The method of substitution is the chain rule in reverse. At first it looks magical, then logical, and then you realize there's an art to choosing the right substitution. We try to demystify with many worked-out examples.

1. Section 5.5
Integration by Substitution
V63.0121, Calculus I
April 27, 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
Substitution for Definite Integrals
Theory
Examples
. . . . . .

3. Office Hours and other help this week
In addition to recitation
Day Time Who/What Where in WWH
M 1:00–2:00 Leingang OH 624
3:30–4:30 Katarina OH 607
5:00–7:00 Curto PS 517
T 1:00–2:00 Leingang OH 624
4:00–5:50 Curto PS 317
W 1:00–2:00 Katarina OH 607
2:00–3:00 Leingang OH 624
R 9:00–10:00am Leingang OH 624
5:00–7:00pm Maria OH 807
F 2:00–4:00 Curto OH 1310
. . . . . .

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

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

6. Outline
Announcements
Last Time: The Fundamental Theorem(s) of Calculus
Substitution for Indefinite Integrals
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
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 f′ (x) = 2x and so
d√ x
1
g′ (x) = √
g(x) = √
dx x2 + 1
2 g(x)
. . . . . .

18. Say what?
Solution (More slowly, now)
Let g(x) = x2 + 1. Then f′ (x) = 2x and so
d√ x
1
g′ (x) = √
g(x) = √
dx x2 + 1
2 g(x)
Thus
∫( )
∫
d√
x
√ dx = g(x) dx
dx
x2 + 1
√
√
= 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
∫ ∫ ∫
1( )
x 1
√ 1 du =
√ √ du
dx = 2
u 2u
x2 + 1
. . . . . .

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
∫ ∫ ∫
1( )
x 1
√ 1 du =
√ √ du
dx = 2
u 2u
x2 + 1 ∫
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
∫ ∫ ∫
1( )
x 1
√ 1 du =
√ √ du
dx = 2
u 2u
x2 + 1 ∫
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
14 1 2
u = (x + 3)4
=
2 2
. . . . . .

27. A polynomial example, the hard way
Compare this to multiplying it out:
∫ ∫
(6 )
2 3
x + 9x4 + 27x2 + 27 4x dx
(x + 3) 4x dx =
∫
(7 )
4x + 36x5 + 108x3 + 108x dx
=
18
x + 6x6 + 27x4 + 54x2
=
2
. . . . . .

28. Compare
We have
∫
12
(x2 + 3)3 4x dx = (x + 3)4
2
∫
1
(x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2
2
Now
1( 8 )
12
(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
∫
12
(x2 + 3)3 4x dx = (x + 3)4 + C
2
∫
1
(x2 + 3)3 4x dx = x8 + 6x6 + 27x4 + 54x2 + C
2
Now
1( 8 )
12
(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
∫
tan x dx.
Find
. . . . . .

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

32. A slick example
Example
∫
sin x
tan x dx. (Hint: tan x =
Find )
cos x
Solution
Let u = cos x. Then du = − sin x dx.
. . . . . .

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

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

35. A slick example
Example
∫
sin x
tan x dx. (Hint: tan x =
Find )
cos x
Solution
Let u = cos x. Then du = − sin x dx. So
∫ ∫ ∫
sin x 1
dx = −
tan x 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
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
cos2 x sin x dx.
Compute
0
. . . . . .

39. Example ∫
π
cos2 x sin x dx.
Compute
0
Solution (Slow Way) ∫
cos2 x sin x dx and then
First compute the indefinite integral
evaluate.
. . . . . .

40. Example ∫
π
cos2 x sin x dx.
Compute
0
Solution (Slow Way) ∫
cos2 x sin x dx and then
First compute the indefinite integral
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 = 2.
3 3
0
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
∫ ∫ −1
π
cos2 x sin x dx = −u2 du
0 1
∫ 1
u2 du
=
−1
2
1 31
3 u −1
= = .
3
. . . . . .

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

45. 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
. . . . . .

46. 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 8
4
. . . . . .

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

48. Solution
1
θ
Let φ = . Then dφ = dθ.
6 6
() ()
∫ 3π/2 ∫ π/4
5θ 2θ
cot5 φ sec2 φ dφ
dθ = 6
cot sec
6 6
π π/6
∫ π/4
sec2 φ dφ
=6
tan5 φ
π/6
. . . . . .

49. Solution
1
θ
Let φ = . Then dφ = dθ.
6 6
() ()
∫ 3π/2 ∫ π/4
5θ 2θ
cot5 φ sec2 φ dφ
dθ = 6
cot sec
6 6
π π/6
∫ π/4
sec2 φ dφ
=6
tan5 φ
π/6
Now let u = tan φ. So du = sec2 φ dφ, and
∫ ∫ 1
sec2 φ dφ
π/4
u−5 du
6 =6 √
tan5 φ 1/ 3
π/6
( ) 1
1
− u−4
=6 √
4 1/ 3
3
[9 − 1] = 12.
=
2
. . . . . .

50. 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
ex and ex
ax and ax (fudge the coefficient)
√ 1
x and √
x
1
ln x and
x
. . . . . .