integration of irrational functions rational substitution is the usual way to integrate them. ex....
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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