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mat 129 exam 1 fa24., Exams of Mathematics

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Math 129 Exam 1 Fall 2024
Name (Please Print.):
8pointseach
You need to show all your work to get a full credit.
sin21cos2, cos21sin2,sec
2tan21, tan 2sec21,
sin21cos 2
2, cos21cos 2
2,sin22sincos,
sin ucos v1
2sinuvsinuv,sinusin v1
2cosuvcosuv
Problem 1 Evaluate each integral.
(a) xe5xdx
(b) cos48xdx
1
Cheawon
kim
=
(
(
"
)
-
Se
"
ds
=
e
"
sC
=
xe
-
S
5
es
"
da
=
xe
-
(
Ses
"
dx
)
U
=
5
K
duzsdx
dn
:
da
.
x
Seda
=
xe
-
Sda
=
x
-
s
Sed
da
=
x
-
(
entc
)
=
xex
-
5
e
+
C
2
=
SEoseEu
38
dy
U
=
8
c
=
S
cos
4
]
du
duz
8
dc
U
2
=
Ih
dy
=
dx
.
=
S
cos
4
Eu
)
du
duz
=
dn
=
Sco
(
u
)
)
'
du
daz
=
du
=
δ
fl
(
tcos
?
]
e
du
.
@
.
(
+
by
]
=
(
olastctScoscusl
du
,
)
+
f
4
durtfssand
=
방호
S
(
tc
0
s
)
"
'
(
a
+
l
(
aHsn
)
catans
=
1
oS
[
(
Ie
cos
(
az
3
]
]
dar
=
a
a
+
apbtazp
=
(
luatCt
sin
(
us
)
+
c
]
]
+
S
4
dustf
cosr
duz
)
=
azt
zabtakb
2
=
6
(
(
ust
+
(
sin
(
us
)
tc
)
]
t
*
µ
2
+
C
+
fcos
danc
)
=
oS
[
π
4
t
4
s
)
+
JIosCuz
)
dus
(
)
=
(
(
uzt
c
+
*
(
sin
(
u
3
)
tc
)
)
++
ct
sinc
)
t
)
:
(
S
화없
.
.
.
다있
dm
+
s
omiaur
+
5
.
)
여뿐
-
(
+
δ
+
이뉴
+
채약
)
tC
.
=
(
S
4
daz
+
(
*
[
*
S
(
t
cos
(
u
)
duz
)
+
Sco
dua
)
=
(
3
o
tin
(
u
)
.
tsin
)
+
c
=
(
Sedut
(
ofl
t
cos
[
2
uz
)
dus
)
+
f
ose
duz
)
=
+
in
(
s
6
+
sin
(
16
tC
합합다이없
.
.
하에이아
)
이끌
)
이쁜하
.
*
에쁜
)
*
다지
.
=
+
바채임등개했에
U
pf3
pf4

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Math 129 Exam 1 Fall 2024 Name (Please Print.): 8 points each You need to show all your work to get a full credit. sin^2  1  cos^2 , cos^2  1  sin^2 , sec^2  tan^2  1, tan^2  sec^2  1, sin^2  1 cos 2 2 , cos^2  1 cos 2 2 , sin 2  2 sin cos , sin u cos v  12 sin uv   sin uv , sin u sin v  12 cos uv   cos uv  Problem 1 Evaluate each integral.

(a)  xe^5 xdx

(b)  cos^4  8 x  dx

Cheawon kim = (^) ( (" (^) )- Se"ds^ = e" sC = xe

  • S 5 es "da = xe
  • ( (^) Ses"^ dx) U =^5 K duzsdx dn:^ da. x Seda = xe^ - Sda = -^ x

s Sed^

da =^ x^ - (entc)

= xex - 5 e + C

2 = SEoseEu 38 dy ∅ U =^8 c = S cos^4 ]^ du duz 8 dc^ U 2 = (^) Ih dy=^ dx. = ☆ S cos^4 Eu) du =^ →^ duz^ =dn ⑤ Sco(u)^ ) ' du daz=^ du = δ fl (^ tcos ?]e du. @.^ (+^ by]^ = (olastctScoscusl du,) +^ f (^4) durtfssand

= S

( (^) tc 0 s) " ' a(

l(aHsn =^ )catans

1 oS [ (Iecos(az^3 ]

]

dar = a a + apbtazp =^ (^ luatCt^ sin^ (us)^ +c^ ]]^ +^ S^4 dustfcosr

duz)

= aztzabtakb^2 = (^6) ( ( ust + (^) ( sin( us)tc )]t * (^) μ 2 + C +fcosdanc) = oS [ π (^4 4) t s)+ JIosCuz )dus ( )

= ((uztc^ +*(sin(u^3 )tc))++^ ct sinc^ )t)

: (^) (

S

..^.

dm +s

omiaur+ 5 .) -^ (^ +δ^

+ )^ tC^.

= (S 4 daz^ + ([ S ( tcos(u) duz) + Sco dua)

= (^) ( o^3 tin (u) .tsin) +^ c

= (Sedut ( ofl tcos[ 2 uz )dus) + f ose duz) =

③ +in( s 6 +^ sin(^16 tC .. )) .

)

.

U

(c)  2 x

x^2  x  2 dx

(d)  ex^ cos xdx

U 1 =^ x-^2 Ua =^ at^1 dul =^ dss^ duz = (^) bl. = (^2) S = x- (^22) J ( - 2 )-

()dx =^2 (^ S (^) - sdctSx^ ←^ ")dx ) = (^2) (* 5 -^ dbu+^5 - +Ddc

Sa,^ du^ +^ S - (+ *) da )^ = 2 (^ (^ n…^ (^ m^ ))^ +^ c? - ( (^) ) .)) = =^ [ (n^ (^ ui^ ))-^ (^3) S du. ) = (^2) (( In lul) +c) - ( inluajtc) = (^2) ( | (|) - In(^ lul))^ +^ c = (^2) CU(|^ )-^ ⑤ u =^ Eos^ (ax) dureaa = Scosx e^ " dx =^ (os (x) e ". Soctsin (xy^ )dc = cos (a)e Ife

" (sinoa))adsc

= (^) cosoc) e^ " t

S sin^ Ese)^

ex ds^ u= sin EsC) dr =^ ex e " cos (a)^ dx^ = (^) cos (sa) extsina )ea^ = Sercosa)ds = rcosa^ )^ e'^4 sinxer : sue

'tsine

Gpr

'tsin

Problem 2 Write the partial fraction decomposition but do not finish the computation. 3 x  5  x  8 ^3  x^2  x  2  Problem 3 Find the following integral by the trigonometric substitution

 1  x^2 dx

=

) ∞.^

∞ ) .

C

since x=^ sin^ (ω) (^) , - W^ Then (^) , dac = Cos tw)^ theefore (^) cos(o)ispositive.

SFsinlw)^ cos(w)^ dw

= S cos (ωldw co^5 =^ ltcos

f * cos dw

SIt cos^ (^2 w)^ dw

= ( Sdwt (^) Scos (aw)^ dw> a =^2 w

:

C+ Scos (^2 w)^ do)

dl= 2 dw Idu =^ dw = (W^ +^ c^ +^ S cos (^) la). *du (^) )

= (weCt Scosta)du)

= ( w + c + * (sin (u) tc))

o sarcsimb)

= (^ Wtsin (al) tc

= Larcsin (oc) + (^) (s ))tC

= Carcsin(x )+ sim))C

= (arcsin(sa) + simaarasin(s))tC

=^ arcsi tin (^2 arcsina +^ c = aresinca) t

Ce^4 sinczaresinab+