This document contains the notes from a calculus class. It provides announcements about the final exam schedule and review sessions. It then discusses the technique of u-substitution for both indefinite and definite integrals. Examples are provided to illustrate how to use u-substitution to evaluate integrals involving trigonometric, polynomial, and other functions. The document emphasizes that u-substitution often makes evaluating integrals much easier than expanding them out directly.

1. Section 5.5
Integration by Substitution
V63.0121.034, Calculus I
December 9, 2009
Announcements
Final Exam: Friday 12/18, 2:00-3:50pm, Tisch UC50
Practice finals on the website. Solutions Friday
. . . . . .

2. Schedule for next week
Monday, class: review, evaluations, movie!
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
Friday, 2:00pm: final in Tisch UC50
. . . . . .

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. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
. . . . . .

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. 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. 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. 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. So far we don’t have any way to find
∫
2x
√ dx
x2 + 1
or ∫
tan x dx.
. . . . . .

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. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
. . . . . .

12. Substitution for Indefinite Integrals
Example
Find ∫
x
√ dx.
x 2+1
. . . . . .

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

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

18. Leibnizian notation FTW
Solution (Same technique, new notation)
√ √
Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u.
. . . . . .

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. 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. 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. 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. 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. A polynomial example
Example ∫
Use the substitution u = x2 + 3 to find (x2 + 3)3 4x dx.
. . . . . .

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. 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. 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. 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. 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. 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. A slick example
Example
∫
Find tan x dx.
. . . . . .

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

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

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. 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. 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. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
. . . . . .

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

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

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. 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. 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. An exponential example
Example√
∫ ln 8 √
Find √ e2x e2x + 1 dx
ln 3
. . . . . .

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. 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. About those limits
√ √ 2
e2(ln 3)
= eln 3
= eln 3 = 3
. . . . . .

55. About those fractional powers
93/2 = (91/2 )3 = 33 = 27
43/2 = (41/2 )3 = 23 = 8
. . . . . .

56. Another way to skin that cat
Example√
∫ ln 8 √
Find √ e2x e2x + 1 dx
ln 3
Solution
Let u = e2x + 1
. . . . . .

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. 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. 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. 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. 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. 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. 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. Example
Find ∫ ( ) ( )
3π/2
5 θ 2 θ
cot sec dθ.
π 6 6
. . . . . .

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. 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. 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. 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. 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. 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
The areas of these two regions are the same. . . . . . .

71. Graphs
∫ π/4 ∫ 1
. 6 cot5 φ sec2 φ dφ . √ 6u−5 du
π/6 1/ 3
y
. y
.
. . . .
φ . ..
u
ππ 1 .1
. . .√
64 3 The areas
of these two regions are the same.
. . . . . .

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
. . . . . .