The document outlines a calculus lecture on integration by substitution. It provides examples of using u-substitution to find antiderivatives of expressions like √(x^2+1) and tan(x). The key ideas are that if u is a function of x, its derivative du/dx can be used to rewrite the integrand and perform a u-substitution integration.

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

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. 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
.
.
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. . . . . .

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.
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.
Image credit: Scott Beale / Laughing Squid
. . . . . .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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. 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. 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. 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. 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. 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
. . . . . .