MAM1021S Lecture Notes Summary

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