Section	5.5
         Integration	by	Substitution

                  V63.0121.027, Calculus	I



                     Decem...
Schedule	for	next	week



      Tuesday, 8:00am: Review	session	for	all	students	with	8:00
      recitations	(Tuesday	or	T...
Resurrection	Policy
        If	your	final	score	beats	your	midterm	score, we	will	add	10%	to
        its	weight, and	subtra...
Outline


  Last	Time: The	Fundamental	Theorem(s)	of	Calculus


  Substitution	for	Indefinite	Integrals
     Theory
     Ex...
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


  Last	Time: The	Fundamental	Theorem(s)	of	Calculus


  Substitution	for	Indefinite	Integrals
     Theory
     Ex...
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	FTW



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




                               ...
Leibnizian	notation	FTW



  Solution	(Same	technique, new	notation)
                                  √                  ...
Leibnizian	notation	FTW



  Solution	(Same	technique, new	notation)
                                  √                  ...
Leibnizian	notation	FTW



  Solution	(Same	technique, new	notation)
                                  √                  ...
Leibnizian	notation	FTW



  Solution	(Same	technique, new	notation)
                                  √                  ...
Useful	but	unsavory	variation

   Solution	(Same	technique, new	notation, more	idiot-proof)
                              ...
Theorem	of	the	Day

  Theorem	(The	Substitution	Rule)
  If u = g(x) is	a	differentiable	function	whose	range	is	an	interva...
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, by	brute	force


   Compare	this	to	multiplying	it	out:
      ∫                    ∫
          2    ...
A polynomial	example, by	brute	force


   Compare	this	to	multiplying	it	out:
      ∫                    ∫
          2    ...
A polynomial	example, by	brute	force


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

  We	have	the	substitution	method, which, when	multiplied	out,
  gives
     ∫
                         1
       (...
Compare

  We	have	the	substitution	method, which, when	multiplied	out,
  gives
     ∫
                         1
       (...
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 =       ...
Can	you	do	it	another	way?

   Example
       ∫
                                     sin x
   Find   tan x dx. (Hint: tan ...
Can	you	do	it	another	way?

   Example
       ∫
                                     sin x
   Find   tan x dx. (Hint: tan ...
Can	you	do	it	another	way?

   Example
       ∫
                                     sin x
   Find   tan x dx. (Hint: tan ...
For	those	who	really	must	know	all


   Solution	(Continued, with	algebra	help)

       ∫             ∫              ∫    ...
Outline


  Last	Time: The	Fundamental	Theorem(s)	of	Calculus


  Substitution	for	Indefinite	Integrals
     Theory
     Ex...
Theorem	(The	Substitution	Rule	for	Definite	Integrals)
If g′ is	continuous	and f is	continuous	on	the	range	of u = g(x),
th...
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...
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
       ...
About	those	limits




                        √                √       2
                e2(ln       3)
                 ...
About	those	fractional	powers




                 93/2 = (91/2 )3 = 33 = 27
                 43/2 = (41/2 )3 = 23 = 8



...
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       ( )      ( )        ...
The	limits	explained

                              √
                  π  sin π/4   2 /2
               tan =         =√ ...
The	limits	explained

                              √
                  π  sin π/4   2 /2
               tan =         =√ ...
Graphs

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

             ∫   π/4                          ∫   1
             .             5       2
                       6 ...
Final	Thoughts




      Antidifferentiation	is	a	“nonlinear”	problem	that	needs
      practice, intuition, and	perservera...
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Lesson 29: Integration by Substition

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Lesson 29: Integration by Substition

  1. 1. Section 5.5 Integration by Substitution V63.0121.027, Calculus I December 10, 2009 Announcements Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50 Practice finals on the website. Solutions Friday . . . . . .
  2. 2. Schedule for next week Tuesday, 8:00am: Review session for all students with 8:00 recitations (Tuesday or Thursday) in CIWW 109 Tuesday, 9:30am: Review session for all students with 9:30 recitations (Tuesday or Thursday) in CIWW 109 Office Hours continue Tuesday, class: review, evaluations, movie! Friday, 2:00pm: final in Tisch UC50 . . . . . .
  3. 3. 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 . . . . . .
  4. 4. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . .
  5. 5. 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 . . . . . .
  6. 6. 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 . . . . . .
  7. 7. 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 . . . . . .
  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 Some are pretty particular, like ∫ 1 √ dx = arcsec x + C. x x2 − 1 What are we supposed to do with that? . . . . . .
  9. 9. So far we don’t have any way to find ∫ 2x √ dx x2 + 1 or ∫ tan x dx. . . . . . .
  10. 10. 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. . . . . . .
  11. 11. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . .
  12. 12. Substitution for Indefinite Integrals Example Find ∫ x √ dx. x 2+1 . . . . . .
  13. 13. Substitution for Indefinite Integrals Example Find ∫ x √ dx. x 2+1 Solution Stare at this long enough and you notice the the integrand is the √ derivative of the expression 1 + x2 . . . . . . .
  14. 14. Say what? Solution (More slowly, now) Let g(x) = x2 + 1. . . . . . .
  15. 15. 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 . . . . . .
  16. 16. 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. . . . . . .
  17. 17. Leibnizian notation FTW Solution (Same technique, new notation) Let u = x2 + 1. . . . . . .
  18. 18. Leibnizian notation FTW Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. . . . . . .
  19. 19. Leibnizian notation FTW Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ 1 ∫ √ x dx 2 du √ 1 √ du = = x2 + 1 u 2 u . . . . . .
  20. 20. Leibnizian notation FTW Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ 1 ∫ √ x dx 2 du √ 1 √ du = = x2 + 1 ∫ u 2 u 1 −1/2 = 2u du . . . . . .
  21. 21. Leibnizian notation FTW Solution (Same technique, new notation) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrand becomes completely transformed into ∫ ∫ 1 ∫ √ x dx 2 du √ 1 √ du = = x2 + 1 ∫ u 2 u 1 −1/2 = 2u du √ √ = u+C= 1 + x2 + C. . . . . . .
  22. 22. Useful but unsavory variation Solution (Same technique, new notation, more idiot-proof) √ √ Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. “Solve for dx:” du dx = 2x So the integrand becomes completely transformed into ∫ ∫ ∫ x x du 1 √ dx = √ · = √ du x2 + 1 u 2x 2 u ∫ 1 −1/2 = 2u du √ √ = u + C = 1 + x2 + C . Mathematicians have serious issues with mixing the x and u like this. However, I can’t deny that it works. . . . . . .
  23. 23. 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 That is, if F is an antiderivative for f, then ∫ f(g(x))g′ (x) dx = F(g(x)) In Leibniz notation: ∫ ∫ du f(u) dx = f(u) du dx . . . . . .
  24. 24. A polynomial example Example ∫ Use the substitution u = x2 + 3 to find (x2 + 3)3 4x dx. . . . . . .
  25. 25. 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 . . . . . .
  26. 26. A polynomial example, by brute force 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 . . . . . .
  27. 27. A polynomial example, by brute force 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 Which would you rather do? . . . . . .
  28. 28. A polynomial example, by brute force 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 Which would you rather do? It’s a wash for low powers But for higher powers, it’s much easier to do substitution. . . . . . .
  29. 29. Compare We have the substitution method, which, when multiplied out, gives ∫ 1 (x2 + 3)3 4x dx = (x2 + 3)4 2 1( 8 ) = x + 12x6 + 54x4 + 108x2 + 81 2 1 81 = x8 + 6x6 + 27x4 + 54x2 + 2 2 and the brute force method ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 2 Is this a problem? . . . . . .
  30. 30. Compare We have the substitution method, which, when multiplied out, gives ∫ 1 (x2 + 3)3 4x dx = (x2 + 3)4 + C 2 1( 8 ) = x + 12x6 + 54x4 + 108x2 + 81 + C 2 1 81 = x8 + 6x6 + 27x4 + 54x2 + +C 2 2 and the brute force method ∫ 1 (x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 + C 2 Is this a problem? No, that’s what +C means! . . . . . .
  31. 31. A slick example Example ∫ Find tan x dx. . . . . . .
  32. 32. A slick example Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x . . . . . .
  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. . . . . . .
  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 . . . . . .
  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 . . . . . .
  36. 36. 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 . . . . . .
  37. 37. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x . . . . . .
  38. 38. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution du Let u = sin x. Then du = cos x dx and so dx = . cos x . . . . . .
  39. 39. Can you do it another way? Example ∫ sin x Find tan x dx. (Hint: tan x = ) cos x Solution du Let u = sin x. Then du = cos x dx and so dx = . cos x ∫ ∫ ∫ sin x u du tan x dx = dx = cos x cos x cos x ∫ ∫ ∫ u du u du u du = = 2 = cos2 x 1 − sin x 1 − u2 At this point, although it’s possible to proceed, we should probably back up and see if the other way works quicker (it does). . . . . . .
  40. 40. For those who really must know all Solution (Continued, with algebra help) ∫ ∫ ∫ ( ) u du 1 1 1 tan x dx = = − du 1 − u2 2 1−u 1+u 1 1 = − ln |1 − u| − ln |1 + u| + C 2 2 1 1 = ln √ + C = ln √ +C (1 − u)(1 + u) 1 − u2 1 = ln + C = ln |sec x| + C |cos x| . . . . . .
  41. 41. Outline Last Time: The Fundamental Theorem(s) of Calculus Substitution for Indefinite Integrals Theory Examples Substitution for Definite Integrals Theory Examples . . . . . .
  42. 42. 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 ) . . . . . .
  43. 43. 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 ) Why the change in the limits? The integral on the left happens in “x-land” The integral on the right happens in “u-land”, so the limits need to be u-values To get from x to u, apply g . . . . . .
  44. 44. Example ∫ π Compute cos2 x sin x dx. 0 . . . . . .
  45. 45. 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. . . . . . .
  46. 46. 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 ∫ ∫ cos x sin x dx = − u2 du 2 = − 1 u3 + C = − 1 cos3 x + C. 3 3 Therefore ∫ π 1 π 1( ) 2 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = . 0 3 0 3 3 . . . . . .
  47. 47. Solution (Fast Way) Do both the substitution and the evaluation at the same time. . . . . . .
  48. 48. 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. . . . . . .
  49. 49. 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 1 1( ) 2 = u3 = 1 − (−1) = 3 −1 3 3 . . . . . .
  50. 50. 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 1 1( ) 2 = u3 = 1 − (−1) = 3 −1 3 3 The advantage to the “fast way” is that you completely transform the integral into something simpler and don’t have to go back to the original x variable. But the slow way is just as reliable. . . . . . .
  51. 51. An exponential example Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 . . . . . .
  52. 52. 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 . . . . . .
  53. 53. 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 . . . . . .
  54. 54. About those limits √ √ 2 e2(ln 3) = eln 3 = eln 3 = 3 . . . . . .
  55. 55. About those fractional powers 93/2 = (91/2 )3 = 33 = 27 43/2 = (41/2 )3 = 23 = 8 . . . . . .
  56. 56. Another way to skin that cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution Let u = e2x + 1 . . . . . .
  57. 57. 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. . . . . . .
  58. 58. 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 . . . . . .
  59. 59. 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 . . . . . .
  60. 60. 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 . . . . . .
  61. 61. A third skinned cat Example√ ∫ ln 8 √ Find √ e2x e2x + 1 dx ln 3 Solution √ Let u = e2x + 1, so that u2 = e2x + 1 . . . . . .
  62. 62. 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 . . . . . .
  63. 63. 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 . . . . . .
  64. 64. Example Find ∫ ( ) ( ) 3π/2 5 θ 2 θ cot sec dθ. π 6 6 . . . . . .
  65. 65. 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? . . . . . .
  66. 66. 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 . . . . . .
  67. 67. 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 . . . . . .
  68. 68. The limits explained √ π sin π/4 2 /2 tan = =√ =1 4 cos π/4 2 /2 π sin π/6 1/2 1 tan = =√ =√ 6 cos π/6 3 /2 3 . . . . . .
  69. 69. The limits explained √ π sin π/4 2 /2 tan = =√ =1 4 cos π/4 2 /2 π sin π/6 1/2 1 tan = =√ =√ 6 cos π/6 3 /2 3 ( ) 1 √ 1 −4 3 [ −4 ]1 √ 3 [ −4 ]1/ 3 6 − u √ = −u 1 / 3 = u 1 4 1/ 3 2 2 3 [ −1/2 −4 ] = (3 ) − (1−1/2 )−4 2 3 3 = [32 − 12 ] = (9 − 1) = 12 2 2 . . . . . .
  70. 70. 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 . . . . . .
  71. 71. 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 . . . . . .
  72. 72. Final Thoughts Antidifferentiation is a “nonlinear” problem that needs practice, intuition, and perserverance Worksheet in recitation (also to be posted) The whole antidifferentiation story is in Chapter 6 . . . . . .
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