BY PARTS
Integration by Parts
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.
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.
6.3 Integration By Parts
Start with the product rule:
d
dv
du
 uv   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
u differentiates to
dv is easy to
integrate.
zero (usually).
The Integration by Parts formula is a “product rule” for
integration.
Choose u in this order:
LIPET
Logs, Inverse trig, Polynomial, Exponential, Trig

Example 1:
 x  cos x dx
polynomial factor
u v   v du
 u dv  uv   v du
LIPET
ux
dv  cos x dx
du  dx
v  sin x
x  sin x   sin x dx
x  sin x  cos x  C

Example:
 ln x dx
logarithmic factor
u v   v du
 u dv  uv   v du
LIPET
u  ln x
dv  dx
1
du  dx
x
vx
1
ln x  x   x  dx
x
x ln x  x  C

Example 4:
 x e dx
u v   v du
x e   e  2 x dx
2 x
2 x
 u dv  uv   v du
u  x2
dv  e x dx
du  2x dx
ve
x
x
x e  2  xe dx
2 x
x

x e  2 xe   e dx
2 x
LIPET
x
x

This is still a product, so we
x
u
x integration
need to
use
by
dv  e dx
parts again.
du  dx
v  ex
x 2 e x  2 xe x  2e x  C

Example 5:
LIPET
u  e x dv  cos x dx
du  e x dx v  sin x
x
e
 cos x dx
u v   v du
ue
dv  sin x dx
x
du  e dx v   cos x
x
e sin x   sin x  e dx
x
x

e sin x  e   cos x    cos x  e dx
x
x
x
uv
v du
e sin x  e cos x   e cos x dx
x
x
x

This is the
expression we
started with!

Example 6:
LIPET
u  e x dv  cos x dx
du  e x dx v  sin x
x
e
 cos x dx
u v   v du
ue
dv  sin x dx
x
du  e dx v   cos x
x
e sin x   sin x  e dx
x
x

e sin x  e   cos x    cos x  e dx
x
x
x

 e cos x dx  e sin x  e cos x   e
2 e cos x dx  e sin x  e cos x
x
x
x
x
x
x
x
x
e
sin
x

e
cos x
x
C
 e cos x dx 
2
x
cos x dx
Example 6:
e
x
This is called “solving
for the unknown
integral.”
cos x dx
u v   v du
It works when both
factors integrate and
differentiate forever.
e sin x   sin x  e dx
x
x

e sin x  e   cos x    cos x  e dx
x
x
x

 e cos x dx  e sin x  e cos x   e
2 e cos x dx  e sin x  e cos x
x
x
x
x
x
x
cos x dx
x
x
x
e
sin
x

e
cos x
x
C
 e cos x dx 
2

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 x
x
 e dx
f  x  & deriv. g  x  & integrals
 x
2
e
 2x
ex
 2
0
x
e
x
e
x
Compare this with
the same problem
done the other way:
2 x
x
x

2
e
x
e

2
xe
C
 x e dx 
2 x

Example 5:
 x e dx
u v   v du
x e   e  2 x dx
2 x
2 x
 u dv  uv   v du
u  x2
dv  e x dx
du  2x dx
ve
x
x
x e  2  xe dx
2 x
x

ux
x e  2 xe   e dx
2 x
LIPET
x
x

x 2 e x  2 xe x  2e x  C
du  dx
dv  e dx
x
v  ex
This is easier and quicker to
do with tabular integration!

x
3
x

3
sin x dx
 3x 2
sin x
 cos x
 6x
 6
 sin x
cos x
0
sin x
 x 3 cos x  3x 2 sin x  6x cos x  6sin x + C
p
Try This
5
x
 ln x dx
• Given
• Choose a u
u
v
and dv
du
dv
• Determine
the v and the du
• Substitute the values, finish integration
u  v   v du  __________________
15
Double Trouble
• Sometimes the second integral must also be
done by parts
x
2
sin x dx
u
x2
du 2x dx
v
-cos x
dv
sin x
 x cos x  2 x  cos x dx
2
u
v
du
dv
16
TRIG INTS
Trigonometric functions
Recall Basic Identities
• Pythagorean Identities
sin 2   cos 2   1
tan 2   1  sec 2 
1  cot   csc 
2
2
• Half-Angle Formulas
1  cos 2
sin  
2
1  cos 2
2
cos  
2
These will be used
to integrate powers
of sin and cos
2
20
Integral of sinn x, n Odd
• Split into product of an even and sin x
5
4
sin
x
dx

sin

 x  sin x dx
• Make the even power a power of sin2 x
 sin
x  sin x dx    sin x  sin x dx
4
2
2
• Use the Pythagorean identity
 sin x 
2
2
sin x dx   1  cos x  sin x dx
2
2
• Let u = cos x, du = -sin x dx
 1  u

2 2
du    1  2u  u du  ...
2
4
21
Integral of sinn x, n Odd
• Integrate and un-substitute
2 3 1 5
  1  2u  u du  u  u  u  C
3
5
2
1
3
  cos x  cos x  cos5  C
3
5
2
4
• Similar strategy with cosn x, n odd
22
Integral of sinn x, n Even
• Use half-angle formulas
1  cos 2
sin  
2
2
4
cos
5x dx Change to power of cos2 x
• Try 
  cos 
2 2
2
1

dx    1  cos10 x   dx
2

• Expand the binomial, then integrate
23
Combinations of sin, cos
• General form
Try with
n
x dx
 sinsin x x cos
 cos x dx

m
2
3
• 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
24
Combinations of sin, cos
• Consider
 sin
3
4 x  cos 4 x dx
2
• Use Pythagorean identity
 sin
3
4 x  1  sin 4 x  dx    sin 4 x  sin 4 x  dx
2
3
5
• Separate and use sinn x strategy for n odd
25
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
• Try
sec
y

tan
y
dy

4
3
26
Combinations of tanm, secn
• When m is odd
– Factor out tan x sec x (for the Note
du) similar strategies for
integrals involving
combinations
of of
– Use identity sec2 x – 1 = tan2 x for even
powers
cotm x and cscn x
tan x
– Let u = sec x, du = sec x tan x
• Try the same integral with this strategy
 sec
4
y  tan y dy
3
27
Integrals of
Even Powers of sec, csc
• Use the identity sec2 x – 1 = tan2 x
4
• Try
 sec 3x dx 
 sec 3x  sec 3x dx 
 1  tan 3x  sec 3x dx 
  sec 3x  tan 3x  sec 3x  dx 
2
2
2
2
2
2
2
1 3
1
tan 3 x  tan 3 x  C
9
3
28
TRIG SUBS
Trigonometric Substitution
We can use right triangles and the pythagorean
theorem to simplify some problems.
1

dx
4 x
2
These are
in the same
form.
a4 xx
ln sec  tan   C
4 x
x
 C
2
2
2
ln
x

2sec2  d
 2sec
 sec d
22
2
a2
4  x2
sec 
2
2sec  4  x
2
x
tan  
2
2 tan  x
2sec 2  d  dx

We can use right triangles and the pythagorean
theorem to simplify some problems.
1

dx
4  x2
2sec2  d
 2sec
4  x2  x
C
2
ln
ln
4  x 2  x  ln 2  C
 sec d
ln sec  tan   C
ln
4  x2 x
 C
2
2
This is a constant.
ln
4  x2  x  C

This method is called Trigonometric Substitution.
a x
2
a x ,
2
If the integral contains
2
2
x

we use the triangle at right.
a
If we need a2  x2 , we
If we need
move a to the hypotenuse.
move x to the hypotenuse.
x
a
x

a x
2
x2  a2 , we
2
x2  a2

a

2

x 2 dx
9  x2
3

9  x2
9sin 2   3cos d

3cos
1  cos 2
9
d
2
9
1  cos 2 d

2
9
9 1
   sin 2  C
2
2 2
x
x sin   3
3sin  x
9  x2
cos 
3
3cos  9  x2
3cos d  dx
x
sin  
3
double angle
x
  sin 1
formula
3
9 1 x 9
sin
  2sin  cos   C
2
3 4
9 1 x 9 x 9  x 2
sin
  
C
2
3 2 3
3

2

x 2 dx
9  x2
3

9  x2
x
x sin   3
3sin  x
9  x2
cos 
3
3cos  9  x2
3cos d  dx
x
sin  
3
double angle
x
  sin 1
formula
3
9 1  x  x
sin   
9  x2  C
2
3 2
9 1 x 9
sin
  2sin  cos   C
2
3 4
9 1 x 9 x 9  x 2
sin
  
C
2
3 2 3
3

5


dx
2x  x2
2x  x 2
dx
1   x  1

2
Let u  x 1


 x2  2x


du  dx
 x2  2x  1  1
du
1 u2
cos  d
 cos
 d
We can get 2x  x 2 into the necessary
form by completing the square.
  x  1  1
2
1   x  1
2
1

sin
u C
  C
 sin 1  x 1  C
1
u

1 u2
1 u2
 1 u2
cos 
1
sin   u cos d  du

6
dx
 4 x2  4 x  2
4 x2  4 x  2
4 x2  4 x  1  1
dx
  2 x  1
Complete the square:
2
1
1 du
2  u2 1
1 sec 2  d
2  sec 2 
 2 x  1
2
1
Let u  2x  1
du  2 dx
1
du  dx
2
1
1
1
1

tan
u C



C
d

2
2
2
1

tan 1  2 x  1  C
2
u2 1
u

1
tan  u
sec 2  d  du
sec  u 2  1
sec2   u 2  1

Here are a couple of shortcuts that are result from
Trigonometric Substitution:
du
1
1 u
 u 2  a 2  a tan a  C

du
u
 sin
C
a
a2  u 2
1
These are on your list of
formulas. They are not
really new.
p
New Patterns for the Integrand
• Now we will look for a different set of patterns
a2  x2
a2  x2
x2  a2
• And we will use them in the context of a right
triangle
a2  x2
a
x
• Draw and label the other two triangles which
show the relationships of a and x
39
Example
• Given

dx
x2  9
3
32  x2
θ
• Consider the labeled triangle
x
– Let x = 3 tan θ
(Why?)
– And dx = 3 sec2 θ dθ
• Then we have

3sec 2  d
9 tan 2   9
Use identity
3sec2  d
 3 sec d  ln sec   tan   C
2x+ 1 =
tan
3sec 
sec2x
40
Finishing Up
• Our results are in terms of θ
– We must un-substitute back into x
ln sec  tan   C
– Use the triangle relationships
ln
3
32  x2
θ
x
9  x2 x
 C
3
3
41
Knowing Which Substitution
u
u
u 2  a2
42
Try It!!
• For each problem, identify which substitution
and which triangle should be used
x
3

x  9 dx
2

1 x
dx
2
x
2
4   x  1 dx
2

x 2  2 x  5 dx
43
PART FRACS
1
5x  3
 x 2  2 x  3 dx
This would be a lot easier if we could
re-write it as two separate terms.
5x  3
A
B


 x  3 x  1 x  3 x  1
These are called nonrepeating linear factors.
You may already know a
short-cut for this type of
problem. We will get to
that in a few minutes.

1
5x  3
 x 2  2 x  3 dx
This would be a lot easier if we could
re-write it as two separate terms.
5x  3
A
B


 x  3 x  1 x  3 x  1
Multiply by the common
denominator.
5x  3  A  x  1  B  x  3
5x  3  Ax  A  Bx  B  3
5x  Ax  Bx
5  A B
Set like-terms equal to
each other.
3  A  B  3
3  A  3B
Solve two equations with
two unknowns.

1
5  A B
5x  3
 x 2  2 x  3 dx
3  A  3B
3   A  3B
8  4B
5x  3
A
B


 x  3 x  1 x  3 x  1
5x  3  A  x  1  B  x  3
5x  3  Ax  A  Bx  B  3
5x  Ax  Bx
5  A B
3  A  B  3
3  A  3B
2B
5  A 2
3 A
3
2
 x  3  x  1 dx
3ln x  3  2ln x  1  C
This
is called
Solvetechnique
two equations
with
Fractions
twoPartial
unknowns.

1
5x  3
 x 2  2 x  3 dx
The short-cut for this type of problem is
called the Heaviside Method, after
English engineer Oliver Heaviside.
5x  3
A
B


 x  3 x  1 x  3 x  1
5x  3  A  x  1  B  x  3
Multiply by the common
denominator.
Let x = - 1
8  A  0  B 4
2B
12  A   4  B  0
3 A
Let x = 3

1
5x  3
 x 2  2 x  3 dx
The short-cut for this type of problem is
called the Heaviside Method, after
English engineer Oliver Heaviside.
5x  3
A
B


 x  3 x  1 x  3 x  1
5x  3  A  x  1  B  x  3
8  A  0  B 4
2B
12  A   4  B  0
3 A
3
2
 x  3  x  1 dx
3ln x  3  2ln x  1  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.

2
6x  7
 x  2
2
A
B


2
x  2  x  2
Repeated roots: we must
use two terms for partial
fractions.
6 x  7  A  x  2  B
6x  7  Ax  2 A  B
6x  Ax
7  2A  B
6 A
7  2 6  B
7  12  B
6
5

x  2  x  2 2
5  B

4
2 x3  4 x 2  x  3
x2  2 x  3
If the degree of the numerator is
higher than the degree of the
denominator, use long division first.
2x
x 2  2 x  3 2 x3  4 x 2  x  3
2 x3  4 x 2  6 x
5x  3
5x  3
2x  2
x  2x  3
(from example one)
5x  3
3
2
2x 
 2x 

 x  3 x  1
 x  3  x  1

A challenging
example:
x
first degree numerator
2 x  4
2

 1  x  1
irreducible
quadratic
factor
2
Ax  B
C
D
 2


x  1 x  1  x  12
repeated root




2 x  4   Ax  B  x  1  C x2  1  x  1  D x 2  1
2




2 x  4   Ax  B  x 2  2 x  1  C x3  x 2  x  1  Dx 2  D
2 x  4  Ax3  2 Ax 2  Ax  Bx 2  2 Bx  B  Cx 3  Cx 2  Cx  C  Dx 2  D

2 x  4  Ax3  2 Ax 2  Ax  Bx 2  2 Bx  B  Cx 3  Cx 2  Cx  C  Dx 2  D
0  A  C 0  2A  B  C  D 2  A  2B  C
4  B C  D
1
0
1
0
0
1
0
1
0
0
2
1
1
1
0 2  r 3
0
1
0
0
1
1
2
1
0
2  r 1
0
3
1
1
4 3  r 2
0
1
1
1
4
0
1
1
1
4
1
0
1
0
0
1
0
1
0
0
0
3
1
1
4
1
0
0
1
0
2
0
0
2   2
0
0
0
1
1
1
0
1
1
1
4
0
0
1
1
3
r2
r3

1
0
1
0
0
1
0
1
0
0
0
1
0
0
1
0
1
0
0
1
0
3
1
1
4 3  r 2
0
0
1
1
1  r 4
0
1
1
1
4
0
0
0
1
1
1
0
1
0
0
1
0
1
0
0
0
1
0
0
1
0
1
0
0
1
0
0
1
1
1
0
0
1
0
2
0
0
1
1
3
0
0
0
1
1
1
0
1
0
0
1
0
0
0
2
0
1
0
0
1
0
1
0
0
1
0
0
1
1
1
0
0
1
0
2
0
0
0
2
2
0
0
0
1
1
r2
r3
2
r3

x
2 x  4
2

 1  x  1
2
Ax  B
C
D
 2


x  1 x  1  x  12
2x 1
2
1
 2


2
x  1 x  1  x  1
We can do this problem on the TI-89:


2  x  4

expand  2
2
  x  1   x  1 


expand ((-2x+4)/((x^2+1)*(x-1)^2))
1 0 0 0 2
03 1 0 0 1
F2
0 0 1 0 2
2 x
1
2
1
Of course with the TI-89, we could



0 integrate
0 0and wouldn’t
1 1 need
just
x 2  1 x 2  1 x  1  x  12
partial fractions!
p
Strategy for Integration
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
from a constant factor.
in the integrand whose
differential also occurs, apart
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.
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
4
1 x
dx
1 x
The function is
undefined at x = 1 .
Since x = 1 is an asymptote, the
function has no maximum.
3
Can we find
the area under
an infinitely
high curve?
2
1
0
1
We could define this integral as:
lim 
b 1
b
0
1 x
dx
1 x
(left hand limit)
We must approach the limit from
inside the interval.

lim 
b 1
0
1 x
dx
1 x
1 x 1 x
dx
1 x 1 x

1+x


b
1 x
1
1  x2
2
dx
x
dx  
1  x2

1
2
1
sin x   u du
2
1
Rationalize the numerator.
dx
u  1  x2
du  2x dx
1
 du  x dx
2

1

1  x2
x
dx  
1  x2

dx
1
2
1
sin x   u du
2
1
sin 1 x  u
1
du  2x dx
1
 du  x dx
2
1
2
lim sin x  1  x
b 1
u  1  x2
2
b
This integral converges
because it approaches a
solution.
0
p
0
0
2
1
lim sin b  1  b 2  sin 1 0  1

b 1

 


p
2
1

Example 2:
4
3
dx
0 x
1
2
1
lim 
dx
b 0 b x
1
1
-1
0
1
-1
1
lim ln x b
(right hand limit)
b 0
We approach the limit from inside
the interval.
lim ln1  ln b
b 0
1
lim ln
b 0
b

This integral diverges.

4
Example 3:

0
The function
approaches 
when x 1 .
dx
3
 x  1
  x 1
3
0

2
3
2
3
3
2
1
dx
0
lim   x  1
b
b 1

2
3
0
lim 3  x  1
b1
dx  lim   x  1
3
c 1
1 b
3
0
c
 lim 3  x  1
c 1

1
2
3
2
3
dx
1 3
3
c

lim   x  1
b
b 1

2
3
0
lim 3  x  1
b1
dx  lim   x  1
3
c 1
1 b
3
c
 lim 3  x  1
0

c 1
2
3
dx
1 3
3
c
0
0
1
1
1
 13


lim 3  b  1 3  3  1 3   lim 3  2  3  c  1 3 
b 1 
 c1 

3  33 2

b P 1
1 P 1
lim

b   P  1
P  1
Example 4:


1
dx
P
x


1
P0
(P is a constant.)
x
P
What happens here?
 P 1
b
If P  1 then
gets bigger
and bigger as b   , therefore
dx
the integral diverges.
b
lim  x  P dx
b  1
b
1
lim
x  P 1
b   P  1
1
If P  1 then b has a negative
exponent and b  P 1  0 ,
therefore the integral converges.
p
Improper Integrals
<
Idea of the Comparison Theorem
Observe that the function to be integrated satisfies
3
3

(1  2sin2 (4p x ))(1  x 2 ) 1  x 2
for all x  0. The following graph illustrates this observation.
3
The blue curve is the graph of the function
while the red curve is the
2
1 x
graph of the function to be integrated.
0

The improper integral 0
3
dx
(1  2sin2 (4p x ))(1  x 2 )
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
To show that the area under the blue curve in the previous
figure is finite, compute as follows:


0
3p
b
b
3
3

lim
3
arctan(
x
)

.
dx  lim 
dx
0
2
2
b

0
b

2
1 x
1 x
3
0 1  x 2 dx converges.

3
Hence also the improper integral 
dx converges.
0 (1  2sin2 (4p x ))(1  x 2 )
This means that the improper integral

Comparison Theorem
Let a, b 
 , , a  b. Assume that the functions f and g satisfy
0  f( x )  g( x ) for all x, a  x  b. Assume also that the integral

b
a
f( x )dx
is improper.
1) If the improper integral
integral

b
a

a
a
g( x )dx converges, then also the improper
b
b
a
a
f( x )dx converges, and 0   f( x )dx   g( x )dx.
2) If the improper integral
b

b

b
a
f( x )dx diverges, then also the improper integral
g( x )dx diverges.
The integral

b
a
f( x )dx is improper if either a  , b   or the
function f has singularities in the interval of integration.
Examples



x
x 1
4
1
x
x4 1

dx 

x
x4


1
1
dx
x
DIVERGES
x 1
 2 
x
x
x
lim
x 
4
x( x)
x4 1
x2
x
 lim
 lim
 lim
1
4
4
x 
x4 1
x 1
x 
x 1
x 
x
Since x is
continuous
for x  1

lim
x 
x4
x4 1
 1 1

0
Examples


1
x 1
3
1
1
x3  1
dx 

1
lim
x 


1
x3
1
x
Since x is
continuous
for x  1
CONVERGES
3
P-test p   1
2
1
x3
x3  1
 lim
1
3
dx
x 

x3
x 1
3
lim
x 
 lim
x3
x3  1
x 
x3
x3  1
 1 1

0
Normal Distribution Function


e
 x2



e
 x2

1

e
1
dx
1
dx   e

 x2
dx
 x2
1
dx   e
Converges
1
 x2

dx  e
1
 x2
dx
Normal Distribution Function


1
e
 x2
dx
x  1   x  x
2
0e
 x2

0 e
e
 x2
1

 e
1
 x2
x
Therefore

dx   e
1
dx
Hence
x
dx  
converges
Normal Distribution Function


1
e
 x2
dx
x  1   x  x
2
0e
 x2

0 e
e
 x2
1

 e
1
 x2
Hence
x
Therefore

dx   e
1
dx
x
dx  
Converges
and so does


e

 x2
dx
Examples
1
0 sin x dx
1
It’s improper
because
1
sin x
It’s undefined for
x  0  sin x  x
and
0  x  1  sin x  0
hence
1
1
0 
x sin x
therefore


1 1
1
0   dx  
dx
0 x
0 sin x
1
DIVERGES
x0


1
Examples


x 1
 1
dx
  dx
4
1 x
x 1
3
DIVERGES
x 1 x
1
 4 
4
x 1 x
x
3
lim
x 
3
x3  1
3
4
4
x
(
x

1
)
x
x
x 1 
 lim 4
1
lim
4
1
x 1
x 
x  x  1
x

0


3

1
dx

x
xe
Examples


3

x
e dx
DIVERGES
1
1
x
 x  e
x
xe
e
L’Hop
lim
x 
L’Hop
1
x
x
x
x
e

e
e
xe 




1
lim
x
lim
lim
x
x
1
x

e
1

e
e
x 
x 
x 

0
x
e
Examples

0


1
2
cos x
 1
P-test p  2  1
dx

dx
0
2
2 x 2
2
x

 1

1
x

dx

e
dx
dx 2 x
0

x
2
3 xe
e


4

1  3 sin x
1
1
p

dx P-test 2  1
dx  1
x
x
x
 e

x
dx
0

e
dx
1
1
x
