Here are the key steps: 1) Choose u and dv based on LIPET: u = ex dv = cos x dx 2) Find du and v: du = ex dx v = sin x 3) Apply integration by parts formula: ∫uex dx = uv - ∫vdu = exsinx - ∫sinxexdx 4) Repeat integration by parts on the second term: ∫sinxexdx = excosx - ∫-cosxexdx 5) Combine like terms: exsinx + excosx - ∫excosxdx 6) The integral on the right is

1. INTEGRAL CALCULUS

2. Integration
or
Anti-derivatives
Definition Indefinite Integral
The set of all antiderivatives of a function f(x) is the
indefinite integral of f with respect to x and is denoted by
∫ f ( x)dx
∫ f ( x)dx = F ( x) + C

3. Definite Integration
b
f
∫
a
'
(x )dx
=f (b ) −f (a )

4. Antiderivatives with Slope Fields
1. ∫ x dx =
n
dx
2. ∫ =
x
x n+1
+ C , n ≠ −1
n +1
ln x + C
3. ∫ e dx = e + C
x
x
4. ∫ sin x dx = − cos x + C

5. Antiderivatives with Slope Fields
5. ∫ cos x dx =
sin x + C
6. ∫ sec x dx =
tan x + C
7. ∫ csc 2 x dx =
− cot x + C
8. ∫ sec x tan x dx =
sec x + C
2
9. ∫ csc x cot x dx = − csc x + C

6. Antiderivatives with Slope Fields
∫ (x
∫x
3
3
− 2 x + 3)dx
x dx
1

∫  t 2 − cos t  dt


4
x
2
− x + 3x + C
4
9
2
2
x +C
9
1
− − sin t + C
t

7. Antiderivatives with Slope Fields
∫ sec x(tan x + cos x)dx
sec x + x + C
∫ (1 + sin x csc x) dx
x − cos x + C
(e 2t + t 2 ) dt
∫
e2t t 3
+ +C
2 3
2

8. Antiderivatives with Slope Fields
Find the position function for v(t) = t3 - 2t2 + t s(0) = 1
4
3
2
t
2t
t
s (t ) = −
+ +C
4
3
2
4
3
2
0 2( 0) ( 0)
1= −
+
+C
4
3
2
4
3
2
t
2t t
s (t ) = −
+ +1
4
3
2
C=1

9. Anti-derivatives with Slope
Fields
∫
d (cabin)
=
cabin
log cabin + C
log cabin
houseboat

10. Integration by Substitution
The chain rule allows us to differentiate a
wide variety of functions, but we are able
to find antiderivatives for only a limited
range of functions. We can sometimes
use substitution to rewrite functions in a
form that we can integrate.

11. Integration by Substitution
∫ ( x + 2)
5
dx
∫ u du
5
1 6
u +C
6
( x + 2)
6
6
+C
Let u = x + 2
du = dx
The variable of integration must match
the variable in the expression.
Don’t forget to substitute the
value for u back into the
problem!

12. Integration by Substitution
∫
1 + x 2 ⋅ 2 x dx
∫u
1
2
One of the clues that we look for is if we can
find a function and its derivative in the
integrand.
The derivative of
du
Let u = 1 + x
du = 2 x dx
3
2
2
u +C
3
2
(1+ x
3
3
2 2
)
1 + x2
+C
is
.
2 x dx
2
Note that this only worked because of the 2x in
the original.
Many integrals can not be done by substitution.

13. Integration by Substitution
∫
4 x − 1 dx
∫u
1
2
1
⋅ du
4
3
2
2
1
u ⋅ +C
3
4
3
2
1
u +C
6
Let u = 4 x − 1
du = 4 dx
1
du = dx
4
3
1
( 4 x − 1) 2 + C
6
Solve for dx.

14. Integration by Substitution
∫ cos ( 7 x + 5) dx
1
∫ cos u ⋅ 7 du
1
sin u + C
7
1
sin ( 7 x + 5 ) + C
7
Let u = 7 x + 5
du = 7 dx
1
du = dx
7

15. Integration by Substitution
x 2 sin ( x 3 ) dx
∫
Let u = x 3
1
∫ sin u du
3
du = 3 x 2 dx
1
− cos u + C
3
1
3
− cos x + C
3
1
du = x 2 dx
3
2
We solve forx dx because
we can find it in the integrand.

16. Integration by Substitution
sin 4 x ⋅ cos x dx
∫
Let u = sin x
∫ ( sin x ) cos x dx
du = cos x dx
4
u 4 du
∫
1 5
u +C
5
1 5
sin x + C
5

17. Integration by Substitution
π
4
0
∫
2
tan x sec x dx
new limit
1
∫ u du
0
new limit
1
1 2
u
2 0
1
2
The technique is a little different for definite
integrals.
Let u = tan x
2
du = sec x dx
u ( 0 ) = tan 0 = 0
π
π 
u   = tan = 1
4
4
We can find new
limits, and then
we don’t have to
substitute back.
We could have substituted back and used the original
limits.

18. Integration by Substitution
π
4
0
∫
π
4
0
∫
Using the original limits:
tan x sec 2 x dx
Let u = tan x
du = sec x dx
2
u du
∫ u du
2
Leave the
limits out until
you substitute
back.
1 2
= u
π
2
1
2 4
= ( tan x )
2
0
1
π 1
2
=  tan  − ( tan 0 )
2
Wrong! 4  2
This is usually
more work than
finding new
limits
The limits don’t 1
match!
1 2 1 2
= ⋅1 − ⋅ 0 =
2
2
2

19. Integration by Substitution
∫
1
−1
∫
2
0
3x
2
x + 1 dx
3
3 2
2
2
⋅2
3
du = 3 x dx
2
1
2
u du
2
u
3
Let u = x 3 + 1
u ( 1) = 2
Don’t forget to use the new limits.
0
3
2
u ( −1) = 0
2
= ⋅2 2
3
4 2
=
3

20. Integration by Substitution
∫ 2 x 1 + x dx
∫ tan t dt
∫
sin 3 θ cosθ dθ
∫
2
3 x + 4 dx
∫ t (5 + 3t ) dt
2 8
∫
cot θ csc 2 θ dθ

21. Integration by Substitution
∫
x−2
dx
x −1
5
2
π
4
π
−
2
∫
−
∫
e2
e
5x
e
∫ 3 + e5 x dx
cos x sin 2 x dx
∫
dx
dx
x ln x
∫
1 − x x 2 dx
1
0
x
e

22. 2. Find
x = sin u
∫
∫
(1 − x 2 ) dx. Let u = sin−1 x
dx = cos u du
Substituting gives,
(1 − x 2 ) dx = ∫ 1 − sin2 u cos u du
= ∫ cos2 u du
We can not integrate this yet. Let us use trig.
1
cos2 u = (1 + cos 2u )
2
1 1
u 1
= ∫ + cos 2u du = + sin2u + c
2 2
2 4
u 1
1
= + .2sin u cos u + c = ( u + sin u cos u ) + c
2 4
2

23. 1
= ( u + sin u cos u ) + c
2
Now for some trig play…..
x = sin u, but what does cos u equal?
sin2 u + cos2 u = 1
cos2 u = 1 − sin2
cos u = 1 − sin2 u
∫
1
(1 − x ) dx = ( u + sin u cos u ) + c
2
2
(
)
1
= sin−1 x + x 1 − x 2 + c
2

24. 3. Find
∫
x2
4−x
2
dx. Let x = 2 sin θ
dx = 2cos θ dθ
∫
x2
4−x
2
dx = ∫
=∫
4 sin2 θ
4 − 4 sin θ
2
2 sin2 θ
cos θ
2
.2cos θ dθ = ∫
4 sin2 θ
2 1 − sin θ
2
.2cos θ dθ
.2cos θ dθ
= ∫ 4 sin θ dθ
2
1 1
 2

sin θ = − cos 2θ ÷

2 2


= ∫ 2 − 2cos 2θ dθ
= 2θ − sin2θ + c
= 2θ − 2 sinθ cosθ + c
We now need to substitute theta
in terms of x.

25. x 2 = 4 sin2 θ
x = 2sinθ
x
sinθ =  ÷
2
−1  x 
θ = sin  ÷
2
= 4 − 4 cos2 θ
4 cos2 θ = 4 − x 2
2cos θ = 4 − x 2
1
cosθ =
4 − x2
2
∫
x2
4−x
2
dx = 2θ − 2 sinθ cos θ + c
x 1
x
= 2 sin  ÷ − 2. .
4 − x2 + c
2 2
2
−1
x x
= 2 sin  ÷ −
4 − x2 + c
2 2
−1

26. Now for some not very obvious substitutions……………….
1. Find
sin5 x dx
∫
sin5 x = sin x sin4 x = sin x (sin2 x )2 = sin x (1 − cos2 x )2
sin5 x dx = ∫ sin x (1 − cos2 x )2 dx
∫
Let u = cos x
du = − sin x dx
sin5 x dx = ∫ −(1 − u 2 )2 du = ∫ −(1 − 2u 2 + u 4 ) du = ∫ −1 + 2u 2 − u 4 du
∫
2 3 1 5
= u − u −u +c
3
5
2
1
3
= cos x − cos5 x − cos x + c
3
5

27. 1
2. Find ∫
dx
1− x
Let u = 1 − x
1
1
1 −2
du = − x dx ⇒ dx = −2 x 2 du ⇒ dx = −2 ( 1 − u ) du
2
1
2(u − 1)
2
dx = ∫
du = ∫ 2 − du
∫ 1− x
u
u
= 2u − 2ln u + c
= 2 − 2 x − 2ln 1 − x + c

28. Integration by Parts

29. Integration By Parts
Start with the product rule:
d
dv
du
( uv ) = u + v
dx
dx
dx
d ( uv ) = u dv + v du
d ( uv ) − v du = u dv
u dv = d ( uv ) − v du
∫ u dv = ∫ ( d ( uv ) − v du )
∫ u dv = ∫ ( d ( uv ) ) − ∫ v du
∫ uv′ dx = uv − ∫ u′v dx
This is the Integration by Parts formula.
→

30. ∫ uv′ dx = uv − ∫ u′v dx
u differentiates to zero
v’ is easy to integrate.
(usually).
The Integration by Parts formula is a “product rule” for integration.
Choose u in this order:
LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig
→

31. Example 1:
∫ x cos x dx
polynomial factor
∫ uv′ dx = uv − ∫ u′v dx
∫ uv′ dx = uv − ∫ u′v dx
LIPET
u=x
v′ = cos x
u′ = 1
v = sin x
x ⋅ sin x − ∫ sin x dx
∫ x cos x dx = x sin x + cos x + c
→

32. Example 2:
∫ uv′ dx = uv − ∫ u′v dx
∫ ln x dx
LIPET
logarithmic factor
∫ uv′ dx = uv − ∫ u′v dx
u = ln x
1
u′ =
x
v′ = 1
v=x
1
ln x ⋅ x − ∫ x ⋅ dx
x
∫ ln x ⋅1dx = x ln x − x + c
→

33. ∫ uv′ dx = uv − ∫ u′v dx
Example 3:
x 2 e x dx
∫
u = x2
= x 2 e x − ∫ e x 2 x dx
x
(
= x e − 2 xe − ∫ e dx
∫x e
2 x
x
v = ex
This is still a product, so we need to use
integration by parts again.
x
= x e − 2 ∫ xe dx
2 x
v′ = e x
u′ = 2 x
= uv − ∫ u′v dx
2 x
LIPET
x
)
u=x
v′ = e
u′ = 1
v = ex
dx = x e − 2 xe + 2e + c
2 x
x
x
→

34. uv′ dx = uv − ∫ u′v dx
∫
The Integration by Parts formula can be written as:
u dv = uv − ∫ v du
∫

35. Example 4:
LIPET
e x cos x dx
∫
u v − ∫ v du
e x sin x − ∫ sin x ×e x dx
(
u = e x dv = cos x dx
du = e x dx v = sin x
dv = sin x dx
u=e
x
du = e dx v = − cos x
x
e sin x − e ×− cos x − ∫ − cos x ×e dx
x
x
x
uv
v du
e x sin x + e x cos x − ∫ e x cos x dx
)
This is the
expression we
started with!
→

36. Example 4 continued …
e x cos x dx
∫
LIPET
u v − ∫ v du
e x sin x − ∫ sin x ×e x dx
(
u = e x dv = cos x dx
du = e x dx v = sin x
dv = sin x dx
u=e
x
du = e dx v = − cos x
x
e sin x − e ×− cos x − ∫ − cos x ×e dx
x
x
x
)
e x cos x dx = e x sin x + e x cos x − ∫ e x cos x dx
∫
2 ∫ e x cos x dx = e x sin x + e x cos x
e x sin x + e x cos x
e x cos x dx =
+C
∫
2

37. Example 4 continued …
e x cos x dx
∫
u v − ∫ v du
e x sin x − ∫ sin x ×e x dx
(
This is called “solving for the
unknown integral.”
It works when both factors
integrate and differentiate
forever.
e sin x − e ×− cos x − ∫ − cos x ×e dx
x
x
x
)
e x cos x dx = e x sin x + e x cos x − ∫ e x cos x dx
∫
2 ∫ e x cos x dx = e x sin x + e x cos x
e x sin x + e x cos x
e x cos x dx =
+C
∫
2
→

38. A Shortcut: Tabular Integration
Tabular integration works for integrals of the form:
∫ f ( x ) g ( x ) dx
where:
Differentiates to zero
in several steps.
Integrates
repeatedly.
→

39. x 2 e x dx
∫
f ( x ) & deriv. g ( x ) & integrals
2
e
− 2x
ex
+ x
+ 2
0
x
e
x
e
Compare this with the
same problem done the
other way:
x
x 2 e x −2 xe x +2e x +C
∫ x e dx =
2 x
→

40. ∫ uv′ dx = uv − ∫ u′v dx
Example 3:
x 2 e x dx
∫
u = x2
= x 2 e x − ∫ e x 2 x dx
x
(
= x e − 2 xe − ∫ e dx
∫x e
2 x
x
v = ex
This is still a product, so we need to use
integration by parts again.
x
= x e − 2 ∫ xe dx
2 x
v′ = e x
u′ = 2 x
= uv − ∫ u′v dx
2 x
LIPET
x
)
u=x
v′ = e
u′ = 1
v = ex
dx = x e − 2 xe + 2e + c
2 x
x
x
This is easier and quicker to do with
tabular integration!
→

41. x3 sin x dx
∫
x3
+
sin x
− 3x 2
− cos x
+ 6x
− 6
− sin x
0
sin x
cos x
− x3 cos x + 3 x 2 sin x + 6 x cos x − 6sin x + C
π

42. Properties of the Definite Integral
1:
2:
3:
4:
5:
6:

43. 7:

44. 8:
9:
10:
11:

45. 12:

46. Substitution and definite integrals
Assuming the function is continuous over the interval, then exchanging the limits for
x by the corresponding limits for u will save you having to substitute back after the
integration process.
2
(2x+4)(x 2 + 4 x )3 dx
∫
1. Evaluate
1
Let u = x 2 + 4 x
∫
2
1
du = 2 x + 4 dx When x = 2, u = 12; x = 1, u = 5
(2x+4)(x + 4 x ) dx = ∫
2
3
12
5
u 3du
12
1 
=  u4 
 4 5
= 5027.75

47. Special (common) forms
Some substitutions are so common that they can be treated as standards and, when
their form is established, their integrals can be written down without further ado.
1
∫ f (ax + b)dx = a F (ax + b) + c
f `( x )
∫ f ( x ) dx = ln f ( x ) + c
1
f `( x )f ( x )dx = (f ( x ))2 + c
∫
2

48. Area under a curve
y = f(x)
b
A = ∫ f ( x ) dx
a
a
b
a
b
b
A = − ∫ f ( x ) dx
a
y = f(x)

49. Area between the curve and y - axis
y = f(x)
b
a
b
A = ∫ f ( y ) dy
a

50. 1. Calculate the area shown in the diagram below.
y = x2 + 1
5
2
5
1
2
A = ∫ ( y − 1) dy
2
y = x2 + 1
x2 = y − 1
x = y −1
5
2

=  ( y − 1) 
3
2
3
2
=
14
units squared.
3