Integration by Substitution

8 Key excerpts on "Integration by Substitution"

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  eBook - PDF

  Calculus

  Single and Multivariable

  • Deborah Hughes-Hallett, Andrew M. Gleason, William G. McCallum, Daniel E. Flath, Patti Frazer Lock, David O. Lomen, David Lovelock, Brad G. Osgood, Douglas Quinney, Karen R. Rhea, Jeff Tecosky-Feldman, Thomas W. Tucker, Otto K. Bretscher, Sheldon P. Gordon, Andrew Pasquale, Joseph Thrash(Authors)
  • 2020(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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  Calculus

  Early Transcendental Single Variable

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

    (Publisher)

  u-SUBSTITUTION The method of substitution can be motivated by examining the chain rule from the view- point of antidifferentiation. For this purpose, suppose that F is an antiderivative of f and that g is a differentiable function. The chain rule implies that the derivative of F(g(x)) can be expressed as d dx [F(g(x))] = F  (g(x))g  (x) which we can write in integral form as  F  (g(x))g  (x) dx = F(g(x)) + C (1) or since F is an antiderivative of f ,  f (g(x))g  (x) dx = F(g(x)) + C (2) For our purposes it will be useful to let u = g(x) and to write du / dx = g  (x) in the differ- ential form du = g  (x) dx. With this notation (2) can be expressed as  f (u) du = F(u) + C (3) The process of evaluating an integral of form (2) by converting it into form (3) with the substitution u = g(x) and du = g  (x) dx is called the method of u-substitution. Here the differential notation serves primarily as a useful “bookkeeping” device for the method of u-substitution. The following example illustrates how the method works. Example 1 Evaluate  (x 2 + 1) 50 · 2x dx. Solution. If we let u = x 2 + 1, then du / dx = 2x, which implies that du = 2x dx. Thus, the given integral can be written as  (x 2 + 1) 50 · 2x dx =  u 50 du = u 51 51 + C = (x 2 + 1) 51 51 + C It is important to realize that in the method of u-substitution you have control over the choice of u, but once you make that choice the value of du is computed. The method of u-substitution will fail if the chosen u and the computed du cannot be used to produce an in- tegrand in which no expressions involving x remain, or if you cannot evaluate the resulting integral. Thus, for example, the substitution u = x 2 , du = 2x dx will not work for the inte- gral  2x sin x 4 dx because this substitution results in the integral  sin u 2 du which still cannot be evaluated in terms of familiar functions. There are no hard and fast rules for choosing u, and in some problems no choice of u will work.

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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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  Calculus Early Transcendentals

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

    (Publisher)

  Again, this may or may not be possible. Step 3. If you are successful in Step 2, then replace u by g(x) to express your final answer in terms of x. Easy-to-Recognize Substitutions The easiest substitutions occur when the integrand is the derivative of a known function, except for a constant added to or subtracted from the independent variable. 296 Chapter 5 / Integration ▶ Example 2 Evaluate ∫ (x − 1) 7 dx Solution If we let u = x − 1, then du∕dx = 1, which implies that du = dx. Thus, the given integral can be written as ∫ (x − 1) 7 dx = ∫ u 7 du = u 8 8 + C = (x − 1) 8 8 + C Another easy u-substitution occurs when the integrand is the derivative of a known function, except for a constant that multiplies or divides the independent variable. The following example illustrates two ways to evaluate such integrals. ▶ Example 3 Evaluate ∫ e 4x dx Solution If we let u = 4x, then du∕dx = 4, which implies that du 4 = dx. Thus, the given integral can be written as ∫ e 4x dx = 1 4 ∫ e u du = e u 4 + C = e 4x 4 + C More generally, if the integrand is a composition of the form f (ax + b), where f (x) is easy to integrate, then the substitution u = ax + b, du = a dx will work. ▶ Example 4 Evaluate ∫ 1 (8 − 7x) 3 dx. Solution If we let u = 8 − 7x, then du dx = −7, which implies that − du 7 = dx. Thus, the given integral can be written as ∫ 1 (8 − 7x) 3 dx = − 1 7 ∫ 1 u 3 du = − 1 7 × u −3+1 −3 + 1 + C = 1 14u 2 + C = 1 14(8 − 7x) 2 + C ▶ Example 5 Evaluate ∫ 1 √ 1 − 2x 2 dx. Solution Substituting u = √ 2x, du √ 2 = dx yields ∫ 1 √ 1 − 2x 2 dx = 1 √ 2 ∫ 1 √ 1 − u 2 du = 1 √ 2 sin −1 u + C = 1 √ 2 sin −1 √ 2x + C With the help of Theorem 5.2.3, a complicated integral can sometimes be computed by expressing it as a sum of simpler integrals. 5.3 Integration by Substitution 297 ▶ Example 6 Evaluate ∫ ( 1 x 2 + 1 1 + 2x 2 ) dx.

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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)

  A most surprising and interesting fact comes to light when we study (the first and the second) fundamental theorems of Calculus, to be introduced later in Chapter 6a. The concept of the Definite Integral (discussed later in Chapter 5) clearly tells that in the development of the idea of the definite integral, the concept of derivatives does not come into play. On the other hand, the fundamental theorems of Calculus prove that computation of Definite Integral(s) can be done very easily using antiderivative(s). [The method of computing definite integrals is otherwise a very complicated process and it cannot be applied to many functions.] It is for this reason that the term ‘integral’, (picked up from the definite integral) is also used to stand for antiderivative. Accordingly, the process of computing both antide- rivative(s) and definite integral(s) is called integration. Introduction to Integral Calculus: Systematic Studies with Engineering Applications for Beginners, First Edition. Ulrich L. Rohde, G. C. Jain, Ajay K. Poddar, and A. K. Ghosh. Ó 2012 John Wiley & Sons, Inc. Published 2012 by John Wiley & Sons, Inc. (1) Generally, such integrals arise in practical applications of integration, namely computation of areas, volumes, and other quantities, using the concept of definite integral. 4a-Method of integration by parts (When the integrand is in the form of product of two functions) 97 4a.2 OBTAINING THE RULE FOR INTEGRATION BY PARTS Let u and v be functions of x possessing continuous derivatives. Then, we have d dx ðu  vÞ ¼ u  dv dx þ v  du dx ð1Þ Also, we can restate (1) in the language of indefinite integrals as u  v ¼ ð u  dv dx  dx þ ð v  du dx  dx ð2Þ or by rearranging, we get ð u  dv dx  dx ¼ u  v  ð v  du dx  dx ð3ðAÞÞ Or ð u  dv ¼ u  v  ð v  du ð3ðBÞÞ For computational purposes, a more convenient way of writing this formula is obtained, if we put, u ¼ f(x) and v ¼ g(x).

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  Calculus: An Applied Approach, Brief, International Metric Edition

  • Ron Larson(Author)
  • 2016(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Copyright 2018 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. WCN 02-300 Section 5.2 Integration by Substitution and the General Power Rule 327 The basic steps for Integration by Substitution are outlined in the guidelines below. Guidelines for Integration by Substitution 1. Let u be a function of x (usually part of the integrand). 2. Rewrite the integral in terms of the variable u . 3. Find the resulting integral in terms of u . 4. Rewrite the antiderivative as a function of x . 5. Check your answer by differentiating. Integration by Substitution Find integral.alt1 x radical.alt2 x 2 -1 dx . SOLUTION Consider the substitution u = x 2 -1, which produces du = 2 x dx . To create 2 x dx as part of the integral, multiply and divide by 2. integral.alt1 x radical.alt2 x 2 -1 dx = 1 2 integral.alt1 ( x 2 -1 ) 1 H20862 2 2 x d x Multiply and divide by 2. = 1 2 integral.alt1 u 1 H20862 2 d u Substitute for x and dx . = 1 2 parenleft.alt4 u 3 H20862 2 3 H20862 2 parenright.alt4 + C Apply Power Rule. = 1 3 u 3 H20862 2 + C Simplify. = 1 3 ( x 2 -1 ) 3 H20862 2 + C Substitute for u . You can check this result by differentiating. d dx bracketleft.alt2 1 3 ( x 2 -1 ) 3 H20862 2 + C bracketright.alt2 = 1 3 parenleft.alt4 3 2 parenright.alt4 ( x 2 -1 ) 1 H20862 2 ( 2 x ) = 1 2 ( 2 x )( x 2 -1 ) 1 H20862 2 = x radical.alt2 x 2 -1 Checkpoint 7 Worked-out solution available at LarsonAppliedCalculus.com Find integral.alt1 x radical.alt2 x 2 + 4 d x by the method of substitution. To become efficient at integration, you should learn to use both techniques discussed in this section. For simpler integrals, you should use pattern recognition and create du H20862 dx by multiplying and dividing by an appropriate constant. For more complicated integrals, you should use a formal change of variables, as shown in Examples 6 and 7.

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  Calculus

  Late Transcendentals

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

    (Publisher)

  This may or may not be possible. Step 2. If you are successful in Step 1, then try to evaluate the resulting integral in terms of u. Again, this may or may not be possible. Step 3. If you are successful in Step 2, then replace u by g(x) to express your final answer in terms of x. EASY TO RECOGNIZE SUBSTITUTIONS The easiest substitutions occur when the integrand is the derivative of a known function, except for a constant added to or subtracted from the independent variable. Example 2  sin(x + 9) dx =  sin u du = − cos u + C = −cos(x + 9) + C u = x + 9 du = 1 · dx = dx  (x − 8) 23 dx =  u 23 du = u 24 24 + C = (x − 8) 24 24 + C u = x − 8 du = 1 · dx = dx Another easy u-substitution occurs when the integrand is the derivative of a known function, except for a constant that multiplies or divides the independent variable. The following example illustrates two ways to evaluate such integrals. Example 3 Evaluate  cos 5x dx. Solution.  cos 5x dx =  ( cos u) · 1 5 du = 1 5  cos u du = 1 5 sin u + C = 1 5 sin 5x + C u = 5x du = 5 dx or dx = 1 5 du 4.3 Integration by Substitution 219 Alternative Solution. There is a variation of the preceding method that some people prefer. The substitution u = 5x requires du = 5 dx. If there were a factor of 5 in the inte- grand, then we could group the 5 and dx together to form the du required by the substitu- tion. Since there is no factor of 5, we will insert one and compensate by putting a factor of 1 5 in front of the integral. The computations are as follows:  cos 5x dx = 1 5  cos 5x · 5 dx = 1 5  cos u du = 1 5 sin u + C = 1 5 sin 5x + C u = 5x du = 5 dx More generally, if the integrand is a composition of the form f (ax + b), where f (x) is an easy to integrate function, then the substitution u = ax + b, du = a dx will work.

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  Anton's Calculus

  Early Transcendentals

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

    (Publisher)

  If we let u = x 2 + 1, then du ∕ dx = 2x, which implies that du = 2x dx. Thus, the given integral can be written as ∫ (x 2 + 1) 50 ⋅ 2x dx = ∫ u 50 du = u 51 51 + C = (x 2 + 1) 51 51 + C It is important to realize that in the method of u-substitution you have control over the choice of u, but once you make that choice the value of du is computed. The method of u-substitution will fail if the chosen u and the computed du cannot be used to produce an integrand in which no expressions involving x remain, or if you cannot evaluate the resulting integral. Thus, for example, the substitution u = x 2 , du = 2x dx will not work for the integral ∫ 2x sin x 4 dx because this substitution results in the integral ∫ sin u 2 du which still cannot be evaluated in terms of familiar functions. There are no hard and fast rules for choosing u, and in some problems no choice of u will work. In such cases other methods need to be used, some of which will be discussed later. Making appropriate choices for u will come with experience, but you may find the following guidelines, combined with a mastery of the basic integrals in Table 5.2.1, helpful. 5.3 Integration by Substitution 269 Guidelines for u-Substitution Step 1. Look for some composition f (g(x)) within the integrand for which the substitution u = g(x), du = g ′ (x) dx produces an integral that is expressed entirely in terms of u and its differential du. This may or may not be possible. Step 2. If you are successful in Step 1, then try to evaluate the resulting integral in terms of u. Again, this may or may not be possible. Step 3. If you are successful in Step 2, then replace u by g(x) to express your final answer in terms of x. EASY TO RECOGNIZE SUBSTITUTIONS The easiest substitutions occur when the integrand is the derivative of a known function, except for a constant added to or subtracted from the independent variable.

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