Integration of irrational functions Rational substitution is the usual way to integrate them. Ex. Evaluate

Sol. Let then

3.

3 1 2

dx

x 3 3 1 ,x t 3 21

( 1), .3

x t dx t dt 2

3

4( 2 )

2 23 1 2

dx tdt t dt

t tx

2 3 331(3 1) 2 3 1 4ln | 3 1 2 | .

2x x x C

212 4ln | 2 |

2t t t C

Example Ex. Evaluate

Sol.

23.

( 1)( 1)

dx

x x

323

1 1 1,

1 1( 1)( 1)

x

x xx x

3 2

33 3 2

1 1 6,

1 1 ( 1)

x t t dtt x dx

x t t

323

3

1( 1)( 1)

dx dt

tx x

21 2 1ln( 1) 3 arctan ln | 1| .

2 3

tt t t C

Strategy for integration First of all, remember basic integration formulae. Then, try the following four-step strategy: 1. Simplify the integrand if possible. For example:

2. Look for an obvious substitution. For example:

2

tan

sec d

1sin cos sin 2

2 d d

2

arcsin

1

xdx

xarcsin arcsin , xd x

2

1x

dxx

2 22 [1 ( 1 ) ] 1x d x

Strategy for integration 3. Classify the integrand according to its form

a. rational functions: partial fractions

b. rational trigonometric functions:

c. product of two different kind of functions: integration

by parts

d. irrational functions: trigonometric substitution, rational

substitution, reciprocal substitution 4. Try again. Manipulate the integrand, use several

methods, relate the problem to known problems

tan2

t

x

Example Integrate

Sol I rational substitution works but complicated

Sol II manipulate the integrand first

1.

1

xdxx

1,

1

x

tx

2 2 2

1 1.

1 1 1

x xdx dx dx

x x x

Example Ex. Find

Sol I. Substitution works but complicated Sol II.

2 2sin cos( 0).

sin cos

c x d x

I dx a ba x b x

tan ,2

x

t

sin cos ( sin cos ) ( cos sin ) c x d x a x b x a x b x

2 2 2 2, .

ac bd ad bc

a b a b

cos sin[ ] ln | sin cos |

sin cos

a x b x

I dx x a x b x Ca x b x

2 2 2 2ln | sin cos | .

ac bd ad bc

x a x b x Ca b a b

Can we integrate all continuous

functions? Since continuous functions are integrable, any continuous

function f has an antiderivative. Unfortunately, we can NOT integrate all continuous

functions. This means, there exist functions whose

integration can not be written in terms of essential functions. The typical examples are:

2 3 2sin 1, , 1 , , sin ,

ln x x

e dx dx x dx dx x dxx x

, cos x

xedx e dx

x

Approximate integration In some situation, we can not find An alternative

way is to find its approximate value.

By definition, the following approximations are obvious:

left endpoint approximation

right endpoint approximation

11

( ) ( )nb

n iai

f x dx L f x x

( ) .b

af x dx

1

( ) ( )nb

n iai

f x dx R f x x

Approximate integration Midpoint rule:

Trapezoidal rule

Simpson’s rule

11

1( ) ( ) , ( )

2

nb

i in i iai

f x dx M f x x x x x

11

1( ) [ ( ) ( )]

2

nb

n i iai

f x dx T f x f x x

11

1( ) [ ( ) 4 ( ) ( )]

3

nb

n i i iai

f x dx S f x f x f x x

Improper integrals The definite integrals we learned so far are defined on a

finite interval [a,b] and the integrand f does not have an

infinite discontinuity. But, to consider the area of the (infinite) region under the

curve from 0 to 1, we need to study the integrability

of the function on the interval [0,1]. Also, when we investigate the area of the (infinite) region

under the curve from 1 to we need to evaluate

1/y x

,

1/y x

21/y x

21

1.

dxx

Improper integral: type I We now extend the concept of a definite integral to the

case where the interval is infinite and also to the case where

the integrand f has an infinite discontinuity in the interval. In

either case, the definite integral is called improper integral. Definition of an improper integral of type I If for any

b>a, f is integrable on [a,b], then

is called the improper integral of type I of f on and

denoted by If the right side limit

exists, we say the improper integral converges.

lim ( )

b

abf x dx

[ , ],a( ) lim ( ) .

b

a abf x dx f x dx

Improper integral: type I Similarly we can define the improper integral

and its convergence.

The improper integral is defined as

only when both and are convergent,

the improper integral converges.

( ) lim ( )

b b

aaf x dx f x dx

( )

f x dx

( ) ( ) ( ) .

c

cf x dx f x dx f x dx

( ) cf x dx ( )

c f x dx

( )

f x dx

Example Ex. Determine whether the integral converges or

diverges. Sol. diverge

Ex. Find

Sol.

1

1

dxx

1 1

1 1lim lim ln

b

b bdx dx bx x

22

1 ln.

xdx

x

2 2 22 2 2 22

1 ln 1 ln 1 1ln

bb b b bx x

dx dx dx xdx x x x x

222 2 2

1 1 1 ln ln 2 1 ln 1ln ln 2.

2 2

b b bb x

x dxx x x b x

Example Ex. Find

Sol.

Remark From the definition and above examples, we see

the New-Leibnitz formula for improper integrals is also true:

0.

xxe dx

0 0lim lim lim ( 1 ) 1.

x x x a a

aaa a axe dx xe e ae e

( ) ( ) lim ( ) ( ).

aa x

f x dx F x F x F a

( ) ( ) lim ( ) lim ( ).

b a

f x dx F x F b F a

Example Ex. Evaluate

Sol.

Ex. For what values of p is the integral convergent? Sol. When

2.

1

dx

x

2arctan .

1

dx

xx

1

p

dx

x1

1

11

11

( lim 1) .111 1

1

p

pp x

pdx x

xpx p p

p

1,p

11ln .

dx

xx

Example All the integration techniques, such as substitution rule,

integration by parts, are applicable to improper integrals.

Especially, if an improper integral can be converted into a

proper integral by substitution, then the improper integral

is convergent. Ex. Evaluate Sol. Let then

20.

(1 )(1 )

dx

Ix x

tan ,x t

2 2

0 0

cos

1 tan sin cos

dt t

I dtt t t

0

2

sin.

sin cos 4

u

duu u

Improper integral: type II Definition of an improper integral of type II If f is

continuous on [a,b) and x=b is a vertical asymptote ( b is

said to be a singular point ), then

is called the improper integral of type II. If the limit exists,

we say the improper integral converges.

( ) lim ( )b t

a at bf x dx f x dx

Improper integral: type II Similarly, if f has a singular point at a, we can define the

improper integral

If f has a singular point c inside the interval [a,b], then the

improper integral

Only when both of the two improper integrals and

converge, the improper integral converge.

( ) ( ) ( ) . b c b

a a cf x dx f x dx f x dx

( )c

af x dx

( )b

cf x dx ( )

b

af x dx

( ) lim ( )b b

a tt af x dx f x dx

Example Ex. Find Sol. x=0 is a singular point of lnx.

Ex. Find

Sol.

1

0ln . xdx

1 1 1

0 0 0ln lim ln lim [ ln ]

aaa a

xdx xdx x x x

0lim ( 1 ln ) 1.

a

a a a

3

0.

1dx

x

3 3

00ln | 1| l .n 2

1

dxx

x

1

00 1 1lim [ln(1 )] lim ln(1 ) .

1

b

b b

dxx b

x

Example Again, Newton-Leibnitz formula, substitution rule and

integration by parts are all true for improper integrals of

type II.

Ex. Find

Sol. x=a is a singular point.

2 20( 0).

a dx

aa x

2 200

arcsin .2

aa dx x

aa x

Example Ex. For what values of p>0 is the improper integral

convergent?

Sol. x=b is the singular point. When

( )( )

b

pa

dxb a

b x

1 1 1( ) ( ) ( )lim

( ) 1 1 1

bp p pb

pa x ba

dx b x b a b x

b x p p p

1,p

ln( ) ln( ) lim ln( ) .

b b

aa x b

dxb x b a b x

b x

1( )1.1

1

pb ap

p

p

Comparison test Comparison principle Suppose that f and g are

continuous functions with for then

(a)If converges, then converges.

(b)If the latter diverges, then the former diverges.

Ex. Determine whether the integral converges.

Sol.

( )

a f x dx ( )

a g x dx

2 21

1

1

xedx

x x

22 2

1 2

1

xe

xx x

( ) ( ) 0f x g x ,x a

Example Determine whether the integral is convergent or divergent

61(1)

1

xdx

x

2

0

1(2)

sindx

x x

Evaluation of improper integrals All integration techniques and Newton-Leibnitz formula

hold true for improper integrals.

Ex. The function defined by the improper integral

is called Gamma function. Evaluate Sol.

1

0( ) ( 0)

xx e dx ( ). n

0 0( 1)

x xx e dx x de 1

0 0( ).

x xx e x e dx

00(1) 1x xe dx e

( ) ( 1)!. n n

Example Ex. Find

Sol.

0.

(1 )(1 )

dxI

x x x

20

2

(1 )(1 )

tdtx t I

t t t

20

1 1( )1 1

tdt

t t

2

0

1ln(1 ) ln(1 ) arctan .

2 2t t t

202

(1 )(1 )

dt

t t

Homework 19 Section 7.4: 37, 38, 46, 48

Section 7.5: 31, 39, 44, 47, 59, 65