Chapter 9 Calculus Techniques for the Elementary Functions.
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Transcript of Chapter 9 Calculus Techniques for the Elementary Functions.
9.2 Integration by parts – A way to integrate Products
Integration by parts is a rule that transforms the integral of products of functions into other, hopefully simpler, integrals.
xdxdxexdxe xx coscos
.sderivativeforruleproducttheatlookmustWe
Integration by parts – A way to integrate Products
udvvdudy
vdudyudv
vduyudv
If where u and v are differentiable functions of x then:
uvy
vduuvudv Formula for integration by parts.
dvuvdudy
Integration by parts – A way to integrate Products
xdxxxxx sinsincos
xvdxdu sin1 xdvxu cos
Suppose we want to find the volume of the solid formed by rotating the region under around the y-axis.
From aside:
xy cos
xdxxV
xdxxdV
ydxxdV
cos2
cos2
2
2
0cossin2
xxxV
Aside
Let
vduuvudv
xxxxx cossincos
Integration by parts – A way to integrate Products
Examples:
1.
2.
xdxx sin2
Let
vduuvudv2xu
xdxdu 2xdxdv sinxv cos xdxxxxxdxx 2coscossin 22
dxxex
9.3 Rapid Repeated Integration by Parts
Example:
xdxx 2cos5
U dV
0
120
120
60
20
5
2
3
4
5
x
x
x
x
x
x
x
x
x
x
x
x
2cos64
1
2sin32
1
2cos16
1
2sin8
1
2cos4
1
2sin2
1
2cos
vduuvudv
+
-
+
-
+
-
+
Reason for the sign change.
vduuvuvuv
vduuvuv
vdu
dvu mult
integrate
U dV
0
120
120
60
20
5
2
3
4
5
x
x
x
x
x
x
x
x
x
x
x
x
2cos64
1
2sin32
1
2cos16
1
2sin8
1
2cos4
1
2sin2
1
2cos
+
-
+
-
+
-
+
Cxxxxxxxxxxxxdxx 2cos8
152sin
4
152cos
4
152sin
4
102cos
4
52sin
2
12cos 23455
Using Trig properties to make original reappear.
dxx 4.0sin2
xx
xx
4.0cos4.0
14.0cos4.0
4.0sin4.0sin
+
-
xxxdxx 4.0cos4.0sin4.0cos4.0
14.0sin 22
xxxdxx 4.0sin14.0sin4.0cos4.0
14.0sin 22
xdxdxxxdxx 4.0sin14.0sin4.0cos4.0
14.0sin 22
xxxdxx 4.0sin4.0cos4.0
14.0sin2 2
Cxxxdxx 5.04.0sin4.0cos8.0
14.0sin2
Reassociation between Steps
dxex x35
3
3
3
13 2
23
x
x
ex
exx
3
3
9
10
3
13 2
x
x
e
ex
reassociate
+
-
+
Integrating with substitution method
Ceexdxex xxx 333
3
1
3
1 35
9.4 Reduction Formulas Objective is to find a generic formula for
xdxncos
xdx6cosxxx
xx
sinsincos5
coscos4
5
xdxxxx 245 sincos5sincos
dxxxxx 245 cos1cos5sincos
xdxxdxxx 645 cos5cos5sincos
xdxxx 45 cos6
5sincos
6
1
xdxn
nxx
nxdx nnn 21 cos
1sincos
1cos
9 – 6 Integration by Trig Substitution9 – 6 Integration by Trig Substitution
249 x
7
Reminder:
1.
xxxxx
xxxxx
222
2
sincos2cos2cos12
1sin
cossin22sin2cos12
1cos
Since this reminds us of Pythagoras, let’s label a triangle and attempt to change into integration of trig.
Key to success is deciding whether the variable x is the leg or the hypotenuse. If we choose x as a leg we can choose which leg. Vertical or horizontal will affect which trig function we will use later on.
dxx249
x
7
49cos
7sin
2xx
d
ddx
xx
cos7cos7
cos7
49cos7sin7 2
dd 2cos12
149cos49 2
C
d
2sin2
1
2
49
2cos12
49
C cossin24
49
2
49
Cxxx
7
49
72
49
7sin
2
49 21
9 – 6 Integration by Trig Substitution9 – 6 Integration by Trig Substitution
2.When changing from must also change the bounds.
dtodx
2
024 x
dx
9 – 7 Integration of Rational Functions by Partial 9 – 7 Integration of Rational Functions by Partial FractionsFractions
101
x
B
x
A
101
1510
xx
x
10
1
1
1
101
15101
x
xB
x
xA
xx
xx
linear
ntconsta
linear
ntconsta
linear
ntconsta
Polynomial
Polynomial ...
1011
15102
xx
x
10
9115
1
925
1011
15102
xxxx
x
We are wanting to ‘undo’ the common denominator
Let x=1 so that we get rid of B and just have A left
10
1
10
1510
x
xBA
x
x
A
101
15110
10
10
1
10
101
151010
x
xB
x
xA
xx
xx
B
1
10
1
1510
x
xA
x
x
B
110
151010
SHORTCUT!!
101
1510
xx
x
101
x
B
x
A
Simply place finger over (x-1) when letting x=1 and solve to get A. Then place finger over (x-10) when solving for B.
Integration of Rational Functions by Partial Fractions
Examples:
1.
This is an example of a rational function in proper form. That is lower degree than .
Improper form
2.
3.
dxxxx
x
6
6223
xqxp xp
xq xqofegreedxpofegreed
dx
x
xxx
1
1116 23
dxxxx
x
11
14
With improper form we need to perform long division first. Partial fractions will work only with proper form.
Integration of Rational Functions by Partial Fractions
B
BAxx
x
Bx
x
Ax
x
xx
6
193
1
1
1
1
1
9312
22
2
2
A
xAx
xAx
Axx
3
113
133
6193
One problem that we may encounter is that when we factor the denominator, we find some factors repeated, that is occurring with multiplicities greater than 1.
Repeated linear factors
4. Trick to partial fractions:
dxx
x21
93
22 111
93
x
B
x
A
x
x
Cx
x
xxdx
x
x
1
61ln3
1
6
1
3
1
9322
Integration of Rational Functions by Partial Fractions
214 22 xCxBAxx
CCxBBxAxAxx 24 222
CBxBAxCAx 24 22
Unfactorable Quadratics:5. Trick:
dxxx
x
12
42
2 1212
422
2
x
C
x
BAx
xx
x
CB
BA
CA
24
0
1
1
1
2
22
12
422
2
xx
xdx
xx
x
dxx
dxx
dxx
x
1
1
2
2
2
222
9 – 8 Integrals of the Inverse Trig Functions9 – 8 Integrals of the Inverse Trig Functions
“Like climbing mount Everest, they are interesting more from the standpoint that they can be done, rather than because they are of great practical use.”
In 4 – 5 we found derivatives of inverse trig by implicit differentiation.
1
1csc
1
1sec
1
1cot
1
1tan
1
1cos
1
1sin
2
1
2
1
21
21
2
1
2
1
xxx
dx
d
xxx
dx
dx
xdx
d
xx
dx
dx
xdx
d
xx
dx
d
9 – 8 Integrals of the Inverse Trig Functions9 – 8 Integrals of the Inverse Trig Functions
Integrate using parts.
dx
x
xxxx
211
1cotcot
xx
x
2
1
1
1
1cot
x1cot+
-
Cxxxx 211 1ln2
1cotcot
Aside: our calc and the back of the book use the notion sgnx
Treat this as a constant even though we do not know the sign.
x
xx sgn
HW 1 – 9 odd
9 – 10 Improper Integrals
Improper Integrals:•Upper or lower limit of integration is infinite.•Integrand is discontinuous for at least one value of x at or between the limits of integration.
An improper integral converges to a certain number if each applicable limit shown below is finite. Otherwise, the integral diverges.
fdxxfdxxfdxxf
dxxfdxxf
dxxfdxxf
k
ack
b
kck
b
a
b
aa
b
a
b
ab
,)(lim)(lim)(
)(lim)(
)(lim)(
Is discontinuous at x=c in ba,
Improper Integrals
** If the integrand seems to approach zero as x gets very large or small, then the integral might converge. Conversely, if the integrand grows without bound as grows without bound, then the integral definitely diverges.
Examples:
1. For the improper integral
a. Graph the integrand, and tell whether or not the integral might converge.
b. If the integral might converge,find out whether or not it really does,and if so, to what limit it converges.
x
,11
07.0 dx
x
Graph looks to be approaching 0 and therefore might converge.
17.0
0
1
07.0
lim1
aa
dxxdxx
Improper Integrals
2. For the improper integral
a. Graph the integrand, and tell whether or not the integral might converge.
b. If the integral might converge, find out whether or not it really does, and if so, to what limit it converges.
3. For the improper integral
a. Graph the integrand, and tell whether or not the integral might
converge.
b. If the integral might converge, find out whether or not it really does, and if so, to what limit it converges.
,13
32
dxx
,3
4
x
dx
9.11 Miscellaneous Integrals and DerivativesDifferentiation Sum Product Quotient
Composite Implicit Power Function Exponential Function Logarithmic function
'vu '' vu 'vu
2
''
v
uvvu '
v
u
'' uvvu
'uf '' uuf
xgyf xgyyf '''
'nx 1 nnx
'xn nn x ln
'log xbbx ln
1
Differentiation Con’t
Logarithmic
Trigonometric Function
Inverse Trig Function – differentiate Implicitly
xfy
xfy lnln 'ln'1
xfyy
'ln' xfyy
xx
xx
xx
2sectan'
sincos'
cossin'
Integration Sum
Product
Reciprocal Function
Power Function
Power of a Function
dxvu vdxudx
udv vduuv
duu 1 Cu ln
duun 11
1 1
nCun
n
dxxf n formulareduction