ADV
integration by parts
.
CalC It
n =
Juar ur-Judu
=
dV =
now to find L I PE
.
T
=
n >
-
. . .
ultraviolet minus van
(xinxdX /180cODO
[xsin(2x)dX ↓
n 1nX du Y/X
9180 (CS(O)do
= =
u du dX
·
XdX
=
X
Ex dV
=
V = =
dv =
sin(2x) v=
-
ECOS(2X)dx 180 du 18
U= =
(inx -x] (2x2 * Ecos(2x) 3-zos(2x)dx DV cs20 v co+8
= =
X
- -
-
↓
X . -
Ecos(2X) tysin(2x) + C
+
-
180 ·
Co + E
-
9 -
10+0 -
18
((nx =]
↓
-
-
* + -
188 .
co+ + 18/n/sinx)
(10x(n(3x)dx
+ I
24x
flixe" (n(3x)du =
u=
dx
dV = 10x v =
5x
u=
du = ex
/7X du = 17
v= eXdx 5x2 (n(3x)
.
-
((5x)dX
17x .
ex _ ]eX17dx 5x2 In (3x) -
EX2 + C
↓
C
17x ex .
-
17ex +
tabular method
[xeX(XSI
↓ E
en
t
xex-2xe
*
+ zex +
(sinvac
+ Exxxx
,1216 2024 28326164
worksheet answers
Y
⑫ I b ⑯
(x" inx u= InX dv =
x4
S
du =
1/Xv =
5X
↓
jix ex .
uX du 5Xinx-95x*
..-x) fe-*
d
14x n
I
.
↓
9 -ye
-
↓
nex
(5X(nx) x - + b
-*
-
4xe -yex + d
↓
⑳
-
-X(4X + 4) + c
In du X-
2
⑭
=
↓ v=
*
-
gax-finzx x -
u= In2x (Inx .
-t)-f-**
an =
(n2x -*)
.
-
Y t+-
dV X =
- 2
↓
v=
- 9 :f *
-
-
(In2x -- *
.
E
d
↓
1nx .
-y ( +
+ D
-
(((n2x 1) + c
⑳
-
+
Ixsinx
u = X dv sinx
dx
=
au = V = - CoSX
⑫
Let O be X
Josecftano do I + Xsex tanx
-
-
xcosx-
C- cost
uX desex an
is
d
sec
-
XcOSX + Sinx + C
(xse(X)-9sexdx
↳ XsecX-In/sex + tanxl + b
,⑪ ⑭ (xcos2xx
[xe
>4 =
⑳
**
(
(e -1x . Y
*
He S D I
&
+
+
C
# [sin2x) +(3x2.cos2x) (6x
-
.
&Sinix)
-
(6 +20s2x)
.
+
C
, Adv . Cak It
partial fraction decomposition
1. Factor denominator
the denominator into
Iz
#writex
>
-
(x 2)(X+ 1)
-
linear/ quadratic
factors
-5) Label constants
>
-
.
2
2
. choose constants for
>+ algebra
- .
3 use to cancel
the numerator
5
(x(x = +is
↓
(x)(x)
to start
.
3 Multiply/cancel
solving for constants >
- +1
& .
4 Evaluate & Zeros
. New
4 equation Z
7 23
-
= A(X 5) + B(X-2) to cancel a constant
B(X 2)
-
1 A(x+ 1) +
-
=
d
- . Selection Method
5 cancels
7x 23 = A(2 5) + B(2 -
-
2)7X 23
- =
A(5 5) + B15 2)
- -
~
(()
A 3
(e + x =
-
=
A(1+1) + B( 2) -
B 4
=
=
.
5 Plug in values for 1 =
-
3BB =
-
t
- 1
&
Let X = 2
constants ! +
B(z 2)
4X -
2x
-
5 1 =
A(2 + 1) + -
1 = 3A A =
5
* Ill terms of integration , just integrate
each part after the decomposition !
.
6
Integration Equation
2(3)( + 11
*X =
-
&x -
↓
↓ ↓ ↓
X 33
-
: A(X)(X+ 3) + B(x + 3) + ((x2) & In (x 21 -
-
-(n(x + 11 + c
let x =
-
3 let x = 0 le + x =
1
36 9) 3B 4A
& =X
33 16
-
= =
- =
C = -
4 B =
-
11 A= 4 x
1 ax +
( tax f-
-
+ x
* + 7x + 5 = A(x2 + 2x +5) + (BX + C)(X)
In
4 (x1 + 11/X -
4(n(x + 31 + c le+ x = 0 ↓ expand to
evaluate identity
5 =
5A
=
(A B)(Xx)
+ +
(2A + c)x + 5A
A= 1
A +B = 1 B= 2
2A +2 7
5
=
c=
(ax +
[x +5
(x + 13 + 22
an" (1) +,
(n(x1 +
+ +
integration by parts
.
CalC It
n =
Juar ur-Judu
=
dV =
now to find L I PE
.
T
=
n >
-
. . .
ultraviolet minus van
(xinxdX /180cODO
[xsin(2x)dX ↓
n 1nX du Y/X
9180 (CS(O)do
= =
u du dX
·
XdX
=
X
Ex dV
=
V = =
dv =
sin(2x) v=
-
ECOS(2X)dx 180 du 18
U= =
(inx -x] (2x2 * Ecos(2x) 3-zos(2x)dx DV cs20 v co+8
= =
X
- -
-
↓
X . -
Ecos(2X) tysin(2x) + C
+
-
180 ·
Co + E
-
9 -
10+0 -
18
((nx =]
↓
-
-
* + -
188 .
co+ + 18/n/sinx)
(10x(n(3x)dx
+ I
24x
flixe" (n(3x)du =
u=
dx
dV = 10x v =
5x
u=
du = ex
/7X du = 17
v= eXdx 5x2 (n(3x)
.
-
((5x)dX
17x .
ex _ ]eX17dx 5x2 In (3x) -
EX2 + C
↓
C
17x ex .
-
17ex +
tabular method
[xeX(XSI
↓ E
en
t
xex-2xe
*
+ zex +
(sinvac
+ Exxxx
,1216 2024 28326164
worksheet answers
Y
⑫ I b ⑯
(x" inx u= InX dv =
x4
S
du =
1/Xv =
5X
↓
jix ex .
uX du 5Xinx-95x*
..-x) fe-*
d
14x n
I
.
↓
9 -ye
-
↓
nex
(5X(nx) x - + b
-*
-
4xe -yex + d
↓
⑳
-
-X(4X + 4) + c
In du X-
2
⑭
=
↓ v=
*
-
gax-finzx x -
u= In2x (Inx .
-t)-f-**
an =
(n2x -*)
.
-
Y t+-
dV X =
- 2
↓
v=
- 9 :f *
-
-
(In2x -- *
.
E
d
↓
1nx .
-y ( +
+ D
-
(((n2x 1) + c
⑳
-
+
Ixsinx
u = X dv sinx
dx
=
au = V = - CoSX
⑫
Let O be X
Josecftano do I + Xsex tanx
-
-
xcosx-
C- cost
uX desex an
is
d
sec
-
XcOSX + Sinx + C
(xse(X)-9sexdx
↳ XsecX-In/sex + tanxl + b
,⑪ ⑭ (xcos2xx
[xe
>4 =
⑳
**
(
(e -1x . Y
*
He S D I
&
+
+
C
# [sin2x) +(3x2.cos2x) (6x
-
.
&Sinix)
-
(6 +20s2x)
.
+
C
, Adv . Cak It
partial fraction decomposition
1. Factor denominator
the denominator into
Iz
#writex
>
-
(x 2)(X+ 1)
-
linear/ quadratic
factors
-5) Label constants
>
-
.
2
2
. choose constants for
>+ algebra
- .
3 use to cancel
the numerator
5
(x(x = +is
↓
(x)(x)
to start
.
3 Multiply/cancel
solving for constants >
- +1
& .
4 Evaluate & Zeros
. New
4 equation Z
7 23
-
= A(X 5) + B(X-2) to cancel a constant
B(X 2)
-
1 A(x+ 1) +
-
=
d
- . Selection Method
5 cancels
7x 23 = A(2 5) + B(2 -
-
2)7X 23
- =
A(5 5) + B15 2)
- -
~
(()
A 3
(e + x =
-
=
A(1+1) + B( 2) -
B 4
=
=
.
5 Plug in values for 1 =
-
3BB =
-
t
- 1
&
Let X = 2
constants ! +
B(z 2)
4X -
2x
-
5 1 =
A(2 + 1) + -
1 = 3A A =
5
* Ill terms of integration , just integrate
each part after the decomposition !
.
6
Integration Equation
2(3)( + 11
*X =
-
&x -
↓
↓ ↓ ↓
X 33
-
: A(X)(X+ 3) + B(x + 3) + ((x2) & In (x 21 -
-
-(n(x + 11 + c
let x =
-
3 let x = 0 le + x =
1
36 9) 3B 4A
& =X
33 16
-
= =
- =
C = -
4 B =
-
11 A= 4 x
1 ax +
( tax f-
-
+ x
* + 7x + 5 = A(x2 + 2x +5) + (BX + C)(X)
In
4 (x1 + 11/X -
4(n(x + 31 + c le+ x = 0 ↓ expand to
evaluate identity
5 =
5A
=
(A B)(Xx)
+ +
(2A + c)x + 5A
A= 1
A +B = 1 B= 2
2A +2 7
5
=
c=
(ax +
[x +5
(x + 13 + 22
an" (1) +,
(n(x1 +
+ +
