INTEGRALSINTEGRALS

5

Indefinite Integrals

INTEGRALS

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 INTEGRAL

INDEFINITE INTEGRALS

For example, we can write

Thus, we can regard an indefinite integral as representing an entire family of functions (one antiderivative for each value of the constant C).

3 32 2because

3 3

x d xx dx C C x

dx

INDEFINITE INTEGRALS

Any formula can be verified by differentiating

the function on the right side and obtaining

the integrand.

For instance, 2

2

sec tan

because

(tan ) sec

x dx x C

dx C x

dx

TABLE OF INDEFINITE INTEGRALS

1

2 2

( ) ( ) [ ( ) ( )]

( ) ( )

( 1)1

sin cos cos sin

sec tan csc cot

sec tan sec csc cot csc

nn

cf x dx c f x dx f x g x dx

f x dx g x dx

xk dx kx C x dx C n

n

x dx x C x dx x C

x dx x C x dx x C

x x dx x C x x dx x C

Table 1

INDEFINITE INTEGRALS

Thus, we write

with the understanding that it is valid on

the interval (0, ∞) or on the interval (-∞, 0).

2

1 1dx C

xx

INDEFINITE INTEGRALS

This is true despite the fact that the general

antiderivative of the function f(x) = 1/x2,

x ≠ 0, is:

1

2

1if 0

( )1

if 0

C xxF x

C xx

INDEFINITE INTEGRALS

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.

Example 1

INDEFINITE INTEGRALS

Evaluate

This indefinite integral isn’t immediately apparent in Table 1.

So, we use trigonometric identities to rewrite the function before integrating:

Example 2

2

cos 1 cos

sin sinsin

csc cot csc

d d

d C

2

cos

sind

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.

INTEGRALS

• Our antidifferentiation formulas don’t tell

us how to evaluate integrals such as

22 1x x dx

Equation 1INTRODUCTION

• To find this integral, we use the problem-

solving 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,

Then, the differential of u is du = 2x dx.

INTRODUCTION

22 1x x dx

• 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

INTRODUCTION

22 1x x dx

• So, formally, without justifying our

calculation, we could write:

2 2

3/ 223

2 3/ 223

2 1 1 2

( 1)

x x dx x x dx

udu

u C

x C

Equation 2INTRODUCTION

• However, now we can check that we have

the correct answer by using the Chain Rule

to differentiate

2 3 2 2 1 232 23 3 2

2

( 1) ( 1) 2

2 1

dx C x x

dx

x x

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

• Observe that, if F’ = f, then

• ∫ F’(g(x))g’(x) dx = F(g(x)) + C

• because, by the Chain Rule,

Equation 3INTRODUCTION

( ( )) '( ( )) '( )dF g x F g x g x

dx

• 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.

INTRODUCTION

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

Equation 4

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

• 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:

Notice that, at the final stage, we had to return to the original variable x.

3 4 1 14 4

14

414

cos( 2) cos cos

sin

sin( 2)

x x dx u du udu

u C

x C

Example 1

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.

• 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.

SUBSTITUTION RULE

SUBSTITUTION RULE

• Evaluate

Let u = 2x + 1.

Then, du = 2 dx.

So, dx = du/2.

2 1x dxE. g. 2—Solution 1

Thus, the rule gives:

1 212

3 2

3 213

3 213

2 12

1

2 3/ 2

(2 1)

dux dx u

u du

uC

u C

x C

SUBSTITUTION RULE E. g. 2—Solution 1

SUBSTITUTION RULE

• Find

Let u = 1 – 4x2. Then, du = -8x dx. So, x dx = -1/8 du and

21 4

xdx

x

1 21 18 82

21 18 4

1

1 4

(2 ) 1 4

xdx du u du

ux

u C x C

Example 3

SUBSTITUTION RULE

• Calculate ∫ cos 5x dx

If we let u = 5x, then du = 5 dx. So, dx = 1/5 du. Therefore,

15

15

15

cos5 cos

sin

sin 5

x dx u du

u C

x C

Example 4