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Solved Assignment for Calculus II | MATH 166, Assignments of Calculus

Iowa State University (ISU)Calculus

Material Type: Assignment; Class: CALCULUS II; Subject: MATHEMATICS; University: Iowa State University; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 09/02/2009

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koofers-user-f0x 🇺🇸

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bg1
Exercises 8.1
4.
1
0
2
1dxxx
Let u = 1 – x2.
du = -2xdx
.
2
1duxdx
.
3
1
)1(
3
1
1
.)1(
3
1
3
2
2
1
2
1
1
1
0
2
1
0
2
22
2
3
2
3
2
3
2
1
xdxxx
cxcuduudxxx
5.
.
)(1
1
4
1
)1(4
4
2
2
4
2
2
dx
dx
x
dx
x
x
Let u =
.
2
x
du
.2
2
1dudxdx
.
2
arctan
2
1
)arctan(
2
1
2
1
1
4
1
4
22
c
x
cudu
ux
dx
7.
.
4
2
x
xdx
Let u =
du
.
2
1
2duxdxxdx
.4ln
2
1
ln
2
1
2
11
4
2
2
cxcudu
u
x
xdx
11.
.)(sec)tan(
)(cos
)tan(
2
2
dzzzdz
z
z
Let u =
).tan(z
du
.)(sec
2
dzz
.tan
2
1
2
1
)(cos
)tan(
22
2
czcuudu
z
z
13.
dt
t
t)sin(
Let u =
.t
du
.2
2
1du
t
dt
dt
t
1
pf3
pf4
pf5
pf8
pf9
pfa

Partial preview of the text

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Exercises 8.

1

0

2

x 1 x dx

Let u = 1 – x

2

du = -2xdx

xdx  du

1

0

2

1

0

2

2 2

2

3

2

3

2

3

2

1

x x dx x

x x dx u du u c x c

2

4 2

2

2

dx

dx

x

dx

x x

Let u =

x

du 2.

dxdxdu

arctan

arctan( )

2 2

c

x

du u c

x u

dx

2

x

xdx

Let u = 4.

2

x 

du.

 2 xdxxdxdu

ln 4.

ln

2

2

du u c x c

x u

xdx

 tan( )sec ( ).

cos ( )

tan( )

2

2

dz z zdz

z

z

Let u =

tan( z ).

du sec ( ).

2

zdz

tan  .

cos ( )

tan( )

2 2

2

udu u c z c

z

z

dt

t

sin( t )

Let u =

t.

du

du

t

dt

dt

t

sin( )( 2 ) 2 cos( ) 2 cos .

sin( )

dt u du u c t c

t

t

    

 

4

0

2

1 sin ( )

cos( )

dx

x

x

Let u =

sin( x ).

du

cos( x ) dx.

arctan   arctan(sin( )).

1 sin ( )

cos( )

2 2

du u c x c

u

dx

x

x

 

 

4 4

0

0 2

arctan(sin( )) arctan

1 sin ( )

cos( )

dx x

x

x

dx

e

e

x

x

2

2

Let u =

2 x

 e

du

2 2

e dx e dx du

x x

2

2

2

2

1

e c

du u c

u

dx

e

e

x

x

x

 

 .

ln 2

ln| |

Letu 2 e.

x

du u c e c

u

dx

e

e

du e dx

dx

e

e

x

x

x

x

x

x

 

 2 2 .

ln( 2 )

ln( 2 )

2 sin( )

ln( 2 )

ln( 2 )

2 sin( )

ln( 2 )

2 ln( 2 ) 2 ln( 2 ).

2 sin( ) 2 ( ).

2

3

6

6

0

cos( )

0

cos(x)

cos() cos( )

(ln( 2 ))

(ln( 2 ))

cos( )

 

 

x

x u x

u u

u u u

u u

x u

x dx

xdx C C

du C

e

du

d

e

xdx du

cos ( 4 1 )

sin( 4 1 )

1 sin ( 4 1 )

sin( 4 1 )

2

2

dt

t

t

dt

t

t

Let u = cos(4t – 1).

du = -4sin(4t - 1)dt.

sec( 4 1 ).

cos( 4 1 )

cos ( 4 1 )

sin( 4 1 )

2 2

t C

C

t

C

u

du

u

dt

t

t

 

dt

t

t t

sin ( 2 )

(cos( 2 ))

2 3

2 3

Let u = sin(t

3

  • 2).

du = cos(t

3

    1. t dt

2

sin( 2 )

sin ( 2 )

(cos( 2 ))

3

2 3 2

2 3

C

t

C

u

du

u

dt

t

t t

 

sin ( 2 )

cos ( 2 )

2 3

2 2 3

dt

t

t t

Let u = t

3

du =

2 2

t dtt dtdu

 

 

(cot( 2 ) ).

(cot( 2 ) ( 2 ))

( cot( ) )

(csc 1 )

cot

sin ( 2 )

cos ( 2 )

3 3

3 3

2

2

2 3

2 2 3

t t C

t t C

u u C

u du

dt u du

t

t t

 

dy

y

y

4

16 9

Let u = 3y

2

du = 6ydy

ydydu

 

du

u

dy

y

y

4 2

16

1

6

1

16 9

 

   

 

 

   

 

   

 

 

     

   

   

 

 

 

ˆ arcsin

5 cos

5 cos

5 cos.

25 3 5 cos.

25 3 25 25 sin 25 1 sin 25 cos.

Let x- 3 5sin.

2 cos 3 1 ˆ.

2 cos( ) ˆ

2 sin( )

6 1 sin 3 1

Letu 3 1.

6 1 sin 3 1

tan

sec sec ( ) tan( )

Letu e.

sec

2

2

2 2 2

2

2

2 2 2

2

2

2

2

2

1

2

2

1

2

2

2

2

2

x

2

 

  

 

c

x

c

d

x x

dx

dx d

x

x

x

dx

x x

dx

x x

dx

x x

dx

t t c

u c

u du

dt

t t

t t t

dt du

t t

t

du t t t dt

t t

dt

t t

t t t

e e dx udu u c e c

du e dx

e e dx

x x x

x

x x

Exercises 8.

 

 

ln 9 18 10.

ln

Letu 9(x 1) 1.

2

2

2

2

2

2

x x C

u C

du

u

dx

x x

x

du x dx x dx du

dx

x

x

dx

x x

x

dx

x x

x

1 sec ( ) 1 tan ( ).

3

2

.

3

2

sec

3

2

Let sec(u)

.

1

3

2

3

1

9

2

9

2 9

2 2

2

1

2

2

2

t u u

t u t

t t

dt

t t

dt

t t

dt

   

  

 

  

sec ( ).

sec( ) tan( )

sec( )tan( )d

du sec( )tan( )d.

1 sec ( )- 1 tan ( ).

Letu sec( ) sec ( ).

sec ( ) 4

tan( )

sec( )tan( ) 2.

sec( )tan( ).

sec( ) 2.

sec(x)

Let u

sec( )

sec( )

sec( )tan( )

sec( )

tan( )

sec ( )

tan( )

sec ( ) 4

tan( )

1

2

2 2 2

  • 1

2

2

2

2

2

2

2

d C u C

u u

du

u

u

u u

du

u u

du

dx

x

x

x xdx du

du x x dx

x u

dx

x

x

x x

dx

x

x

dx

x

x

dx

x

x

 

sec ( ) 4

tan( )

sec( )tan( )

sec ( ) 4

sec( )

sec( )tan( )

sec ( ) 4

sec( )

sec( )tan( )

sec ( ) 1

sec( )

sec( )

sec( ) 1

sec( )

sec( )

sec

sec ( )

Check

sec( )).

sec (

sec ( )

sec ( ) 4

tan( )

2

2

2

2

2

1

2

1

1

1

2

x

x

x x

x x

x x

x x

x x

x x

x

dx

d

x x

x C

dx

d

dx

dw

w w

w

dx

d

x C

u C

dx

x

x

1 sec ( ) 1 tan ( ).

3

2

.

3

2

sec

3

2

Let sec(u)

.

1

3

2

3

1

9

2

9

2 9

2 2

2

1

2

2

2

t u u

t u t

t t

dt

t t

dt

t t

t

   

  

 