The document is a summary of lecture notes for a Calculus I class. It discusses integration by substitution, providing theory, examples, and objectives. Key points covered include the substitution rule for indefinite integrals, working through examples like finding the integral of √x2+1 dx, and noting substitution can transform integrals into simpler forms. Definite integrals using substitution are also briefly mentioned.

1. Sec on 5.5
Integra on by Subs tu on
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University
May 4, 2011
.

2. Announcements
Today: 5.5
Monday 5/9: Review in class
Tuesday 5/10: Review Sessions by TAs
Wednesday 5/11: TA office hours:
Adam 10–noon (WWH 906)
Jerome 3:30–5:30 (WWH 501)
Soohoon 6–8 (WWH 511)
Thursday 5/12: Final Exam,
2:00–3:50pm, CANT 200

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: Sco Beale / Laughing Squid

4. Objectives
Given an integral and a
subs tu on, transform the
integral into an equivalent one
using a subs tu on
Evaluate indefinite integrals
using the method of
subs tu on.
Evaluate definite integrals using
the method of subs tu on.

5. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite Integrals
Theory
Examples
Subs tu on for Definite Integrals
Theory
Examples

6. Differentiation and Integration as
reverse processes
Theorem (The Fundamental Theorem of Calculus)
1. Let f be con nuous on [a, b]. Then
∫
d x
f(t) dt = f(x)
dx a
2. Let f be con nuous on [a, b] and f = F′ for some other func on
F. Then ∫ b
f(x) dx = F(b) − F(a).
a

7. Techniques of antidifferentiation?
So far we know only a few rules for an differen a on. Some are
general, like
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx

8. Techniques of antidifferentiation?
So far we know only a few rules for an differen a on. Some are
general, like
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
Some are pre y par cular, like
∫
1
√ dx = arcsec x + C.
x x2 − 1

9. Techniques of antidifferentiation?
So far we know only a few rules for an differen a on. Some are
general, like
∫ ∫ ∫
[f(x) + g(x)] dx = f(x) dx + g(x) dx
Some are pre y par cular, like
∫
1
√ dx = arcsec x + C.
x x2 − 1
What are we supposed to do with that?

10. No straightforward system of
antidifferentiation
So far we don’t have any way to find
∫
2x
√ dx
x2 + 1
or ∫
tan x dx.

11. No straightforward system of
antidifferentiation
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 “an ” version of one of the
most important rules of differen a on: the chain rule.

12. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite Integrals
Theory
Examples
Subs tu on for Definite Integrals
Theory
Examples

13. Substitution for Indefinite
Integrals
Example
Find ∫
x
√ dx.
x2 + 1

14. Substitution for Indefinite
Integrals
Example
Find ∫
x
√ dx.
x2 + 1
Solu on
Stare at this long enough and you no ce the the integrand is the
√
deriva ve of the expression 1 + x2 .

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

16. Say what?
Solu on (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

17. Say what?
Solu on (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.

18. Leibnizian notation FTW
Solu on (Same technique, new nota on)
Let u = x2 + 1.

19. Leibnizian notation FTW
Solu on (Same technique, new nota on)
√ √
Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u.

20. Leibnizian notation FTW
Solu on (Same technique, new nota on)
√ √
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

21. Leibnizian notation FTW
Solu on (Same technique, new nota on)
√ √
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

22. Leibnizian notation FTW
Solu on (Same technique, new nota on)
√ √
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.

23. Useful but unsavory variation
Solu on (Same technique, new nota on, more idiot-proof)
√ √
Let u = x2 + 1. Then du = 2x dx and 1 + x2 = u. “Solve for dx:”
du
dx =
2x

24. Useful but unsavory variation
Solu on (Same technique, new nota on, 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
√ dx = √ ·
x2 + 1 u 2x

25. Useful but unsavory variation
Solu on (Same technique, new nota on, 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

26. Useful but unsavory variation
Solu on (Same technique, new nota on, 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

27. Useful but unsavory variation
Solu on (Same technique, new nota on, 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
√
= 2 u du = u + C = 1 + x2 + C.

28. Theorem of the Day
Theorem (The Subs tu on Rule)
If u = g(x) is a differen able func on whose range is an interval I
and f is con nuous on I, then
∫ ∫
′
f(g(x))g (x) dx = f(u) du

29. Theorem of the Day
Theorem (The Subs tu on Rule)
If u = g(x) is a differen able func on whose range is an interval I
and f is con nuous on I, then
∫ ∫
′
f(g(x))g (x) dx = f(u) du
That is, if F is an an deriva ve for f, then
∫
f(g(x))g′ (x) dx = F(g(x))

30. Theorem of the Day
Theorem (The Subs tu on Rule)
If u = g(x) is a differen able func on whose range is an interval I
and f is con nuous on I, then
∫ ∫
′
f(g(x))g (x) dx = f(u) du
In Leibniz nota on:
∫ ∫
du
f(u) dx = f(u) du
dx

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

32. A polynomial example
Example
∫
Use the subs tu on u = x2 + 3 to find (x2 + 3)3 4x dx.
Solu on
If u = x2 + 3, then du = 2x dx, and 4x dx = 2 du. So
∫
(x2 + 3)3 4x dx

33. A polynomial example
Example
∫
Use the subs tu on u = x2 + 3 to find (x2 + 3)3 4x dx.
Solu on
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

34. A polynomial example
Example
∫
Use the subs tu on u = x2 + 3 to find (x2 + 3)3 4x dx.
Solu on
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
= u4
2

35. A polynomial example
Example
∫
Use the subs tu on u = x2 + 3 to find (x2 + 3)3 4x dx.
Solu on
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 1
= u4 = (x2 + 3)4
2 2

36. A polynomial example (brute force)
Solu on

37. A polynomial example (brute force)
Solu on
∫
(x2 + 3)3 4x dx

38. A polynomial example (brute force)
Solu on
∫ ∫
2 3
( )
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx

39. A polynomial example (brute force)
Solu on
∫ ∫
2 3
( )
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫
( )
= 4x7 + 36x5 + 108x3 + 108x dx

40. A polynomial example (brute force)
Solu on
∫ ∫
2 3
( )
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫
( )
= 4x7 + 36x5 + 108x3 + 108x dx
1
= x8 + 6x6 + 27x4 + 54x2
2

41. A polynomial example (brute force)
Solu on
∫ ∫
2 3
( )
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫
( )
= 4x7 + 36x5 + 108x3 + 108x dx
1
= x8 + 6x6 + 27x4 + 54x2
2
Which would you rather do?

42. A polynomial example (brute force)
Solu on
∫ ∫
2 3
( )
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫
( )
= 4x7 + 36x5 + 108x3 + 108x dx
1
= x8 + 6x6 + 27x4 + 54x2
2
Which would you rather do?
It’s a wash for low powers

43. A polynomial example (brute force)
Solu on
∫ ∫
2 3
( )
(x + 3) 4x dx = x6 + 9x4 + 27x2 + 27 4x dx
∫
( )
= 4x7 + 36x5 + 108x3 + 108x dx
1
= x8 + 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 subs tu on.

44. Compare
We have the subs tu on method, which, when mul plied 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 there a difference? Is this a problem?

45. Compare
We have the subs tu on method, which, when mul plied 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 there a difference? Is this a problem? No, that’s what +C means!

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

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

48. A slick example
Example
∫
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solu on
Let u = cos x . Then du = − sin x dx . So
∫ ∫
sin x
tan x dx = dx
cos x

49. A slick example
Example
∫
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solu on
Let u = cos x . Then du = − sin x dx . So
∫ ∫
sin x
tan x dx = dx
cos x

50. A slick example
Example
∫
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solu on
Let u = cos x . Then du = − sin x dx . So
∫ ∫
sin x
tan x dx = dx
cos x

51. A slick example
Example
∫
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solu on
Let u = cos x . Then du = − sin x dx . So
∫ ∫ ∫
sin x 1
tan x dx = dx = − du
cos x u

52. A slick example
Example
∫
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solu on
Let u = cos x . Then du = − sin x dx . So
∫ ∫ ∫
sin x 1
tan x dx = dx = − du
cos x u
= − ln |u| + C

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

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

55. Can you do it another way?
Example
∫
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solu on
du
Let u = sin x. Then du = cos x dx and so dx = .
cos x

56. Can you do it another way?
Example
∫
sin x
Find tan x dx. (Hint: tan x = )
cos x
Solu on
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
= = =
cos2 x 1 − sin2 x 1 − u2

57. For those who really must know all
Solu on (Con nued, with algebra help)
Let y = 1 − u2 , so dy = −2u du. Then
∫ ∫ ∫
u du u dy
tan x dx = =
1∫− u2 y −2u
1 dy 1 1
=− = − ln |y| + C = − ln 1 − u2 + C
2 y 2 2
1 1
= ln √ + C = ln √ +C
1 − u2 1 − sin2 x
1
= ln + C = ln |sec x| + C
|cos x|

58. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Subs tu on for Indefinite Integrals
Theory
Examples
Subs tu on for Definite Integrals
Theory
Examples

59. Substitution for Definite Integrals
Theorem (The Subs tu on Rule for Definite Integrals)
If g′ is con nuous and f is con nuous on the range of u = g(x), then
∫ b ∫ g(b)
′
f(g(x))g (x) dx = f(u) du.
a g(a)

60. Substitution for Definite Integrals
Theorem (The Subs tu on Rule for Definite Integrals)
If g′ is con nuous and f is con nuous 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 le 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

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

62. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (Slow Way)

63. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (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

64. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (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

65. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (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

66. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (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

67. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (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 = − 3 cos3 x + C.
3
1

68. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (Slow Way)
Let u = cos x . Then du = − sin x dx and
∫ ∫
cos x sin x dx = − u2 du = − 1 u3 + C = − 3 cos3 x + C.
2
3
1
Therefore
∫ π π
1
cos2 x sin x dx = − cos3 x
0 3 0

69. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (Slow Way)
Let u = cos x . Then du = − sin x dx and
∫ ∫
cos x sin x dx = − u2 du = − 1 u3 + C = − 3 cos3 x + C.
2
3
1
Therefore
∫ π
1( )
π
1
cos2 x sin x dx = − cos3 x =− (−1)3 − 13
0 3 0 3

70. Example
∫ π
Compute cos2 x sin x dx.
0
Solu on (Slow Way)
Let u = cos x . Then du = − sin x dx and
∫ ∫
cos x sin x dx = − u2 du = − 1 u3 + C = − 3 cos3 x + C.
2
3
1
Therefore
∫ π
1( ) 2
π
1
cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = .
0 3 0 3 3

71. Definite-ly Quicker
Solu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
∫ π
cos2 x sin x dx
0

72. Definite-ly Quicker
Solu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
∫ π
cos2 x sin x dx
0

73. Definite-ly Quicker
Solu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
∫ π
cos2 x sin x dx
0

74. Definite-ly Quicker
Solu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
∫ π ∫ −1 ∫ 1
2
cos x sin x dx = −u du =
2
u2 du
0 1 −1

75. Definite-ly Quicker
Solu on (Fast Way)
Do both the subs tu on and the evalua on at the same me. Let
u = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So
∫ π ∫ −1 ∫ 1
2
cos x sin x dx = −u du =
2
u2 du
0 1 −1
1( ) 2
1
1 3
= u = 1 − (−1) =
3 −1 3 3

76. Compare
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 variable (x).
But the slow way is just as reliable.

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

78. An exponential example
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
Let u = e2x , so du = 2e2x dx. We have
∫ √ ∫
ln 8
2x
√ 1 8√
√ e e2x + 1 dx = u + 1 du
ln 3 2 3

79. About those limits
Since
√ √ 2
e2(ln 3)
= eln 3
= eln 3 = 3
we have ∫ √ ∫
ln 8 √ 1 8√
√ e2x e2x + 1 dx = u + 1 du
ln 3 2 3

80. An exponential example
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
Now let y = u + 1, dy = du. So
∫ ∫
1 8√
9
1 9√ 1 2 1 19
u + 1 du = y dy = · y3/2 = (27 − 8) =
2 3 2 4 2 3 4 3 3

81. About those fractional powers
We have
93/2 = (91/2 )3 = 33 = 27
43/2 = (41/2 )3 = 23 = 8
so ∫ 9 9
1 1 2 1 19
1/2
y dy = · y3/2 = (27 − 8) =
2 4 2 3 4 3 3

82. An exponential example
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
Now let y = u + 1, dy = du. So
∫ ∫
1 8√
9
1 9√ 1 2 1 19
u + 1 du = y dy = · y3/2 = (27 − 8) =
2 3 2 4 2 3 4 3 3

83. Another way to skin that cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
Let u = e2x + 1,

84. Another way to skin that cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
Let u = e2x + 1,so that du = 2e2x dx.

85. Another way to skin that cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
Let u = e2x + 1,so that du = 2e2x dx. Then
∫ √ ∫
ln 8 √ 1 9√
√ e2x e2x + 1 dx = u du
ln 3 2 4

86. Another way to skin that cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
Let u = e2x + 1,so that du = 2e2x dx. Then
∫ √ ∫
ln 8 √ 1 9√ 1
9
√ e2x e2x + 1 dx = u du = u3/2
ln 3 2 4 3 4

87. Another way to skin that cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
Let u = e2x + 1,so that du = 2e2x dx. Then
∫ √ ∫
ln 8 √ 1 9√ 1
9
1 19
√ e2x e2x + 1 dx = u du = u3/2 = (27 − 8) =
ln 3 2 4 3 4 3 3

88. A third skinned cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
√
Let u = e2x + 1, so that u2 = e2x + 1

89. A third skinned cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
√
Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx

90. A third skinned cat
Example
∫ ln √8 √
Find √ e2x e2x + 1 dx
ln 3
Solu on
√
Let u = e2x + 1, so that u2 = e2x + 1 =⇒ 2u du = 2e2x dx Thus
∫ √ ∫ 3
ln 8 √ 1
3
19
√ e2x e2x + 1 dx = u · u du = u3 =
ln 3 2 3 2 3

91. A Trigonometric Example
Example
Find ∫ ( ) ( )
3π/2
θ θ
cot5 sec2 dθ.
π 6 6

92. A Trigonometric Example
Example
Find ∫ ( ) ( )
3π/2
θ θ
cot5 sec2 dθ.
π 6 6
Before we dive in, think about:
What “easy” subs tu ons might help?
Which of the trig func ons suggests a subs tu on?

93. Solu on
θ 1
Let φ = . Then dφ = dθ.
6 6
∫ 3π/2 ( ) ( ) ∫ π/4
θ θ
cot5 sec2 dθ = 6 cot5 φ sec2 φ dφ
π 6 6 π/6
∫ π/4
sec2 φ dφ
=6
π/6 tan5 φ

94. Solu on
θ 1
Let φ = . Then dφ = dθ.
6 6
∫ 3π/2 ( ) ( ) ∫ π/4
θ θ
cot5 sec2 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

95. Solu on
Now let u = tan φ. So du = sec2 φ dφ, and
∫ π/4 ∫ 1
sec2 φ dφ
6 5φ
= 6 √ u−5 du
π/6 tan 1/ 3
( )1
1 −4 3
=6 − u √
= [9 − 1] = 12.
4 1/ 3 2

96. 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

97. The limits explained
( )
3 [ −4 ]1 √ 3 [ −4 ]1/√3
1
1 −4
6 − u √
= −u 1/ 3 = u 1
4 1/ 3 2 2
3 [ −1/2 −4 −1/2 −4
]
= (3 ) − (1 )
2
3 3
= [32 − 12 ] = (9 − 1) = 12
2 2

98. Graphs
∫ 3π/2 ( ) ( ) ∫ π/4
θ θ
cot5 sec2 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.

99. ∫ ∫
Graphs π/4
5 2
6 cot φ sec φ dφ
1
√ 6u
−5
du
π/6 1/ 3
y y
. φ u
ππ 11
√
64 3
The areas of these two regions are the same.

100. u/du pairs
When deciding on a subs tu on, look for sub-expressions where
one is (a constant mul ple of) the deriva ve of the other. Such as:
√
u xn ln x sin x cos x tan x x ex
1 1
constant × du xn−1 cos x sin x sec2 x √ ex
x x

101. Summary
If F is an an deriva ve for f, then:
∫
f(g(x))g′ (x) dx = F(g(x))
If F is an an deriva ve for f, which is con nuous on the range
of g, then:
∫ b ∫ g(b)
′
f(g(x))g (x) dx = f(u) du = F(g(b)) − F(g(a))
a g(a)
An differen a on in general and subs tu on in par cular is a
“nonlinear” problem that needs prac ce, intui on, and
perserverance.
The whole an differen a on story is in Chapter 6.