Lesson 27: Integration by Substitution (Section 041 slides)
1. Section 5.5
Integration by Substitution
V63.0121.041, Calculus I
New York University
December 13, 2010
Announcements
”Wednesday”, December 15: Review, Movie
Monday, December 20, 12:00pm–1:50pm: Final Exam
2. Announcements
”Wednesday”, December 15:
Review, Movie
Monday, December 20,
12:00pm–1:50pm: Final
Exam
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 2 / 37
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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 3 / 37
4. Objectives
Given an integral and a
substitution, transform the
integral into an equivalent
one using a substitution
Evaluate indefinite integrals
using the method of
substitution.
Evaluate definite integrals
using the method of
substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 4 / 37
5. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 5 / 37
6. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 6 / 37
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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
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
What are we supposed to do with that?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
10. No straightforward system of antidifferentiation
So far we don’t have any way to find
2x
√ dx
x2 + 1
or
tan x dx.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
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 “anti” version of one of the most
important rules of differentiation: the chain rule.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
12. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 9 / 37
13. Substitution for Indefinite Integrals
Example
Find
x
√ dx.
x2 +1
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
14. 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 + x 2 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
15. Say what?
Solution (More slowly, now)
Let g (x) = x 2 + 1.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
16. Say what?
Solution (More slowly, now)
Let g (x) = x 2 + 1. Then g (x) = 2x and so
d 1 x
g (x) = g (x) = √
dx 2 g (x) x2 +1
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
17. Say what?
Solution (More slowly, now)
Let g (x) = x 2 + 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 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
18. Leibnizian notation FTW
Solution (Same technique, new notation)
Let u = x 2 + 1.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
19. Leibnizian notation FTW
Solution (Same technique, new notation)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
20. Leibnizian notation FTW
Solution (Same technique, new notation)
√
Let u = x 2 + 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
21. Leibnizian notation FTW
Solution (Same technique, new notation)
√
Let u = x 2 + 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
22. Leibnizian notation FTW
Solution (Same technique, new notation)
√
Let u = x 2 + 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 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
23. Useful but unsavory variation
Solution (Same technique, new notation, more idiot-proof)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
du
dx =
2x
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
24. Useful but unsavory variation
Solution (Same technique, new notation, more idiot-proof)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:”
du
dx =
2x
So the integrand becomes completely transformed into
x x du
√ dx = √ ·
x2 +1 u 2x
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
25. Useful but unsavory variation
Solution (Same technique, new notation, more idiot-proof)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
26. Useful but unsavory variation
Solution (Same technique, new notation, more idiot-proof)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
27. Useful but unsavory variation
Solution (Same technique, new notation, more idiot-proof)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = 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 .
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
28. Useful but unsavory variation
Solution (Same technique, new notation, more idiot-proof)
√
Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = 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.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
29. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 14 / 37
30. A polynomial example
Example
Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
31. A polynomial example
Example
Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx.
Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So
(x 2 + 3)3 4x dx
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
32. A polynomial example
Example
Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx.
Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So
(x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
33. A polynomial example
Example
Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx.
Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So
(x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du
1
= u4
2
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
34. A polynomial example
Example
Use the substitution u = x 2 + 3 to find (x 2 + 3)3 4x dx.
Solution
If u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So
(x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du
1 1
= u 4 = (x 2 + 3)4
2 2
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
35. A polynomial example, by brute force
Compare this to multiplying it out:
(x 2 + 3)3 4x dx
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
36. A polynomial example, by brute force
Compare this to multiplying it out:
(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
37. A polynomial example, by brute force
Compare this to multiplying it out:
(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx
= 4x 7 + 36x 5 + 108x 3 + 108x dx
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
38. A polynomial example, by brute force
Compare this to multiplying it out:
(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx
= 4x 7 + 36x 5 + 108x 3 + 108x dx
1
= x 8 + 6x 6 + 27x 4 + 54x 2
2
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
39. A polynomial example, by brute force
Compare this to multiplying it out:
(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx
= 4x 7 + 36x 5 + 108x 3 + 108x dx
1
= x 8 + 6x 6 + 27x 4 + 54x 2
2
Which would you rather do?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
40. A polynomial example, by brute force
Compare this to multiplying it out:
(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx
= 4x 7 + 36x 5 + 108x 3 + 108x dx
1
= x 8 + 6x 6 + 27x 4 + 54x 2
2
Which would you rather do?
It’s a wash for low powers
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
41. A polynomial example, by brute force
Compare this to multiplying it out:
(x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx
= 4x 7 + 36x 5 + 108x 3 + 108x dx
1
= x 8 + 6x 6 + 27x 4 + 54x 2
2
Which would you rather do?
It’s a wash for low powers
But for higher powers, it’s much easier to do substitution.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
42. Compare
We have the substitution method, which, when multiplied out, gives
1
(x 2 + 3)3 4x dx = (x 2 + 3)4
2
1 8
= x + 12x 6 + 54x 4 + 108x 2 + 81
2
1 81
= x 8 + 6x 6 + 27x 4 + 54x 2 +
2 2
and the brute force method
1
(x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2
2
Is there a difference? Is this a problem?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
43. Compare
We have the substitution method, which, when multiplied out, gives
1
(x 2 + 3)3 4x dx = (x 2 + 3)4 + C
2
1 8
= x + 12x 6 + 54x 4 + 108x 2 + 81 + C
2
1 81
= x 8 + 6x 6 + 27x 4 + 54x 2 + +C
2 2
and the brute force method
1
(x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2 + C
2
Is there a difference? Is this a problem? No, that’s what +C means!
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
44. A slick example
Example
Find tan x dx.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
45. A slick example
Example
sin x
Find tan x dx. (Hint: tan x = )
cos x
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
46. 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
tan x dx = dx
cos x
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
47. 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
tan x dx = dx
cos x
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
48. 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
tan x dx = dx
cos x
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
49. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
50. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
51. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
52. Can you do it another way?
Example
sin x
Find tan x dx. (Hint: tan x = )
cos x
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
53. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
54. 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
= = =
cos 2x 1 − sin2 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).
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
55. For those who really must know all
Solution (Continued, with algebra help)
Let y = 1 − u 2 , so dy = −2u du. Then
u du u dy
tan x dx = 2
=
1−u y −2u
1 dy 1 1
=− = − ln |y | + C = − ln 1 − u 2 + 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|
There are other ways to do it, too.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 20 / 37
56. Outline
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
Theory
Examples
Substitution for Definite Integrals
Theory
Examples
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 21 / 37
57. Substitution for Definite Integrals
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)
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37
58. Substitution for Definite Integrals
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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37
59. Example
π
Compute cos2 x sin x dx.
0
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
60. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
61. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
62. 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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
63. 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 = − u 2 du
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
64. 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 = − u 2 du
= − 3 u 3 + C = − 1 cos3 x + C .
1
3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
65. 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 = − u 2 du
= − 3 u 3 + C = − 1 cos3 x + C .
1
3
Therefore
π π
1
cos2 x sin x dx = − cos3 x
0 3 0
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
66. 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 = − u 2 du
= − 3 u 3 + C = − 1 cos3 x + C .
1
3
Therefore
π π
1 1
cos2 x sin x dx = − cos3 x =− (−1)3 − 13
0 3 0 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
67. 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 = − u 2 du
= − 3 u 3 + C = − 1 cos3 x + C .
1
3
Therefore
π π
1 1 2
cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = .
0 3 0 3 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
68. Definite-ly Quicker
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
cos2 x sin x dx
0
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
69. Definite-ly Quicker
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
cos2 x sin x dx
0
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
70. Definite-ly Quicker
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
cos2 x sin x dx
0
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
71. Definite-ly Quicker
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 1
cos2 x sin x dx = −u 2 du = u 2 du
0 1 −1
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
72. Definite-ly Quicker
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 1
cos2 x sin x dx = −u 2 du = u 2 du
0 1 −1
1
1 3 1 2
= u = 1 − (−1) =
3 −1 3 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
73. Definite-ly Quicker
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 1
cos2 x sin x dx = −u 2 du = u 2 du
0 1 −1
1
1 3 1 2
= u = 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 variable (x).
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
74. Definite-ly Quicker
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 1
cos2 x sin x dx = −u 2 du = u 2 du
0 1 −1
1
1 3 1 2
= u = 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 variable (x).
But the slow way is just as reliable.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
75. An exponential example
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
76. An exponential example
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x , so du = 2e 2x dx. We have
√
ln 8 8√
1
√ e 2x e 2x + 1 dx = u + 1 du
ln 3 2 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
77. About those limits
Since
√ √ 2
e 2(ln 3)
= e ln 3
= e ln 3 = 3
we have √
ln 8 8√
1
√ e 2x e 2x + 1 dx = u + 1 du
ln 3 2 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 26 / 37
78. An exponential example
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x , so du = 2e 2x dx. We have
√
ln 8 8√
1
√ e 2x e 2x + 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 = y 1/2 dy
2 3 2 4 2 4
9
1 2 1 19
= · y 3/2 = (27 − 8) =
2 3 4 3 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 27 / 37
79. 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 3/2 1 19
y 1/2 dy = · y = (27 − 8) =
2 4 2 3 4 3 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 28 / 37
80. An exponential example
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x , so du = 2e 2x dx. We have
√
ln 8 8√
1
√ e 2x e 2x + 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 = y 1/2 dy
2 3 2 4 2 4
9
1 2 1 19
= · y 3/2 = (27 − 8) =
2 3 4 3 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 29 / 37
81. Another way to skin that cat
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x + 1,
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
82. Another way to skin that cat
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x + 1,so that du = 2e 2x dx.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
83. Another way to skin that cat
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x + 1,so that du = 2e 2x dx. Then
√
ln 8 9√
1
√ e 2x e 2x + 1 dx = u du
ln 3 2 4
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
84. Another way to skin that cat
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x + 1,so that du = 2e 2x dx. Then
√
ln 8 9√
1
√ e 2x e 2x + 1 dx = u du
ln 3 2 4
9
1
= u 3/2
3 4
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
85. Another way to skin that cat
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x + 1,so that du = 2e 2x dx. Then
√
ln 8 9√
1
√ e 2x e 2x + 1 dx = u du
ln 3 2 4
1 3/2 9
= u
3 4
1 19
= (27 − 8) =
3 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
86. A third skinned cat
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x + 1, so that
u 2 = e 2x + 1
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
87. A third skinned cat
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x + 1, so that
u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
88. A third skinned cat
Example
√
ln 8
Find √ e 2x e 2x + 1 dx
ln 3
Solution
Let u = e 2x + 1, so that
u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx
Thus √
ln 8 3 3
1 3 19
√ = u · u du = u =
ln 3 2 3 2 3
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
89. A Trigonometric Example
Example
Find
3π/2
θ θ
cot5 sec2 dθ.
π 6 6
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37
90. A Trigonometric Example
Example
Find
3π/2
θ θ
cot5 sec2 dθ.
π 6 6
Before we dive in, think about:
What “easy” substitutions might help?
Which of the trig functions suggests a substitution?
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37
92. Solution
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
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
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 33 / 37
93. Graphs
3π/2 π/4
θ θ
cot 5
sec 2
dθ 6 cot5 ϕ sec2 ϕ dϕ
π 6 6 π/6
y y
θ ϕ
3π π ππ
2 64
The areas of these two regions are the same.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 35 / 37
94. Graphs
π/4 1
−5
6 cot5 ϕ sec2 ϕ dϕ √ 6u du
π/6 1/ 3
y y
ϕ u
ππ 1 1
64 √
3
The areas of these two regions are the same.
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 36 / 37
95. Summary
If F is an antiderivative for f , then:
f (g (x))g (x) dx = F (g (x))
If F is an antiderivative for f , which is continuous 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)
Antidifferentiation in general and substitution in particular is a
“nonlinear” problem that needs practice, intuition, and perserverance
The whole antidifferentiation story is in Chapter 6
V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 37 / 37