Maths notes
Semester 2
, me
Integrals
imme
Review of Integration
An ANTIDERIVATIVE of f is a function F
>
such that -
dx
(F(x)] =
f(x)
The INDEFINITE INTEGRAL of f is the infinite
family of antiderivatives F(x) + C
(f(x)dx =
f(x) + C
The DEFINITE INTEGRAL of f from a to ↳
is the
signed bounded
by c= x b
=
area a
;
y
=
f(x) ; y
=
0
1
I'll a
Area-lim f(x, &x
n+ 0k = 1
This evaluated the FTC
can be
using
Area =
"f(x) dx
=
F(b) -
F(c)
E
, imme
Review of Integration
EX
5x
.
fxdx = + C
jxdx [5 1·
3
=
x + C
I =
5(2) + -
(5(k Fe)
=
E
meet
Integration by Substitution
Ex .
[20c . cos(s) dx LET x =
u
=
a
Scos(u) -
↳ CHAIN RULE
=
Scos(u) du
=
sin(u) + C
:
sin(x) + C
* CHAIN RULE :
Scos(u) dxc =
/ [sin (2)] o
e
=> (sin(u)] (u
dx
=
sin
, meet
Integration by Substitution
I 2
EX -
- x
S >c2 dx let u
= -
x
au=
O
value of u
- A change
S -
-C ·
"oc du
s
-
dx
=
Is
- du
- -Je
-
I
-F(c I
U
au
=
-
O
O
-
=
H -
el
-
=
E(t -
1)
Ex .
Soc Vect2 do let V =
x + 2
A - I
It
Sx
= an dx
du =
doc
I
2)
S(v ~ av
-
=
=
Sve -zu av
-
z -A + C
-
E(x
+ 2) -
(x+ 2) + c
, immense
Integration by Parts
Product rule
(fg)' =
fg +
fg)
d
fg =
(f'gax +
Sfg'dx
=>
Sf'gdx =
fg-Sfg'dx * FORMULA
6
Ex .
(xc .
cos(oc) doc
↑ ↑
9 >
-
g =
1 f =
since
=
xsin(x) /sinx I
-
.
do
=
xsin(x) +
cosx + C -
ADD + C WHEN No
MORE
S
!!! CHECK
[xsinxc + cosx + c]
=
Six + inx !!!
=
xCossa occosoc # SAME
, immense
Integration by Parts
IF ...
So cosoc doc
↓ ↓
- 9
-x g'
&
- = =
-sino
= Cossa +
Sjxsinx dx
-
MORE COMPLEX THAN
ORIGINAL
↳
DO OTHER WAY
Ex
.
Sarctan (c). I doc
↓ >
- add x 1 B MUST
g MAKE FI
g' =
1 +
I
x
>
- = xC
=
Sarctan(x) -J x 1 +
doc
-5) The
2
=
arcton u =
1 + 0
x .
(c) -
- = Zoc
=
xc .
arcton(x) -[Inful + C daC
=
arctan(x) (n)1 + C
x =
+
x . -
+ x
, imme
Integration by Parts
3t
S
2
Ex .
t e
O d d
g
-' g' =
2t f =
523t
want
MAKE SIMPLER
=
[5 rest] !
+
-
=teat
↓
>
- Do IBP AGAIN
g
O
fl
g 2 =
== 5
- -
0 -
[te't' Ect g
at
=+
* /estjo
3
-
+
3
O
-ja - -
, mens
Integrals with Trig functions
Stasc f f
-
Cos sins COSOC
sinx COSOC -
sinoc
-
In(cosod tanoc seco
? seco secoctanx
TRIG IDENTITIES
cos2 x + sin2 x
= I
sin 2x =
asinkcoss
cos2x = 1 -Isin2x = 2cos - I
1 + tan(x) =
Seco
Ex .
[secos doc
secx + tanx
=
x
sec
Seco + tans
&Seco
+ Secostanza
=
do let u =
Seco + tand
Secoct fansc
du = Secostano + seco
du doc
S -
= dx
*
u
=
Stau =
(n(u) + C
=
In/secx + tanxl + C
, mens
Integrals with Trig functions
Ex .
/sinGcosodo let u = sin O
=
cost
Su
=
=
sin C
Ex .
(secit)ton(t) de let u
=
Sec(t)
n -
sec(t)tcn(t)
=
Su
-
dt
=
↳ us + C
=
-se(t) + c
Ex
.
(cos(0) do * cos20 =
2 coo -
I
Cos20 + 1 =
cosO
= (cos (20) + 1 do Z
-
(tsin(20) + 0) + C
-
* sin(20) +
4 + C
, mens
Integrals with Trig functions
Ex .
[sin 0 cos'6 ao
-
u =
=
sin O
CosO =
I
du
-2
3 I
Sus COSO
=
cos O do dO
COS O
-
Su (1 Sus-
7
=>
-
sir 8) du
=
u du
-jut -**
I
+
C
-sin'o -Tsin8 + C
We can use this whenever we want to
[sin" (0). cos" (8)
integrate 90 B p or
q
is Odd
Ex .
(sin" (x) .
cos" (3) as
*
sin" (c)
S . . cos(c)
(x)
=
cos dx
=
(sin "(x)
*
. (cos"(x))" . cos(o) do
=
S sin" (x). (1 -
sinpcl)". cos() doc
let u =
sin(x) = cossc
=
S 434(1 -
12)"du
Semester 2
, me
Integrals
imme
Review of Integration
An ANTIDERIVATIVE of f is a function F
>
such that -
dx
(F(x)] =
f(x)
The INDEFINITE INTEGRAL of f is the infinite
family of antiderivatives F(x) + C
(f(x)dx =
f(x) + C
The DEFINITE INTEGRAL of f from a to ↳
is the
signed bounded
by c= x b
=
area a
;
y
=
f(x) ; y
=
0
1
I'll a
Area-lim f(x, &x
n+ 0k = 1
This evaluated the FTC
can be
using
Area =
"f(x) dx
=
F(b) -
F(c)
E
, imme
Review of Integration
EX
5x
.
fxdx = + C
jxdx [5 1·
3
=
x + C
I =
5(2) + -
(5(k Fe)
=
E
meet
Integration by Substitution
Ex .
[20c . cos(s) dx LET x =
u
=
a
Scos(u) -
↳ CHAIN RULE
=
Scos(u) du
=
sin(u) + C
:
sin(x) + C
* CHAIN RULE :
Scos(u) dxc =
/ [sin (2)] o
e
=> (sin(u)] (u
dx
=
sin
, meet
Integration by Substitution
I 2
EX -
- x
S >c2 dx let u
= -
x
au=
O
value of u
- A change
S -
-C ·
"oc du
s
-
dx
=
Is
- du
- -Je
-
I
-F(c I
U
au
=
-
O
O
-
=
H -
el
-
=
E(t -
1)
Ex .
Soc Vect2 do let V =
x + 2
A - I
It
Sx
= an dx
du =
doc
I
2)
S(v ~ av
-
=
=
Sve -zu av
-
z -A + C
-
E(x
+ 2) -
(x+ 2) + c
, immense
Integration by Parts
Product rule
(fg)' =
fg +
fg)
d
fg =
(f'gax +
Sfg'dx
=>
Sf'gdx =
fg-Sfg'dx * FORMULA
6
Ex .
(xc .
cos(oc) doc
↑ ↑
9 >
-
g =
1 f =
since
=
xsin(x) /sinx I
-
.
do
=
xsin(x) +
cosx + C -
ADD + C WHEN No
MORE
S
!!! CHECK
[xsinxc + cosx + c]
=
Six + inx !!!
=
xCossa occosoc # SAME
, immense
Integration by Parts
IF ...
So cosoc doc
↓ ↓
- 9
-x g'
&
- = =
-sino
= Cossa +
Sjxsinx dx
-
MORE COMPLEX THAN
ORIGINAL
↳
DO OTHER WAY
Ex
.
Sarctan (c). I doc
↓ >
- add x 1 B MUST
g MAKE FI
g' =
1 +
I
x
>
- = xC
=
Sarctan(x) -J x 1 +
doc
-5) The
2
=
arcton u =
1 + 0
x .
(c) -
- = Zoc
=
xc .
arcton(x) -[Inful + C daC
=
arctan(x) (n)1 + C
x =
+
x . -
+ x
, imme
Integration by Parts
3t
S
2
Ex .
t e
O d d
g
-' g' =
2t f =
523t
want
MAKE SIMPLER
=
[5 rest] !
+
-
=teat
↓
>
- Do IBP AGAIN
g
O
fl
g 2 =
== 5
- -
0 -
[te't' Ect g
at
=+
* /estjo
3
-
+
3
O
-ja - -
, mens
Integrals with Trig functions
Stasc f f
-
Cos sins COSOC
sinx COSOC -
sinoc
-
In(cosod tanoc seco
? seco secoctanx
TRIG IDENTITIES
cos2 x + sin2 x
= I
sin 2x =
asinkcoss
cos2x = 1 -Isin2x = 2cos - I
1 + tan(x) =
Seco
Ex .
[secos doc
secx + tanx
=
x
sec
Seco + tans
&Seco
+ Secostanza
=
do let u =
Seco + tand
Secoct fansc
du = Secostano + seco
du doc
S -
= dx
*
u
=
Stau =
(n(u) + C
=
In/secx + tanxl + C
, mens
Integrals with Trig functions
Ex .
/sinGcosodo let u = sin O
=
cost
Su
=
=
sin C
Ex .
(secit)ton(t) de let u
=
Sec(t)
n -
sec(t)tcn(t)
=
Su
-
dt
=
↳ us + C
=
-se(t) + c
Ex
.
(cos(0) do * cos20 =
2 coo -
I
Cos20 + 1 =
cosO
= (cos (20) + 1 do Z
-
(tsin(20) + 0) + C
-
* sin(20) +
4 + C
, mens
Integrals with Trig functions
Ex .
[sin 0 cos'6 ao
-
u =
=
sin O
CosO =
I
du
-2
3 I
Sus COSO
=
cos O do dO
COS O
-
Su (1 Sus-
7
=>
-
sir 8) du
=
u du
-jut -**
I
+
C
-sin'o -Tsin8 + C
We can use this whenever we want to
[sin" (0). cos" (8)
integrate 90 B p or
q
is Odd
Ex .
(sin" (x) .
cos" (3) as
*
sin" (c)
S . . cos(c)
(x)
=
cos dx
=
(sin "(x)
*
. (cos"(x))" . cos(o) do
=
S sin" (x). (1 -
sinpcl)". cos() doc
let u =
sin(x) = cossc
=
S 434(1 -
12)"du
