Section	5.5
                         Integration	by	Substitution

                                V63.0121, Calculus	I



...
Outline


  Announcements

  Last	Time: The	Fundamental	Theorem(s)	of	Calculus

  Substitution	for	Indefinite	Integrals
   ...
Office	Hours	and	other	help



      Day        Time      Who/What          Where	in	WWH
       M      1:00–2:00    Leingan...
Final	stuff

               Final	is	May	8, 2:00–3:50pm	in	19W4	101/102
               Old	finals	online, including	Fall	20...
Resurrection	Policy
        If	your	final	score	beats	your	midterm	score, we	will	add	10%	to
        its	weight, and	subtra...
Outline


  Announcements

  Last	Time: The	Fundamental	Theorem(s)	of	Calculus

  Substitution	for	Indefinite	Integrals
   ...
Differentiation	and	Integration	as	reverse	processes


   Theorem	(The	Fundamental	Theorem	of	Calculus)
    1. Let f be	co...
Techniques	of	antidifferentiation?


   So	far	we	know	only	a	few	rules	for	antidifferentiation. Some	are
   general, like...
Techniques	of	antidifferentiation?


   So	far	we	know	only	a	few	rules	for	antidifferentiation. Some	are
   general, like...
Techniques	of	antidifferentiation?


   So	far	we	know	only	a	few	rules	for	antidifferentiation. Some	are
   general, like...
So	far	we	don’t	have	any	way	to	find
                         ∫
                               2x
                         ...
So	far	we	don’t	have	any	way	to	find
                         ∫
                               2x
                         ...
Outline


  Announcements

  Last	Time: The	Fundamental	Theorem(s)	of	Calculus

  Substitution	for	Indefinite	Integrals
   ...
Substitution	for	Indefinite	Integrals



   Example
   Find                ∫
                                x
            ...
Substitution	for	Indefinite	Integrals



   Example
   Find                     ∫
                                     x
  ...
Say	what?


  Solution	(More	slowly, now)
  Let g(x) = x2 + 1.




                                .   .   .   .   .   .
Say	what?


  Solution	(More	slowly, now)
  Let g(x) = x2 + 1. Then g′ (x) = 2x and	so

                d√         1      ...
Say	what?


  Solution	(More	slowly, now)
  Let g(x) = x2 + 1. Then g′ (x) = 2x and	so

                 d√         1     ...
Leibnizian	notation	wins	again



   Solution	(Same	technique, new	notation)
   Let u = x2 + 1.




                      ...
Leibnizian	notation	wins	again



   Solution	(Same	technique, new	notation)
                                   √         ...
Leibnizian	notation	wins	again



   Solution	(Same	technique, new	notation)
                                   √         ...
Leibnizian	notation	wins	again



   Solution	(Same	technique, new	notation)
                                   √         ...
Leibnizian	notation	wins	again



   Solution	(Same	technique, new	notation)
                                   √         ...
Theorem	of	the	Day



  Theorem	(The	Substitution	Rule)
  If u = g(x) is	a	differentiable	function	whose	range	is	an	inter...
A polynomial	example


  Example                                  ∫
  Use	the	substitution u = x2 + 3 to	find       (x2 + 3...
A polynomial	example


  Example                                  ∫
  Use	the	substitution u = x2 + 3 to	find       (x2 + 3...
A polynomial	example, the	hard	way



  Compare	this	to	multiplying	it	out:
     ∫                    ∫
         2       3...
Compare

  We	have
        ∫
                             1 2
            (x2 + 3)3 4x dx =  (x + 3)4
                    ...
Compare

  We	have
        ∫
                             1 2
            (x2 + 3)3 4x dx =  (x + 3)4 + C
                ...
A slick	example


   Example
       ∫
   Find   tan x dx.




                      .   .   .   .   .   .
A slick	example


   Example
       ∫
                                     sin x
   Find   tan x dx. (Hint: tan x =       ...
A slick	example


   Example
       ∫
                                     sin x
   Find   tan x dx. (Hint: tan x =       ...
A slick	example


   Example
       ∫
                                     sin x
   Find   tan x dx. (Hint: tan x =       ...
A slick	example


   Example
       ∫
                                     sin x
   Find   tan x dx. (Hint: tan x =       ...
A slick	example


   Example
       ∫
                                     sin x
   Find   tan x dx. (Hint: tan x =       ...
Outline


  Announcements

  Last	Time: The	Fundamental	Theorem(s)	of	Calculus

  Substitution	for	Indefinite	Integrals
   ...
Theorem	(The	Substitution	Rule	for	Definite	Integrals)
If g′ is	continuous	and f is	continuous	on	the	range	of u = g(x),
th...
Example ∫
              π
Compute           cos2 x sin x dx.
          0




                                     .   .   ...
Example ∫
                π
Compute             cos2 x sin x dx.
            0

Solution	(Slow	Way)                    ∫
F...
Example ∫
                π
Compute             cos2 x sin x dx.
            0

Solution	(Slow	Way)                       ...
Solution	(Fast	Way)
Do	both	the	substitution	and	the	evaluation	at	the	same	time.




                                    ...
Solution	(Fast	Way)
Do	both	the	substitution	and	the	evaluation	at	the	same	time. Let
u = cos x. Then du = − sin x dx, u(0...
Solution	(Fast	Way)
Do	both	the	substitution	and	the	evaluation	at	the	same	time. Let
u = cos x. Then du = − sin x dx, u(0...
An	exponential	example
  Example√
         ∫   ln       8         √
  Find        √           e2x       e2x + 1 dx
       ...
An	exponential	example
  Example√
         ∫   ln       8         √
  Find        √           e2x       e2x + 1 dx
       ...
An	exponential	example
  Example√
         ∫   ln       8         √
  Find        √           e2x       e2x + 1 dx
       ...
Another	way	to	skin	that	cat

   Example√
          ∫   ln       8         √
   Find        √           e2x       e2x + 1 ...
Another	way	to	skin	that	cat

   Example√
          ∫   ln       8         √
   Find        √           e2x       e2x + 1 ...
Another	way	to	skin	that	cat

   Example√
          ∫   ln       8         √
   Find        √           e2x          e2x +...
Another	way	to	skin	that	cat

   Example√
          ∫   ln       8         √
   Find        √           e2x          e2x +...
Another	way	to	skin	that	cat

   Example√
          ∫   ln       8         √
   Find        √           e2x          e2x +...
A third	skinned	cat


   Example√
          ∫   ln       8         √
   Find        √           e2x       e2x + 1 dx
     ...
A third	skinned	cat


   Example√
          ∫   ln       8         √
   Find        √           e2x       e2x + 1 dx
     ...
A third	skinned	cat


   Example√
          ∫   ln       8         √
   Find        √           e2x        e2x + 1 dx
    ...
Example
Find      ∫              ( )      ( )
              3π/2
                       5  θ      2 θ
                    ...
Example
Find                ∫              ( )      ( )
                        3π/2
                                 5  θ...
Solution
        θ              1
Let φ =   . Then dφ = dθ.
        6              6
    ∫ 3π/2       ( )      ( )        ...
Solution
        θ              1
Let φ =   . Then dφ = dθ.
        6              6
    ∫ 3π/2       ( )      ( )        ...
Graphs

        ∫   3π/2       ( )      ( )            ∫   π/4
                        θ      2 θ
        .            5
 ...
Graphs

             ∫   π/4                          ∫   1
             .             5       2
                       6 ...
Summary: What	do	we	substitute?


      Linear	factors (ax + b) are	easy	substitutions: u = ax + b,
      du = a dx
      ...
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Lesson 27: Integration by Substitution (Section 4 version)

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The method of substitution is the chain rule in reverse. At first it looks magical, then logical, and then you realize there's an art to choosing the right substitution. We try to demystify with many worked-out examples and graphics.

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Lesson 27: Integration by Substitution (Section 4 version)

  1. 1. Section 5.5 Integration by Substitution V63.0121, Calculus I April 28, 2009 Announcements Quiz 6 this week covering 5.1–5.2 Practice finals on the website. Solutions Friday . . Image credit: kchbrown . . . . . .
  2. 2. Outline Announcements Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . .
  3. 3. Office Hours and other help Day Time Who/What Where in WWH M 1:00–2:00 Leingang OH 624 5:00–7:00 Curto PS 517 T 1:00–2:00 Leingang OH 624 4:00–5:50 Curto PS 317 W 2:00–3:00 Leingang OH 624 R 9:00–10:00am Leingang OH 624 F 2:00–4:00 Curto OH 1310 . . . . . .
  4. 4. Final stuff Final is May 8, 2:00–3:50pm in 19W4 101/102 Old finals online, including Fall 2008 Review sessions: May 5 and 6, 6:00–8:00pm, SILV 703 . . Image credit: Pragmagraphr . . . . . .
  5. 5. Resurrection Policy If your final score beats your midterm score, we will add 10% to its weight, and subtract 10% from the midterm weight. . . Image credit: Scott Beale / Laughing Squid . . . . . .
  6. 6. Outline Announcements Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . .
  7. 7. Differentiation and Integration as reverse processes Theorem (The Fundamental Theorem of Calculus) 1. Let f be continuous on [a, b]. Then ∫ x d f(t) dt = f(x) dx a 2. Let f be continuous on [a, b] and f = F′ for some other function F. Then ∫ b f(x) dx = F(b) − F(a). a . . . . . .
  8. 8. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx . . . . . .
  9. 9. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx Some are pretty particular, like ∫ 1 √ dx = arcsec x + C. x x2 − 1 . . . . . .
  10. 10. Techniques of antidifferentiation? So far we know only a few rules for antidifferentiation. Some are general, like ∫ ∫ ∫ [f(x) + g(x)] dx = f(x) dx + g(x) dx Some are pretty particular, like ∫ 1 √ dx = arcsec x + C. x x2 − 1 What are we supposed to do with that? . . . . . .
  11. 11. So far we don’t have any way to find ∫ 2x √ dx x2 + 1 or ∫ tan x dx. . . . . . .
  12. 12. So far we don’t have any way to find ∫ 2x √ dx x2 + 1 or ∫ tan x dx. Luckily, we can be smart and use the “anti” version of one of the most important rules of differentiation: the chain rule. . . . . . .
  13. 13. Outline Announcements Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . .
  14. 14. Substitution for Indefinite Integrals Example Find ∫ x √ dx. x2 +1 . . . . . .
  15. 15. Substitution for Indefinite Integrals Example Find ∫ x √ dx. x2 +1 Solution Stare at this long enough and you notice the the integrand is the √ derivative of the expression 1 + x2 . . . . . . .
  16. 16. Say what? Solution (More slowly, now) Let g(x) = x2 + 1. . . . . . .
  17. 17. Say what? Solution (More slowly, now) Let g(x) = x2 + 1. Then g′ (x) = 2x and so d√ 1 x g(x) = √ g′ (x) = √ dx 2 g(x) x2 + 1 . . . . . .
  18. 18. Say what? Solution (More slowly, now) Let g(x) = x2 + 1. Then g′ (x) = 2x and so d√ 1 x g(x) = √ g′ (x) = √ dx 2 g(x) x2 + 1 Thus ∫ ∫ ( ) x d√ √ dx = g(x) dx x2 + 1 dx √ √ = g(x) + C = 1 + x2 + C. . . . . . .
  19. 19. Leibnizian notation wins again Solution (Same technique, new notation) Let u = x2 + 1. . . . . . .
  20. 20. Leibnizian notation wins again Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. . . . . . .
  21. 21. Leibnizian notation wins again Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ ∫ x 1 ( ) 1 √ dx = √ 1 du = 2 √ du x2 + 1 u 2 u . . . . . .
  22. 22. Leibnizian notation wins again Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ ∫ x 1 ( ) 1 √ dx = √ 1 du = 2 √ du x2 + 1 u 2 u ∫ 1 −1/2 = 2u du . . . . . .
  23. 23. Leibnizian notation wins again Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ ∫ x 1 ( ) 1 √ dx = √ 1 du = 2 √ du x2 + 1 u 2 u ∫ 1 −1/2 = 2u du √ √ = u + C = 1 + x2 + C. . . . . . .
  24. 24. Theorem of the Day Theorem (The Substitution Rule) If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then ∫ ∫ ′ f(g(x))g (x) dx = f(u) du or ∫ ∫ du f(u) dx = f(u) du dx . . . . . .
  25. 25. A polynomial example Example ∫ Use the substitution u = x2 + 3 to find (x2 + 3)3 4x dx. . . . . . .
  26. 26. A polynomial example Example ∫ Use the substitution u = x2 + 3 to find (x2 + 3)3 4x dx. Solution If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So ∫ ∫ ∫ (x + 3) 4x dx = u 2du = 2 u3 du 2 3 3 1 4 1 2 = u = (x + 3)4 2 2 . . . . . .
  27. 27. A polynomial example, the hard way Compare this to multiplying it out: ∫ ∫ 2 3 ( 6 ) (x + 3) 4x dx = x + 9x4 + 27x2 + 27 4x dx ∫ ( 7 ) = 4x + 36x5 + 108x3 + 108x dx 1 8 = x + 6x6 + 27x4 + 54x2 2 . . . . . .
  28. 28. Compare We have ∫ 1 2 (x2 + 3)3 4x dx = (x + 3)4 2 ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 2 Now 1 2 1( 8 ) (x + 3)4 = x + 12x6 + 54x4 + 108x2 + 81 2 2 1 81 = x8 + 6x6 + 27x4 + 54x2 + 2 2 Is this a problem? . . . . . .
  29. 29. Compare We have ∫ 1 2 (x2 + 3)3 4x dx = (x + 3)4 + C 2 ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 + C 2 Now 1 2 1( 8 ) (x + 3)4 = x + 12x6 + 54x4 + 108x2 + 81 2 2 1 81 = x8 + 6x6 + 27x4 + 54x2 + 2 2 Is this a problem? No, that’s what +C means! . . . . . .
  30. 30. A slick example Example ∫ Find tan x dx. . . . . . .
  31. 31. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x . . . . . .
  32. 32. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = − sin x dx. . . . . . .
  33. 33. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = − sin x dx. So ∫ ∫ ∫ sin x 1 tan x dx = dx = − du cos x u . . . . . .
  34. 34. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = − sin x dx. So ∫ ∫ ∫ sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C . . . . . .
  35. 35. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution Let u = cos x. Then du = − sin x dx. So ∫ ∫ ∫ sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C = − ln | cos x| + C = ln | sec x| + C . . . . . .
  36. 36. Outline Announcements Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . .
  37. 37. Theorem (The Substitution Rule for Definite Integrals) If g′ is continuous and f is continuous on the range of u = g(x), then ∫ ∫ b g(b) f(g(x))g′ (x) dx = f(u) du. a g(a) . . . . . .
  38. 38. Example ∫ π Compute cos2 x sin x dx. 0 . . . . . .
  39. 39. Example ∫ π Compute cos2 x sin x dx. 0 Solution (Slow Way) ∫ First compute the indefinite integral cos2 x sin x dx and then evaluate. . . . . . .
  40. 40. Example ∫ π Compute cos2 x sin x dx. 0 Solution (Slow Way) ∫ First compute the indefinite integral cos2 x sin x dx and then evaluate. Let u = cos x. Then du = − sin x dx and ∫ ∫ cos2 x sin x dx = − u2 du = − 1 u3 + C = − 1 cos3 x + C. 3 3 Therefore ∫ π π cos2 x sin x dx = − 1 cos3 x 3 0 = 2. 3 0 . . . . . .
  41. 41. Solution (Fast Way) Do both the substitution and the evaluation at the same time. . . . . . .
  42. 42. Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. . . . . . .
  43. 43. Solution (Fast Way) Do both the substitution and the evaluation at the same time. Let u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1. So ∫ π ∫ −1 cos2 x sin x dx = −u2 du 0 1 ∫ 1 = u2 du −1 1 3 1 2 = 3 u −1 = 3 . . . . . . .
  44. 44. An exponential example Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 . . . . . .
  45. 45. An exponential example Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have ∫ √ ∫ ln 8 √ 1 8√ √ e2x e2x + 1 dx = u + 1 du ln 3 2 3 . . . . . .
  46. 46. An exponential example Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x , so du = 2e2x dx. We have ∫ √ ∫ ln 8 √ 1 8√ √ e2x e2x + 1 dx = u + 1 du ln 3 2 3 Now let y = u + 1, dy = du. So ∫ 8√ ∫ 9 ∫ 9 1 1 √ 1 u + 1 du = y dy = y1/2 dy 2 3 2 4 2 4 9 1 2 1 19 = · y3/2 = (27 − 8) = 2 3 4 3 3 . . . . . .
  47. 47. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1 . . . . . .
  48. 48. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. . . . . . .
  49. 49. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ 2x √ e e2x + 1 dx = u du ln 3 2 4 . . . . . .
  50. 50. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ 2x √ e e2x + 1 dx = u du ln 3 2 4 9 1 3/2 = u 3 4 . . . . . .
  51. 51. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1, so that du = 2e2x dx. Then ∫ √ ∫ ln 8 √ 1 9√ 2x √ e e2x + 1 dx = u du ln 3 2 4 1 3/2 9 = u 3 4 1 19 = (27 − 8) = 3 3 . . . . . .
  52. 52. A third skinned cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 . . . . . .
  53. 53. A third skinned cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx . . . . . .
  54. 54. A third skinned cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx Thus ∫ √ ∫ ln 8 3 3 1 3 19 √ = u · u du = u = ln 3 2 3 2 3 . . . . . .
  55. 55. Example Find ∫ ( ) ( ) 3π/2 5 θ 2 θ cot sec dθ. π 6 6 . . . . . .
  56. 56. Example Find ∫ ( ) ( ) 3π/2 5 θ 2 θ cot sec dθ. π 6 6 Before we dive in, think about: What “easy” substitutions might help? Which of the trig functions suggests a substitution? . . . . . .
  57. 57. Solution θ 1 Let φ = . Then dφ = dθ. 6 6 ∫ 3π/2 ( ) ( ) ∫ π/4 5 θ 2 θ cot sec dθ = 6 cot5 φ sec2 φ dφ π 6 6 π/6 ∫ π/4 sec2 φ dφ =6 π/6 tan5 φ . . . . . .
  58. 58. Solution θ 1 Let φ = . Then dφ = dθ. 6 6 ∫ 3π/2 ( ) ( ) ∫ π/4 5 θ 2 θ cot sec dθ = 6 cot5 φ sec2 φ dφ π 6 6 π/6 ∫ π/4 sec2 φ dφ =6 π/6 tan5 φ Now let u = tan φ. So du = sec2 φ dφ, and ∫ π/4 ∫ 1 sec2 φ dφ −5 6 =6 √ u du π/6 tan5 φ 1/ 3 ( ) 1 1 3 =6 − u−4 √ = [9 − 1] = 12. 4 1/ 3 2 . . . . . .
  59. 59. Graphs ∫ 3π/2 ( ) ( ) ∫ π/4 θ 2 θ . 5 cot sec dθ . 6 cot5 φ sec2 φ dφ π 6 6 π/6 y . y . . . . . θ . . . φ . π 3π ππ . . . 2 64 . . . . . .
  60. 60. Graphs ∫ π/4 ∫ 1 . 5 2 6 cot φ sec φ dφ . √ 6u−5 du π/6 1/ 3 y . y . . . . . φ . .. u ππ 1 . 1 . . .√ 64 3 . . . . . .
  61. 61. Summary: What do we substitute? Linear factors (ax + b) are easy substitutions: u = ax + b, du = a dx Look for function/derivative pairs in the integrand, one to make u and one to make du: xn and xn−1 (fudge the coefficient) sine and cosine (fudge the minus sign) ex and ex ax and ax (fudge the coefficient) √ 1 x and √ (fudge the factor of 2) x 1 ln x and x . . . . . .

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