, p
·
LESSON 1 .
INTEGRATION
*
infinite number of
x
> of of
thought
-
can e7 a sum an
-
sections infinite small
=
objects or that are i
* Antiderivative w/limits
>
-
approximately equal to the reverse of derivative
8
S'(x3 + x)dx
=
* TWO KINDS
. Indefinite
Integral genform of integration the antiderivate
(i =) ( E
-
+ -
+
of a function s written .
=
S -
Integral Sign
= -I
(f(x)dx =
F(x) +
CFx-Integrad sfunction I .
Jsin*y cosydy
·variableof
DefiniteIntegrarepresent anumberpirtainto
one of the curves.
t
I
re
Integration
e
undler
+
a
C
Limits of Integration
-
The endpoint values for definite Integral .
I
in +
b
eX .
( xndX
5 .
(sin7xcos7xdx
&
⑮ V =
Sin7x ③ v = cos7x
a
=
lower limit b =
upper limit du =
((0S7x)7dx du = -
Sin7x .
7dX
Antidifferentiation Examples =
Sx7x osExx E)7x
:
-
:
Sax
=
1
= Sudv
.
=
Sxdx
= +
= E(z) + c
=
X + E
= (sin 27x) + C
.
2 ((x2 + x + 3)dX
=
(x
=
dx + (xdx + (3dx
=
x + + 3x + c
, )
I
Sxix
6 .
.
8
JsinExcosxdx
I
=
=
V =
SinX
du =
cosXX
Suf du Su"du
=
-3
=
v
=
+a
-
3
In x
=
+
=x -
sin3x +E
J sin" (e** )cos(e2)e2dX
inx < a
= +
sin(e x)
=
u=
du cos(e )(e2)(2)dX
**
sint/
=
.
7
edx (2)
#J sin" ecos ex .
~ du
*
-)(7 cost)
= + )-sintat) = Su"du
v =
7 + cost
-sintat
= (E) + c
du =
=
to sine + h
-Su du
=
J excx
-
10 .
-y "z
)
=
5)(1
+2
e dx( 5)
*
5e
=
-
-
= u 1 52x
=
+ C
-
du = -
5edX
11-5
2
-
7cost =
-
z)1 52x) + 2
-
=
, p
I
ANTIDIFFENTATION EXAMPLE
4 .
Stanxdx
logarithms =
J =
Inu1C
-J-snx
1) 4
V
du
=
=
COSX
sinxdX
-J
= Fy -
Inv + C
IncosX + C
Ed
-
=
8r
-
ninx Inx-n
=
3(n(7y + 4) + C =
+
In (20sX)
In =
cosX
.
2 (2x =
Insex + [
x
)ax
u
= -
5x + 3
6 .
du =
(2x 5)dX-
u = eX + 1
=
(n(x2 -
5x + 3) + C du = edX
+
(1 2x
) + )1 x+
+
3 .
x =
=
) dx +
)4dx (4ax +
-
RIGONOMETRIC IDENTITIES
= (ax + 4)ax +
4)xdx lant =
sinf
cost
sect cost
4
=
=
(nx + 4x + + C
csat = isn't
=
(nx + 4x + 2x + C
.
7 Join 20
)xdx if asa-multiply by
*
5 .
as + cot
x2 + x +1
2 s + cot
X -
1X3 if
-
(x3 xz) -
* sea
by
X2
sect tan
(x2
Y
- -
sect tan
1)
(X
Jasc28dt (2sc2 + co+
20)
- -
=
I
.
(2528 + cot 27
((x2 + x 1)dx (sc2t co+ 2t)dt( 2)
=)(2s
x +1 + -
28 + -
=
[S2* + cot 20
(x dx (xdx (dx )x + + let v csc28 + not 20
+
=
=
du =
-
csc2cot2 ·
It)-cs2 2) do .
du =
-
2 (asc2cot 2 + cs20) &f
=
X + x + X + (n(x 1)-
+ 2 =
In (2s2 + cot 2) + C
·
LESSON 1 .
INTEGRATION
*
infinite number of
x
> of of
thought
-
can e7 a sum an
-
sections infinite small
=
objects or that are i
* Antiderivative w/limits
>
-
approximately equal to the reverse of derivative
8
S'(x3 + x)dx
=
* TWO KINDS
. Indefinite
Integral genform of integration the antiderivate
(i =) ( E
-
+ -
+
of a function s written .
=
S -
Integral Sign
= -I
(f(x)dx =
F(x) +
CFx-Integrad sfunction I .
Jsin*y cosydy
·variableof
DefiniteIntegrarepresent anumberpirtainto
one of the curves.
t
I
re
Integration
e
undler
+
a
C
Limits of Integration
-
The endpoint values for definite Integral .
I
in +
b
eX .
( xndX
5 .
(sin7xcos7xdx
&
⑮ V =
Sin7x ③ v = cos7x
a
=
lower limit b =
upper limit du =
((0S7x)7dx du = -
Sin7x .
7dX
Antidifferentiation Examples =
Sx7x osExx E)7x
:
-
:
Sax
=
1
= Sudv
.
=
Sxdx
= +
= E(z) + c
=
X + E
= (sin 27x) + C
.
2 ((x2 + x + 3)dX
=
(x
=
dx + (xdx + (3dx
=
x + + 3x + c
, )
I
Sxix
6 .
.
8
JsinExcosxdx
I
=
=
V =
SinX
du =
cosXX
Suf du Su"du
=
-3
=
v
=
+a
-
3
In x
=
+
=x -
sin3x +E
J sin" (e** )cos(e2)e2dX
inx < a
= +
sin(e x)
=
u=
du cos(e )(e2)(2)dX
**
sint/
=
.
7
edx (2)
#J sin" ecos ex .
~ du
*
-)(7 cost)
= + )-sintat) = Su"du
v =
7 + cost
-sintat
= (E) + c
du =
=
to sine + h
-Su du
=
J excx
-
10 .
-y "z
)
=
5)(1
+2
e dx( 5)
*
5e
=
-
-
= u 1 52x
=
+ C
-
du = -
5edX
11-5
2
-
7cost =
-
z)1 52x) + 2
-
=
, p
I
ANTIDIFFENTATION EXAMPLE
4 .
Stanxdx
logarithms =
J =
Inu1C
-J-snx
1) 4
V
du
=
=
COSX
sinxdX
-J
= Fy -
Inv + C
IncosX + C
Ed
-
=
8r
-
ninx Inx-n
=
3(n(7y + 4) + C =
+
In (20sX)
In =
cosX
.
2 (2x =
Insex + [
x
)ax
u
= -
5x + 3
6 .
du =
(2x 5)dX-
u = eX + 1
=
(n(x2 -
5x + 3) + C du = edX
+
(1 2x
) + )1 x+
+
3 .
x =
=
) dx +
)4dx (4ax +
-
RIGONOMETRIC IDENTITIES
= (ax + 4)ax +
4)xdx lant =
sinf
cost
sect cost
4
=
=
(nx + 4x + + C
csat = isn't
=
(nx + 4x + 2x + C
.
7 Join 20
)xdx if asa-multiply by
*
5 .
as + cot
x2 + x +1
2 s + cot
X -
1X3 if
-
(x3 xz) -
* sea
by
X2
sect tan
(x2
Y
- -
sect tan
1)
(X
Jasc28dt (2sc2 + co+
20)
- -
=
I
.
(2528 + cot 27
((x2 + x 1)dx (sc2t co+ 2t)dt( 2)
=)(2s
x +1 + -
28 + -
=
[S2* + cot 20
(x dx (xdx (dx )x + + let v csc28 + not 20
+
=
=
du =
-
csc2cot2 ·
It)-cs2 2) do .
du =
-
2 (asc2cot 2 + cs20) &f
=
X + x + X + (n(x 1)-
+ 2 =
In (2s2 + cot 2) + C
