5
INTEGRALS
INTEGRALS
Indefinite Integrals
INDEFINITE INTEGRAL
The notation
∫ f(x) dx is traditionally used for an
antiderivative of f and is called an indefinite
integral.
Thus, ∫ f(x) dx = F(x) means F’(x) = f(x)
INDEFINITE INTEGRALS
For example, we can write
3


x
d x
2
2
 x dx  3  C because dx  3  C   x
3
 Thus, we can regard an indefinite integral
as representing an entire family of functions
(one antiderivative for each value of the constant C).
INDEFINITE INTEGRALS
Any formula can be verified by differentiating
the function on the right side and obtaining
the integrand.
For instance, sec 2 x dx  tan x  C

because
d
(tan x  C )  sec 2 x
dx
TABLE OF INDEFINITE INTEGRALS
 cf ( x) dx  c  f ( x) dx
Table 1
 [ f ( x)  g ( x)] dx
  f ( x) dx   g ( x) dx
 k dx  kx  C
n 1
x
n
x
 dx  n  1  C (n  1)
 sin x dx   cos x  C
 cos x dx  sin x  C
 sec
 csc
2
x dx  tan x  C
 sec x tan x dx  sec x  C
2
x dx   cot x  C
 csc x cot x dx   csc x  C
INDEFINITE INTEGRALS
Thus, we write
1
1
 x 2 dx   x  C
with the understanding that it is valid on
the interval (0, ∞) or on the interval (-∞, 0).
INDEFINITE INTEGRALS
This is true despite the fact that the general
antiderivative of the function f(x) = 1/x2,
x ≠ 0, is:
 1


C
if
x

0
1
 x
F ( x)  
1
   C if x  0
2
 x
INDEFINITE INTEGRALS
Example 1
Find the general indefinite integral
∫ (10x4 – 2 sec2x) dx
 Using our convention and Table 1, we have:
∫(10x4 – 2 sec2x) dx = 10 ∫ x4 dx – 2 ∫ sec2x dx
= 10(x5/5) – 2 tan x + C
= 2x5 – 2 tan x + C
 You should check this answer by differentiating it.
INDEFINITE INTEGRALS
Example 2
cos

Evaluate
 sin 2  d
 This indefinite integral isn’t immediately apparent
in Table 1.
 So, we use trigonometric identities to rewrite
the function before integrating:
cos 
 1  cos 
 sin 2  d    sin  
 sin 

 d

  csc  cot  d   csc   C
INTEGRALS
The Substitution Rule
In this section, we will learn:
To substitute a new variable in place of an existing
expression in a function, making integration easier.
Equation 1
INTRODUCTION
• Our antidifferentiation formulas don’t tell
us how to evaluate integrals such as
2
x
1

x
dx

2
INTRODUCTION
• To find this integral, we use the problemsolving strategy of introducing something
extra.
 The ‘something extra’ is a new variable.
 We change from the variable x to a new variable u.
INTRODUCTION
• Suppose we let u = 1 + x2 be the quantity
under
the root sign in the integral below,
 2x
1  x dx
2
 Then, the differential of u is du = 2x dx.
INTRODUCTION
• Notice that, if the dx in the notation for
an integral were to be interpreted as
a differential, then the differential 2x dx
would occur in
2
x
1

x
dx

2
Equation 2
INTRODUCTION
• So, formally, without justifying our
calculation, we could write:
2
x
1

x
dx

1

x
2
x
dx


2
2
  udu
 u
2
3
3/ 2
C
 ( x  1)
2
3
2
3/ 2
C
INTRODUCTION
• However, now we can check that we have
the correct answer by using the Chain Rule
to differentiate
d 2 2
 3 ( x  1)3 2  C   32  32 ( x 2  1)1 2  2 x
dx
 2x x  1
2
INTRODUCTION
• In general, this method works whenever
we have an integral that we can write in
the form
∫ f(g(x))g’(x) dx
INTRODUCTION
Equation 3
• Observe that, if F’ = f, then
• ∫ F’(g(x))g’(x) dx = F(g(x)) + C
• because, by the Chain Rule,
d
 F ( g ( x))  F '( g ( x)) g '( x)
dx
INTRODUCTION
• If we make the ‘change of variable’ or
‘substitution’ u = g(x),
we have:
F
'(
g
(
x
))
g
'(
x
)
dx

F
(
g
(
x
))

C

 F (u )  C
  F '(u ) du
INTRODUCTION
• Writing F’ = f, we get:
∫ f(g(x))g’(x) dx = ∫ f(u) du
 Thus, we have proved the following rule.
SUBSTITUTION RULE
Equation 4
• 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
SUBSTITUTION RULE
• Notice that the Substitution Rule for
integration was proved using the Chain Rule
for differentiation.
• Notice also that, if u = g(x), then du = g’(x)
dx.
 So, a way to remember the Substitution Rule is
to think of dx and du in Equation 4 as differentials.
SUBSTITUTION RULE
• Thus, the Substitution Rule says:
• It is permissible to operate with
dx and du after integral signs as if
they were differentials.
SUBSTITUTION RULE
Example 1
• Find ∫ x3 cos(x4 + 2) dx
 We make the substitution u = x4 + 2.
 This is because its differential is du = 4x3 dx,
which, apart from the constant factor 4,
occurs in the integral.
Example 1
SUBSTITUTION RULE
• Thus, using x3 dx = du/4 and the
Substitution Rule, we have:
1
1
x
cos(
x

2)
dx

cos
u

du

4
4  cos udu


3
4
 14 sin u  C
 sin( x  2)  C
1
4
4
 Notice that, at the final stage, we had to return to
the original variable x.
SUBSTITUTION RULE
• The idea behind the Substitution Rule is to
replace a relatively complicated integral by
a simpler integral.
 This is accomplished by changing from the original
variable x to a new variable u that is a function of x.
 Thus, in Example 1, we replaced the integral
∫ x3cos(x4 + 2) dx by the simpler integral ¼ ∫ cos u du.
SUBSTITUTION RULE
• The main challenge in using the rule is
to think of an appropriate substitution.
 You should try to choose u to be some function in
the integrand whose differential also occurs—except
for a constant factor.
 This was the case in Example 1.
SUBSTITUTION RULE
• If that is not possible, try choosing u to be
some complicated part of the integrand—
perhaps the inner function in a composite
function.
SUBSTITUTION RULE
• Finding the right substitution is
a bit of an art.
 It’s not unusual to guess wrong.
 If your first guess doesn’t work, try another
substitution.
E. g. 2—Solution 1
SUBSTITUTION RULE
• Evaluate
 Let u = 2x + 1.
 Then, du = 2 dx.
 So, dx = du/2.

2 x  1 dx
E. g. 2—Solution 1
SUBSTITUTION RULE
 Thus, the rule gives:

2 x  1 dx  

1
2
du
u
2
u
12
du
32
1 u
 
C
2 3/ 2
1 32
 3u C
 (2 x  1)
1
3
32
C
Example 3
SUBSTITUTION RULE
• Find

x
1  4x2
dx
 Let u = 1 – 4x2.
 Then, du = -8x dx.
 So, x dx = -1/8 du and

x
1 4x
2
dx  
1
8

1
du   18  u 1 2 du
u
  18 (2 u )  C   14 1  4 x 2  C
SUBSTITUTION RULE
Example 4
• Calculate ∫ cos 5x dx
 If we let u = 5x, then du = 5 dx.
 So, dx = 1/5 du.
 Therefore,
1
cos
5
x
dx

5  cos u du

 15 sin u  C
 15 sin 5 x  C