The document is the notes for a Calculus I class. It provides announcements for the upcoming class on Monday, which will involve reviewing course material rather than new topics. Examples are given of integration by substitution, including exponential, odd, and even functions. Multiple methods for substitutions are presented. The properties of odd and even functions are defined, and examples are shown graphically. Symmetric functions and their behavior under combinations are discussed.

1. Section 5.5
Integration by Substitution, Part Deux
V63.0121, Calculus I
April 29, 2009
Announcements
Class on Monday will be review
. . . . . .

2. Yes, there is class on Monday
No new material
We will review the course
We will answer questions, so bring some
. . . . . .

3. Final stuff
Old finals online, including Fall 2008
Review sessions: May 5 and 6, 6:00–8:00pm, SILV 703
Final is May 8, 2:00–3:50pm in CANT 101/200
.
.
Image credit: Pragmagraphr
. . . . . .

4. 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
. . . . . .

5. Outline
Recall: The method of substitution
Multiple substitutions
Odd and even functions
Examples
More examples and advice
Course Evaluations
. . . . . .

6. Last Time: The Substitution Rule
Theorem
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
. . . . . .

7. Last Time: The Substitution Rule for Definite Integrals
Theorem
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)
. . . . . .

8. Last Time: The Substitution Rule for Definite Integrals
Theorem
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)
The integral on the left happens in “x-land”, so its limits are
values of x
The integral on the right happens in “u-land”, so its limits
need to be values of u
To convert x to u, simply apply the substitution u = g(x).
. . . . . .

9. Outline
Recall: The method of substitution
Multiple substitutions
Odd and even functions
Examples
More examples and advice
Course Evaluations
. . . . . .

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

11. An exponential example
Example√
∫ √
ln 8
e2x e2x + 1 dx
Find √
ln 3
Solution
Let u = e2x , so du = 2e2x dx. We have
√
∫ ∫
√ 8√
ln 8
1
e2x e2x + 1 dx = u + 1 du
√
2
ln 3 3
. . . . . .

12. An exponential example
Example√
∫ √
ln 8
e2x e2x + 1 dx
Find √
ln 3
Solution
Let u = e2x , so du = 2e2x dx. We have
√
∫ ∫
√ 8√
ln 8
1
e2x e2x + 1 dx = u + 1 du
√
2
ln 3 3
Now let y = u + 1, dy = du. So
∫ ∫ ∫
8√ 9 9
√
1 1 1
y1/2 dy
u + 1 du = y dy =
2 2 2
3 4 4
9
12 1 19
= · y3/2 = (27 − 8) =
23 3 3
4
. . . . . .

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

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

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

16. Another way to skin that cat
Example√
∫ √
ln 8
e2x e2x + 1 dx
Find √
ln 3
Solution
Let u = e2x + 1, so that du = 2e2x dx. Then
√
∫ ∫
√
ln 8 9√
1
2x
e2x
e + 1 dx = u du
√
2
ln 3 4
9
1 3/2
u
=
3 4
. . . . . .

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

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

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

20. A third skinned cat
Example√
∫ √
ln 8
e2x e2x + 1 dx
Find √
ln 3
Solution
√
e2x + 1, so that
Let u =
u2 = e2x + 1 =⇒ 2u du = 2e2x dx
Thus
√
∫ ∫
√ 3
ln 8 3
13 19
2x
e2x u · u du =
e + 1 dx = u =
√
3 3
ln 3 2 2
. . . . . .

21. Outline
Recall: The method of substitution
Multiple substitutions
Odd and even functions
Examples
More examples and advice
Course Evaluations
. . . . . .

22. Example
∫ π
sin(x) dx
Find
−π
. . . . . .

23. Example
∫ π
sin(x) dx
Find
−π
Solution
∫ π
sin(x) = − cos(x)|π = cos(x)|−π = cos(−π) − cos(π) = 0
−π π
−π
. . . . . .

24. Example
∫ π
sin(x) dx
Find
−π
Solution
∫ π
sin(x) = − cos(x)|π = cos(x)|−π = cos(−π) − cos(π) = 0
−π π
−π
This is obvious from the graph:
y
.
. x
.
. . . . . .

25. Even and Odd Functions
Definition
A function f is even if for all x,
f(−x) = f(x)
A function f is odd if for all x,
f(−x) = −f(x).
. . . . . .

26. Even and Odd Functions
Definition
A function f is even if for all x,
f(−x) = f(x)
A function f is odd if for all x,
f(−x) = −f(x).
. . . . . .

27. Even and Odd Functions
Definition
A function f is even if for all x,
f(−x) = f(x)
A function f is odd if for all x,
f(−x) = −f(x).
These properties are revealed in the graph.
. . . . . .

28. Even and Odd Functions
Definition
A function f is even if for all x,
f(−x) = f(x)
A function f is odd if for all x,
f(−x) = −f(x).
These properties are revealed in the graph.
An odd function has rotational symmetry about the origin.
. . . . . .

29. Even and Odd Functions
Definition
A function f is even if for all x,
f(−x) = f(x)
A function f is odd if for all x,
f(−x) = −f(x).
These properties are revealed in the graph.
An odd function has rotational symmetry about the origin.
An even function has reflective symmetry in the y-axis
. . . . . .

30. Even and Odd functions pictured
y
.
o
. dd
. x
.
y
.
e
. ven
x
.
. . . . . .

31. Examples of symmetric functions
Even and odd functions abound.
x → xn is odd when n is odd and even when n is even.
Funny, that!
sin is odd and cos is even.
. . . . . .

32. Combining symmetric functions
Theorem
(a) The sum of even functions is even. The sum of odd functions
is odd.
(b) The product of even functions is even. The product of odd
functions is even. The product of an odd function and an
even function is an odd function.
(c) If g is even, then f ◦ g is even. The composition of two odd
functions is odd. The composition of an even function and
an odd function is even.
. . . . . .

33. Integrating symmetric functions
Theorem
Let a be any number.
(a) If f is odd, then ∫ a
f(x) dx = 0.
−a
. . . . . .

34. Integrating symmetric functions
Theorem
Let a be any number.
(a) If f is odd, then ∫ a
f(x) dx = 0.
−a
(b) If f is even, then
∫ ∫
a a
f(x) dx = 2 f(x) dx.
−a 0
. . . . . .

35. Proof (odd f).
∫ a
f(x) dx, let u = −x. Then du = −dx and we have
To compute
−a
∫ ∫ −a
a
f(x) dx = − f(−u) du
−a a
∫ −a
f(u) du
=
a
∫ a
=− f(u) du.
−a
The only number which is equal to its own negative is zero.
. . . . . .

36. Proof (even f).
With the same substitution we have
∫0 ∫ 0
f(x) dx = − f(−u) du
−a a
∫0
=− f(u) du
a
∫a
f(u) du.
=
0
So
∫ ∫ ∫ ∫
a a a
0
f(x) dx = f(x) dx + f(x) dx = 2 f(x) dx.
−a −a 0 0
. . . . . .

37. Example
Compute
∫ √
√
e π +1
sin(x) 1 + cos3 (x) dx.
√
−e π −1
. . . . . .

38. Example
Compute
∫ √
√
e π +1
sin(x) 1 + cos3 (x) dx.
√
−e π −1
Solution
The integrand is odd! So the answer is zero.
. . . . . .

39. Example
Compute
∫
( )
2
x4 + x2 + 3 dx.
−2
. . . . . .

40. Solution
Because the integrand is even we can simplify our arithmetic. It’s
especially nice to plug in zero since the result is often zero.
∫ ∫
( ) 2( )
2
4 2
x4 + x2 + 3 dx
x + x + 3 dx = 2
−2 0
[ ]2
x5 x3
=2 + 3x
+
5 3 0
[5 ]
23
2
=2 + 3(2)
+
5 3
[ ]
32 8
=2 + +6
5 3
2
(32 · 3 + 8 · 5 + 6 · 15)
=
15
2 · 226
=
15
. . . . . .

41. Outline
Recall: The method of substitution
Multiple substitutions
Odd and even functions
Examples
More examples and advice
Course Evaluations
. . . . . .

42. Example
∫
x3
dx
(5x4 + 2)2
. . . . . .

43. Example
∫
x3
dx
(5x4 + 2)2
Solution
Let u = 5x4 + 2, so du = 20x3 dx. Then
∫ ∫
x3 1 1
dx = du
4 + 2)2 u2
20
(5x
11
=− · +C
20 u
1
=− +C
20(5x4 + 2)
. . . . . .

44. Example
∫
sin(sin(θ)) cos(θ) dθ
. . . . . .

45. Example
∫
sin(sin(θ)) cos(θ) dθ
Solution
Let u = sin(θ), so du = cos(θ) dθ. Then
∫ ∫
sin(sin(θ)) cos(θ) dθ = sin(u) du
= − cos(u) + C
= − cos(sin(θ)) + C
. . . . . .

46. Example
∫
ex + e−x
dx
ex − e−x
. . . . . .

47. Example
∫
ex + e−x
dx
ex − e−x
Solution
The numerator is the derivative of the denominator! Let
( )
u = ex − e−x , so du = ex + e−x dx. Then
∫ ∫
ex + e−x 1
dx = du
x − e−x
e u
= ln |u| + C = ln ex − e−x + C
. . . . . .

48. Example
∫
3x
dx
1 + 9x
. . . . . .

49. Example
∫
3x
dx
1 + 9x
Solution
Notice 9x = (32 )x = 32x = (3x )2 . So let u = 3x ,
du = (ln 3) · 3x dx. Then
∫ ∫
3x 1 1
x dx = du
1 + u2
(1 + 9 ) ln 3
1
arctan(u) + C
=
ln 3
1
arctan(3x ) + C
=
ln 3
. . . . . .

50. Example
√
∫
sec2 x
√ dx
x
. . . . . .

51. Example
√
∫
sec2 x
√ dx
x
Solution
√ 1
x, so du = √ du. Then
Let u =
2x
√
∫ ∫
sec2 x
dx = 2 sec2 (u) du
√
x
= 2 tan(u) + C
(√ )
= 2 tan x + C
. . . . . .

52. Example
∫
dx
x ln x
. . . . . .

53. Example
∫
dx
x ln x
Solution
1
Let u = ln x, so du = dx. Then
x
∫ ∫
dx 1
du
=
x ln x u
= ln |u| + C
= ln |ln x| + C
. . . . . .

54. 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
. . . . . .

55. Outline
Recall: The method of substitution
Multiple substitutions
Odd and even functions
Examples
More examples and advice
Course Evaluations
. . . . . .

56. Course Evaluations
Please fill out CAS and departmental evaluations
CAS goes to SILV 909 (need a volunteer)
departmental goes to WWH 627 (need another volunteer)
Thank you for your input!
. . . . . .