 bx _ a + c Using the formulae with or without the b factor 4.5: Trigonometric Substitution Suitable substitution Substitution after long division 4.6: Integration of algebraic fractions Integrating the following types of fractions using partial fractions: o ∫ f (x) ___________ (ax ± b)(cx ± d) dx o ∫ f (x) _ (ax ± b) n dx, n = 2 or 3 4.7: Integration using partial fractions Integrating products of two functions that are not the derivatives of each other using the formula: ∫ f (x) g'(x) dx = f (x) g(x) − ∫ f '(x) g(x) dx 4.8: Integration by parts Figure 4.1: Integration is an essential tool for engineers

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  Calculus

  Single Variable

  • Deborah Hughes-Hallett, William G. McCallum, Andrew M. Gleason, Eric Connally, Daniel E. Flath, Selin Kalaycioglu, Brigitte Lahme, Patti Frazer Lock, David O. Lomen, David Lovelock, Guadalupe I. Lozano, Jerry Morris, Brad G. Osgood, Cody L. Patterson, Douglas Quinney, Karen R. Rhea, Ayse Arzu Sahin, Adam H. Spiegler(Authors)
  • 2017(Publication Date)
  • Wiley

    (Publisher)

  Warning We saw in the preceding example that we can apply the substitution method when a constant factor is missing from the derivative of the inside function. However, we may not be able to use substitution if anything other than a constant factor is missing. For example, setting  =  4 + 5 to find   2 √  4 + 5  does us no good because  2  is not a constant multiple of  = 4 3 . Substitution works if the integrand contains the derivative of the inside function, to within a constant factor. 7.1 INTEGRATION BY SUBSTITUTION 345 Some people prefer the substitution method over guess-and-check since it is more systematic, but both methods achieve the same result. For simple problems, guess-and-check can be faster. Example 5 Find   cos  sin  . Solution We let  = cos  since its derivative is − sin  and there is a factor of sin  in the integrand. This gives  =   ()  = − sin  , so − = sin  . Thus   cos  sin   =    (−) = (−1)     = −  +  = − cos  + . Example 6 Find    1 +   . Solution Observing that the derivative of 1 +   is   , we see  = 1 +   is a good choice. Then  =   , so that    1 +    =  1 1 +      =  1   = ln  +  = ln 1 +    +  = ln(1 +   ) + . (Since (1 +   ) is always positive.) Since the numerator is   , we might also have tried  =   . This substitution leads to the integral ∫ (1∕(1+)), which is better than the original integral but requires another substitution,  = 1+, to finish. There are often several different ways of doing an integral by substitution. Notice the pattern in the previous example: having a function in the denominator and its deriva- tive in the numerator leads to a natural logarithm. The next example follows the same pattern. Example 7 Find  tan  . Solution Recall that tan  = (sin )∕(cos ). If  = cos , then  = − sin  , so  tan   =  sin  cos   =  −  = − ln  +  = − ln  cos  + .

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