8.1 Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of...
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Transcript of 8.1 Integration by parts Product Rule:. Integration by parts Let dv be the most complicated part of...
8.1 Integration by parts
udv uv vdu
d dv duuv u v
dx dx dx Product Rule:
d dv duuv dx u dx v dx
dx dx dx
uv udv vdu
Integration by parts
udv uv vdu Let dv be the most complicated part of theoriginal integrand that fits a basic integrationRule (including dx). Then u will be the remaining factors.
Let u be a portion of the integrand whose derivative is a function simpler than u. Thendv will be the remaining factors (including dx).
OR
Integration by parts
udv uv vdu xxe dx u = x dv= exdx
du = dx v = ex
x x xxe dx xe e dx x x xxe dx xe e C
Integration by parts
udv uv vdu 2 lnx xdx u = lnx dv= x2dx
du = 1/x dx v = x3 /3
3 3 3 22 1
ln ln ln3 3 3 3
x x x xx xdx x dx x dx
x
3 32 ln ln
3 9
x xx xdx x C
Integration by parts
udv uv vdu arcsin xdx u = arcsin x dv= dx
2arcsin arcsin
1
xxdx x x dx
x
v = x2
1
1du dx
x
2arcsin arcsin 1xdx x x x C
Integration by partsudv uv vdu
2 sinx xdx u = x2 dv = sin x dx
2 2sin cos 2 cosx xdx x x xdx du = 2x dx v = -cos x
u = 2x dv = cos x dx
du = 2dx v = sin x2 2sin cos 2 sin 2sinx xdx x x x x xdx
2 2sin cos 2 sin 2cosx xdx x x x x x C
8.2 Trigonometric Integrals Powers of Sine and Cosine
sin cosn mu udusin cos cos sinn nu udu u udu
2 2sin 1 cosu u 1. If n is odd, leave one sin u factor and use
for all other factors of sin.
2 2cos 1 sinu u 2. If m is odd, leave one cos u factor and use
for all other factors of cos.
2 21sin (1 cos
1cos (1 cos2 ) 2 )
22oru u uu
3. If neither power is odd, use power reducing formulas:
Powers of sin and cos
2 2sin ( )cos ( )d
3sin (2 )d
2 3sin ( )cos ( )d
Powers of sin and cos3 2 2sin (2 ) sin 2 sin 2 (1 cos 2 )sin 2d d d
2 3 2 2 2 2sin cos sin cos cos sin (1 sin )cosd d d
2 3(sin 2 cos 2 sin 21 1
cos 2 cos2 6
) 2 Cd
2 3 2 3(sin sin )cos (sin cos sin cos )d d 3 41 1
sin sin3 4
C
Powers of sin and cos
2 2 1 1sin ( )cos ( ) (1 cos 2 ) (1 cos 2 )
2 2d d
21 1 1(1 cos 2 ) (1 (1 cos 4 )
4 4 21 1 1 1
(1 cos 4 ) ( 4 )4 2 4 2
d d
d cos d
1 1sin 4
8 16C
Tangents and secants
sec tann mu udu
2tan sec sec sec tann nu udu u u udu
Create an integral that is shown above.2 2 2 2tan sec 1 tan 1 secor
4 4sec tan d
8.3 Eliminating radicals by trig substitution.
2 2a u 2 2a u 2 2u aPythagorean identities:
2 2cos 1 sin
2 2tan sec 1
2 2sec 1 tan
2 2a u Let u = a sin θ
2 2 2 2 2 2sin 1 sin cosa u a a a a
Trig Substitutions2 2a u2 2a u 2 2u a
2 29
dx
x x Let x = a sin θ = 3 sin θ
dx = 3 cos θ dθ
22 2 2 2
3cos 3cos
9sin 3cos9 9sin 9 9sin
dx d d
x x
22
1 1 1csc cot
9 9 9sin
dd C
Ex:
29 x
x3
21 9
9
xC
x
24 1
dx
x Let u=2x, a=1 so 2x = tan θ
dx = ½ sec2 θ dθ2
2
1 sec 1sec
2 sec 24 1
dx dd
x
1
ln sec tan2
C
Ex:
24 1x 2x
1
21ln 4 1 2
2x x C
8.4 Partial Fractions2 5 6
dx
x x 2
1
3 25 6
A B
x xx x
2
1 ( 2) ( 3)
3 25 6
A x B x
x xx x
1 ( 2) ( 3)A x B x
If x = 2: 1=-B so B = -1If x =3: 1=A
2
1 1
3 25 6
dxdx
x xx x
3ln | 3 | ln | 2 | ln
2
xx x C or C
x
Partial Fractions-Repeated linear factors
2
3 2 2
5 20 6 6 1 9
12 ( 1)
x xdx dx
x xx x x x
2 25 20 6 ( 1) ( 1)x x A x Bx x Cx If x =0: 6= A
If x = 1: 31=6(4)+2B+9, B = - 1
2
3 2 2
5 20 6
12 1
x x A B C
x xx x x x
If x = -1: -9 = -C, so C = 9
16ln | | ln | 1| 9( 1)x x x C
6 9ln
1 1
xC
x x
2
3 2
5 20 6
2
x xdx
x x x
Quadratic Factors
3
2
2 4 8
( 1)( 4)
x xdx
x x x
3
2 2
2 4 8
1( 1)( 4) 4
x x A B Cx D
x xx x x x
3 2 22 4 8 ( 1)( 4) ( 4) ( ) ( 1)x x A x x Bx x Cx D x x
If x = 0 then A = 2If x = 1 then B = -2
If x = -1 2 = -C +DIf x = 2 8 = 2C+D
Solving the system of equations you findC = 2 and D = 4.
3
2 2 2
2 4 8 2 2 2 4
1( 1)( 4) 4 4
x x xdx dx
x xx x x x x
22ln | | 2ln | 1| ln( 4) 2arctan2
xx x x C
3
2 2
8 13
( 2)
x xdx
x
Repeated quadratic Factors3
2 2 2 2 2
8 13
( 2) 2 ( 2)
x x Ax B Cx D
x x x
3 28 13 ( )( 2)x x Ax B x Cx D 3 3 28 13 2 2x x Ax Ax Bx B Cx D
3 2 3 200 13 2 28 Bx x x x x AxA Cx D B
A=8
13=2A+C
For third degree: For second degree: B=0
For first degree:
For constant: D+2B=0
3
2 2
8 13
( 2)
x xdx
x
Repeated quadratic Factors
3
2 2 2 2 2
8 13 8 3
( 2) 2 ( 2)
x x x xdx dx
x x x
32
2 2 2
8 13 34ln( 2)
( 2) 2( 2)
x xdx x C
x x
A=8
13=2A+C
B=0
D+2B=0So, D=0 and C = -3
8.8 Areas under curves with infinite domain or range
21
ln xdx
x
1
0
1dx
x
Improper Integrals with infinite limits
Upper limit infinite
Lower limit infinite
Both limit infinite
2 21 1
ln lnlim
b
bx xdx
x x
Infinite limits
2 21 1
ln lnlim
b
bx xdx dx
x x
2
ln
1
dxu x du
xdx
dv vxx
2 2
ln ln ln 1x x dx xdx
x x xx x
We say the improper integral CONVERGES toThe value of 1. (The area is finite.)
Evaluation
Use L’Hôpital’s rule
211
ln ln 1lim lim
b b
b bx xdx
x xx
ln 1
lim (0 1) 0 0 0 1bb
b b
When both limits are infinite
20
2
0
2
2 00
2 2
2
0 00
0
1
lim lim arctan 02
1
lim lim arctan 02 21
2
1
2
2
1
1 1
a a a
bb
a
b
b
dx
x
dx
x
dx dxx
x
dx dxx
x x
x
x
dx
Improper Integrals-integrand becomes infinite
upper endpoint
lowerendpoint
interior point
1 1
00
10
0
1 1lim
lim 2
lim 2 2 2
aa
a a
a
dx dxx x
x
a
Integrals with Infinite discontinuities.
The integral converges to 2.
1
20 3
1 131 12 2 0
0 03 3
131
3
32 2
3
0 3 11
lim lim 3 1
1 1
lim 3( )
1
1 3 3
1
bb
b b
b
dx
x
dx dxx
x x
b
d d
x x
x x
33 3 131 12 2
1 3 3
133
1
3
lim lim 3 1
1 1
1lim 3(3 1) 3(
3
1) 3 23
3 2The total integral is
c cc c
c
dx dxx
x x
c
Calculation with infinite discontinuity
Area is finite
Integral converges to 1
21
1t
dxx
Area is infinite
Integral diverges
1
1t
dxx
21
)dx
xa
1
)bdx
x
11 2
)cdx
x
Integrals of the form 1
p
dx
x
211
1 1) lim lim lim ( 1) 1
b b
b b bdx
ax bx
11
) lim lim ln lim ln ln1b
bb b b
dxb x b
x 1 12 2
11 2 1
) lim lim lim (1)
bb
b b bdx
c x b
x
Convergence or divergence
Integrals of the form 1
p
dx
x
Converge if p > 1 and diverge if p = 1 or p < 1.
31 2
dx
x
21 3
dx
x
1
3dx
x
51
4dx
x
Which of the following converge and which diverge?
Direct comparison testIf f and g are continuous functions with f(x) g(x)For all x a. Then…..
( )a
f x dx
Converges if ( )a
g x dx
Converges
( )a
g x dx
Diverges if ( )a
f x dx
Diverges
A function converges if its values are smaller than another function known to converge.
A function diverges if its values are larger than another function known to diverge.
Limit Comparison test for convergence
If f and g are positive and continuous on [a, )And if ( )
lim , 0( )x
f xL L
g x
Then the integrals ( )a
f x dx
( )a
g x dx
and
If
If
both converge or both diverge:
( )a
g x dx
diverges and
then
( )lim
( )xf x
g x
( )a
f x dx
also diverges.