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BY PARTS
Integration by Parts
Although integration by parts is used most of the time on products of the form described above, it is sometimes effective on single functions. Looking at the following example.
If is a product of a power of x (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing according to the type of function.
6.3 Integration By Parts
Start with the product rule:
d dv duuv 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
u dv uv v du This is the Integration by Parts formula.
u dv uv v du
The Integration by Parts formula is a “product rule” for integration.
u differentiates to zero (usually).
dv is easy to integrate.
Choose u in this order: LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig
Example 1:
cos x x dxpolynomial factor u x
du dx
cos dv x dx
sinv x
u dv uv v du LIPET
sin cosx x x C
u v v du
sin sin x x x dx
Example:
ln x dxlogarithmic factor lnu x
1du dx
x
dv dx
v x
u dv uv v du LIPET
lnx x x C
1ln x x x dx
x
u v v du
This is still a product, so we need to use integration by parts again.
Example 4:2 xx e dx
u dv uv v du LIPET
2u x xdv e dx
2 du x dx xv e u v v du
2 2 x xx e e x dx 2 2 x xx e xe dx u x xdv e dx
du dx xv e 2 2x x xx e xe e dx
2 2 2x x xx e xe e C
Example 5:
cos xe x dxLIPET
xu e sin dv x dx xdu e dx cosv x
u v v du sin sinx xe x x e dx
sin cos cos x x xe x e x x e dx
xu e cos dv x dx xdu e dx sinv x
sin cos cos x x xe x e x e x dx This is the expression we started with!
uv v du
Example 6:
cos xe x dxLIPET
u v v du
cos xe x dx 2 cos sin cosx x xe x dx e x e x
sin coscos
2
x xx e x e x
e x dx C
sin sinx xe x x e dx xu e sin dv x dx
xdu e dx cosv x
xu e cos dv x dx xdu e dx sinv x
sin cos cos x x xe x e x e x dx
sin cos cos x x xe x e x x e dx
Example 6:
cos xe x dx u v v du
This is called “solving for the unknown integral.”
It works when both factors integrate and differentiate forever.
cos xe x dx 2 cos sin cosx x xe x dx e x e x
sin coscos
2
x xx e x e x
e x dx C
sin sinx xe x x e dx
sin cos cos x x xe x e x e x dx
sin cos cos x x xe x e x x e dx
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.
2 xx e dx & deriv.f x & integralsg x
2x
2x
2
0
xexexexe
2 xx e dx 2 xx e 2 xxe 2 xe C
Compare this with the same problem done the other way:
Example 5:2 xx e dx
u dv uv v du LIPET
2u x xdv e dx
2 du x dx xv e u v v du
2 2 x xx e e x dx 2 2 x xx e xe dx u x xdv e dx
du dx xv e 2 2x x xx e xe e dx
2 2 2x x xx e xe e C This is easier and quicker to do with tabular integration!
3 sin x x dx3x
23x
6x
6
sin x
cos x
sin xcos x
0
sin x
3 cosx x 2 3 sinx x 6 cosx x 6sin x + C
15
Try This
• Given
• Choose a u and dv
• Determinethe v and the du
• Substitute the values, finish integration
5 lnx x dxu v
du dv
__________________u v v du
16
Double Trouble
• Sometimes the second integral must also be done by parts
2 sinx x dxu x2 v -cos x
du 2x dx dv sin x
u v
du dv
2 cos 2 cosx x x x dx
TRIG INTS
Trigonometric functions
20
Recall Basic Identities
• Pythagorean Identities
• Half-Angle Formulas
2 2
2 2
2 2
sin cos 1
tan 1 sec
1 cot csc
2
2
1 cos 2sin
21 cos 2
cos2
These will be used to integrate powers
of sin and cos
These will be used to integrate powers
of sin and cos
21
Integral of sinn x, n Odd
• Split into product of an even and sin x
• Make the even power a power of sin2 x
• Use the Pythagorean identity
• Let u = cos x, du = -sin x dx
5 4sin sin sinx dx x x dx
24 2sin sin sin sinx x dx x x dx
2 22 2sin sin 1 cos sinx x dx x x dx
22 2 41 1 2 ...u du u u du
22
Integral of sinn x, n Odd
• Integrate and un-substitute
• Similar strategy with cosn x, n odd
2 4 3 5
3 5
2 11 2
3 52 1
cos cos cos3 5
u u du u u u C
x x C
23
Integral of sinn x, n Even
• Use half-angle formulas
• Try Change to power of cos2 x
• Expand the binomial, then integrate
2 1 cos 2sin
2
4cos 5x dx
2
22 1cos 1 cos10
2dx x dx
24
Combinations of sin, cos
• General form
• If either n or m is odd, use techniques as before– Split the odd power into an even power and power
of one– Use Pythagorean identity– Specify u and du, substitute– Usually reduces to a polynomial– Integrate, un-substitute
sin cosm nx x dxTry withTry with
2 3sin cosx x dx
25
Combinations of sin, cos
• Consider
• Use Pythagorean identity
• Separate and use sinn x strategy for n odd
3 2 3 5sin 4 1 sin 4 sin 4 sin 4x x dx x x dx
dxxx 4cos4sin 23
26
Combinations of tanm, secn
• When n is even– Factor out sec2 x– Rewrite remainder of integrand in terms of
Pythagorean identity sec2 x = 1 + tan2 x– Then u = tan x, du = sec2x dx
• Try4 3sec tany y dy
27
Combinations of tanm, secn
• When m is odd– Factor out tan x sec x (for the du)– Use identity sec2 x – 1 = tan2 x for even powers of
tan x– Let u = sec x, du = sec x tan x
• Try the same integral with this strategy
4 3sec tany y dy
Note similar strategies for integrals involving combinations ofcotm x and cscn x
Note similar strategies for integrals involving combinations ofcotm x and cscn x
28
Integrals of Even Powers of sec, csc
• Use the identity sec2 x – 1 = tan2 x
• Try 4sec 3x dx
2 2
2 2
2 2 2
3
sec 3 sec 3
1 tan 3 sec 3
sec 3 tan 3 sec 3
1 1tan 3 tan 3
3 9
x x dx
x x dx
x x x dx
x x C
TRIG SUBS
Trigonometric Substitution
We can use right triangles and the pythagorean theorem to simplify some problems.
a
x
2 2a x1
24
dx
x
These are in the same form.
2
24 x
24sec
2
x
22sec 4 x
tan2
x
2 tan x
22sec d dx
22sec
2sec
d
sec d ln sec tan C
24ln
2 2
x xC
We can use right triangles and the pythagorean theorem to simplify some problems.
1
24
dx
x
22sec
2sec
d
sec d ln sec tan C
24ln
2 2
x xC
24ln
2
x xC
2ln 4 ln 2x x C This is a constant.
2ln 4 x x C
a
x
2 2a x
This method is called Trigonometric Substitution.
If the integral contains ,
we use the triangle at right.
2 2a x
If we need , we
move a to the hypotenuse.
2 2a x If we need , we
move x to the hypotenuse.
2 2x a
a
x
2 2a x
a
x2 2x a
2 2
29
x dx
x 3
x
29 x
sin3
x
3sin x
3cos d dx
29cos
3
x
23cos 9 x 29sin 3cos
3cos
d
1 cos 29
2d
91 cos 2
2d
9 9 1sin 2
2 2 2C
sin3
x
1sin3
x
19 9sin 2sin cos
2 3 4
xC
double angle formula
219 9 9
sin2 3 2 3 3
x x xC
2 2
29
x dx
x 3
x
29 x
sin3
x
3sin x
29cos
3
x
23cos 9 x
sin3
x
1sin3
x
19 9sin 2sin cos
2 3 4
xC
double angle formula
219 9 9
sin2 3 2 3 3
x x xC
1 29sin 9
2 3 2
x xx C
3cos d dx
522
dx
x x We can get into the necessary
form by completing the square.
22x x
22x x
2 2 x x
2 2 1 1x x
21 1x
21 1x
21 1
dx
x
Let 1u x
du dx
21
du
u 1
u
21 ucos
cos
d
d C 1 sin u C
1 sin 1x C
21cos
1
u 21 u
sin u cos d du
624 4 2
dx
x x Complete the square:24 4 2x x
24 4 1 1x x
22 1 1x 22 1 1
dx
x
Let 2 1u x 2 du dx
2
1
2 1
du
u
1
u
2 1u
2
2
1 sec
2 sec
d
1
2d
1
2C 11
tan2
u C
11 tan 2 1
2x C
tan u
2sec 1u 1
2du dx
2sec d du
2 2sec 1u
Here are a couple of shortcuts that are result from Trigonometric Substitution:
12 2
1tan
du uC
u a a a
1
2 2sin
du uC
aa u
These are on your list of formulas. They are not really new.
39
New Patterns for the Integrand
• Now we will look for a different set of patterns
• And we will use them in the context of a right triangle
• Draw and label the other two triangles which show the relationships of a and x
2 2 2 2 2 2a x a x x a
a
x
2 2a x
40
Example
• Given
• Consider the labeled triangle– Let x = 3 tan θ (Why?)– And dx = 3 sec2 θ dθ
• Then we have
2 9
dx
x 3
x
2 23 xθ
2
2
3sec
9 tan 9
d
Use identitytan2x + 1 =
sec2x
Use identitytan2x + 1 =
sec2x
23sec3 sec ln sec tan
3sec
dd C
41
Finishing Up
• Our results are in terms of θ– We must un-substitute back into x
– Use the triangle relationships
ln sec tan C 3
x
2 23 xθ
29ln
3 3
x xC
42
Knowing Which Substitution
u
u
2 2u a
43
Try It!!
• For each problem, identify which substitution and which triangle should be used
3 2 9x x dx
2
2
1 xdx
x
2 2 5x x dx
24 1x dx
PART FRACS
2
5 3
2 3
xdx
x x
This would be a lot easier if we could
re-write it as two separate terms.
5 3
3 1
x
x x
3 1
A B
x x
1
These are called non-repeating linear factors.
You may already know a short-cut for this type of problem. We will get to that in a few minutes.
2
5 3
2 3
xdx
x x
This would be a lot easier if we could
re-write it as two separate terms.
5 3
3 1
x
x x
3 1
A B
x x
Multiply by the common denominator.
5 3 1 3x A x B x
5 3 3x Ax A Bx B Set like-terms equal to each other.
5x Ax Bx 3 3A B
5 A B 3 3A B Solve two equations with two unknowns.
1
2
5 3
2 3
xdx
x x
5 3
3 1
x
x x
3 1
A B
x x
5 3 1 3x A x B x
5 3 3x Ax A Bx B
5x Ax Bx 3 3A B
5 A B 3 3A B Solve two equations with two unknowns.
5 A B 3 3A B
3 3A B
8 4B
2 B 5 2A
3 A
3 2
3 1dx
x x
3ln 3 2ln 1x x C
This technique is calledPartial Fractions
1
2
5 3
2 3
xdx
x x
The short-cut for this type of problem is
called the Heaviside Method, after English engineer Oliver Heaviside.
5 3
3 1
x
x x
3 1
A B
x x
Multiply by the common denominator.
5 3 1 3x A x B x
8 0 4A B
1
Let x = - 1
2 B
12 4 0A B Let x = 3
3 A
2
5 3
2 3
xdx
x x
The short-cut for this type of problem is
called the Heaviside Method, after English engineer Oliver Heaviside.
5 3
3 1
x
x x
3 1
A B
x x
5 3 1 3x A x B x
8 0 4A B
1
2 B
12 4 0A B 3 A
3 2
3 1dx
x x
3ln 3 2ln 1x x C
Good News!
The AP Exam only requires non-repeating linear factors!
The more complicated methods of partial fractions are good to know, and you might see them in college, but they will not be on the AP exam or on my exam.
26 7
2
x
x
Repeated roots: we must use two terms for partial fractions.
22 2
A B
x x
6 7 2x A x B
6 7 2x Ax A B
6x Ax 7 2A B
6 A 7 2 6 B
7 12 B
5 B
26 5
2 2x x
2
3 2
2
2 4 3
2 3
x x x
x x
If the degree of the numerator is higher than the degree of the denominator, use long division first.
2 3 22 3 2 4 3x x x x x 2x
3 22 4 6x x x 5 3x
2
5 32
2 3
xx
x x
5 3
23 1
xx
x x
3 2
23 1
xx x
4
(from example one)
22
2 4
1 1
x
x x
irreduciblequadratic
factor
repeated root
22 1 1 1
Ax B C D
x x x
first degree numerator
2 2 22 4 1 1 1 1x Ax B x C x x D x
2 3 2 22 4 2 1 1x Ax B x x C x x x Dx D
3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
A challenging example:
3 2 2 3 2 22 4 2 2x Ax Ax Ax Bx Bx B Cx Cx Cx C Dx D
0 A C 0 2A B C D 2 2A B C 4 B C D
1 0 1 0 0
2 1 1 1 0
1 2 1 0 2
0 1 1 1 4
2 r 3
r 1
1 0 1 0 0
0 3 1 1 4
0 2 0 0 2
0 1 1 1 4
2
1 0 1 0 0
0 1 0 0 1
0 3 1 1 4
0 1 1 1 4
3 r 2
r 2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 1 1 3
r 3
1 0 1 0 0
0 1 0 0 1
0 3 1 1 4
0 1 1 1 4
3 r 2
r 2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 1 1 3
r 3
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 0 2 2
2
1 0 1 0 0
0 1 0 0 1
0 0 1 1 1
0 0 0 1 1
r 4
1 0 1 0 0
0 1 0 0 1
0 0 1 0 2
0 0 0 1 1
r 3
1 0 0 0 2
0 1 0 0 1
0 0 1 0 2
0 0 0 1 1
1 0 0 0 2
0 1 0 0 1
0 0 1 0 2
0 0 0 1 1
22
2 4
1 1
x
x x
22 1 1 1
Ax B C D
x x x
22
2 1 2 1
1 1 1
x
x x x
We can do this problem on the TI-89:
22
2 4expand
1 1
x
x x
expand ((-2x+4)/((x^2+1)*(x-1)^2))
22 2
2 1 2 1
1 1 1 1
x
x x x x
Of course with the TI-89, we could just integrate and wouldn’t need partial fractions!
3F2
1. Using Table of Integration Formulas
2. Simplify the Integrand if Possible Sometimes the use of algebraic manipulation or trigonometric identities will simplify the integrand and
make the method of integration obvious.
3. Look for an Obvious Substitution Try to find some function in the integrand whose differential also occurs, apart
from a constant factor.
3. Classify the Integrand According to Its Form Trigonometric functions, Rational functions, Radicals, Integration by parts.
4. Manipulate the integrand.
Algebraic manipulations (perhaps rationalizing the denominator or using trigonometric identities) may be useful in transforming the integral into an easier form.
5. Relate the problem to previous problems When you have built up some experience in integration, you may be able to use a method on a given
integral that is similar to a method you have already used on a previous integral. Or you may even be able to express the given integral in terms of a previous one.
6. Use several methods
Sometimes two or three methods are required to evaluate an integral. The evaluation could involve several successive substitutions of different types, or it might combine integration by parts with one or more substitutions.
Strategy for Integration
IMPROPER INTS
Until now we have been finding integrals of continuous functions over closed intervals.
Sometimes we can find integrals for functions where the function or the limits are infinite. These are called improper integrals.
Example 1:
1
0
1
1
xdx
x
The function is
undefined at x = 1 .
Since x = 1 is an asymptote, the function has no maximum.
Can we find the area under an infinitely high curve?
We could define this integral as:
01
1lim
1
b
b
xdx
x
(left hand limit)
We must approach the limit from inside the interval.
01
1lim
1
b
b
xdx
x
1
1
1
1
x
x
xdx
x
Rationalize the numerator.
2
1+x
1dx
x
2 2
1 x
1 1dx dx
x x
21u x
2 du x dx1
2
du x dx 1
1 21
sin2
x u du
2 2
1 x
1 1dx dx
x x
21u x
2 du x dx1
2
du x dx 1
1 21
sin2
x u du
11 2sin x u
1 2
1 0lim sin 1
b
bx x
1 2 1
1lim sin 1 sin 0 1b
b b
1
2
2
0 0
This integral converges because it approaches a solution.
Example 2:
1
0
dx
x
1
0lim ln
bbx
0lim ln1 lnb
b
0
1lim lnb b
This integral diverges.
(right hand limit)
We approach the limit from inside the interval.
1
0
1lim
bbdx
x
Example 3:
3
2031
dx
x
The function approacheswhen .
1x
233
01 x dx
2 233 3
01 1lim 1 lim 1
b
cb cx dx x dx
31 1
3 31 1
0
lim 3 1 lim 3 1b
b cc
x x
2 233 3
01 1lim 1 lim 1
b
cb cx dx x dx
31 1
3 31 1
0
lim 3 1 lim 3 1b
b cc
x x
11 1 133 3 3
1 1lim 3 1 3 1 lim 3 2 3 1b c
b c
0 0
33 3 2
Example 4:
1 P
dx
x
0P
1 Px dx
1lim
b P
bx dx
1
1
1lim
1
bP
bx
P
1 11lim
1 1
P P
b
b
P P
What happens here?
If then gets bigger and bigger as , therefore the integral diverges.
1P 1Pb
b
If then b has a negative exponent and ,therefore the integral converges.
1P 1 0Pb
(P is a constant.)
Improper Integrals
<
Idea of the Comparison Theorem
2 2 2
Observe that the function to be integrated satisfies
3 30
(1 2sin (4 ))(1 ) 1
for all 0. The following graph illustrates this observation.
x x x
x
2
3The blue curve is the graph of the function while the red curve is the
1graph of the function to be integrated.
x
2 20
3
(1 2sin (4 ))(1 )dx
x x
The improper integral converges if the area
under the red curve is finite. We show that this is true by showing that the area under the blue curve is finite. Since the area under the red curve is smaller than the area under the blue curve, it must then also be finite. This means that the complicated improper integral converges.
Examples
2 20 0
3 3lim
1 1
b
bdx dx
x x
20
2 20
3This means that the improper integral converges.
13
Hence also the improper integral converges.(1 2sin (4 ))(1 )
dxx
dxx x
0
3lim 3arctan( ) .
2
b
bx
To show that the area under the blue curve in the previous figure is finite, compute as follows:
Comparison Theorem
a
Let , , , . Assume that the functions f and g satisfy
0 f( ) g( ) for all , . Assume also that the integral f( )
is improper.
1) If the improper integral g( ) converges,
b
a
b
a b a b
x x x a x b x dx
x dx
a
a
then also the improper
integral f( ) converges, and 0 f( ) g( ) .
2) If the improper integral f( ) diverges, then also the improper integral
g( ) diverges.
b b b
a a
b
b
a
x dx x dx x dx
x dx
x dx
The integral f( ) is improper if either , or the
function f has singularities in the interval of integration.
b
ax dx a b
Examples
dxx
x
1 4 1dx
x
1
1
DIVERGES
44 1 x
x
x
x
xx
x 12
x
x
x
x 114
lim 1
)(4lim
x
xx
x 14
2
lim
x
x
x 14
4
lim
x
x
x
14
4
lim
x
x
x
Since is continuous for
x
1x
110
Examples
dxx
1 3 1
1dx
x
1 3
1CONVERGES
33
1
1
1
xx
3
3
11
1
limx
xx
13
3
lim
x
x
x 13
3
lim
x
x
x
13
3
lim
x
x
x
Since is continuous for
x
1x
110
P-test 12
3p
Normal Distribution Function
dxe x
2
dxedxedxedxe xxxx
1
1
1
1 2222
Convergesdxe x1
1
2
Normal Distribution Function
dxe x
1
2
xxx 21 Hence
xx ee 2
0 Therefore
dxedxe xx
11
2
0
dxe x
1
2
converges
Normal Distribution Function
dxe x
1
2
xxx 21 Hence
xx ee 2
0 Therefore
dxedxe xx
11
2
0
dxe x
1
2
Converges and so does
dxe x
2
Examples
dxx
1
0 sin
1It’s improper
because xsin
1 It’s unde-fined for
0x
xxx sin00sin10 xx
and
hence
xx sin
110 therefore
dxx
dxx
1
0
1
0 sin
110
DIVERGES
Examples
1 4
3
1
1dx
x
x
xx
x
x
x 1
1
14
3
4
3
111
)1(1
11
4
4
4
34
3
limlimlim
x
xx
x
xx
x
xx
xxx 0
dxx
1
1DIVERGES
Examples
dxex x
3
1
xxx
eeex
11
x
x
xx
x
x ex
e
e
ex
limlim 1
1
0
dxex
3
DIVERGES
11 limlim
x
x
xx
x
x e
e
e
e
L’HopL’Hop
Examples
dxx
e
dxx
x
dxex
dxx
x
x
x
1
1
4
3
2 2
2
sin31
1
cos
0
0
0
0 dxx
2 2
1
dxedxe
xx
22
1
dxx1
1
dxe x
1
P-test 12 p
P-test 1
2
1p
