Calculus 2 Chapter 7 Test Practice 1) Identify the integration method you would use for each of the following problems. Do NOT integrate!

a.

€

1x 2 + 8x + 25

dx∫ b.

€

3x + 4x 2 + 2x −15

dx∫ c.

€

64 − x 2dx∫

________________ ________________ ________________ d.

€

x 2 cos(x)dx∫ e.

€

cos4 (x)sin3(x)dx∫ f.

€

5x 25 − x 2dx∫ ________________ ________________ ________________

2) Find the limits using L’Hospital’s Rule, when necessary.

a.

€

limx→ 0

sin(3x)4x

________________

b.

€

limx→ 0

sin−1(2x)x

________________

c.

€

limx→ 0

xx( ) ________________

d.

€

x→+∞lim 1− 2

x%

& '

(

) * 3x

________________

e.

€

x→∞lim x e

1x −1( )

________________

Name _________________________ Period _____

3) Integrate each indefinite integral: a.

€

x ln(x)dx∫ ________________

b.

€

x 2exdx∫ ________________

c.

€

x 2 4 − x 3dx∫ ________________

d.

€

x 3 4 − x 2dx∫ ________________

e.

€

x 5 4x 2 −16∫ dx ________________

f.

€

e2x cos(3x)dx∫ ________________

g.

€

5cos8(x)sin3(x)dx∫ ________________

h.

€

6sec4 (2x)tan3(2x)dx∫ ________________

i.

€

4(x − 3)(x +1)2

dx∫

________________

Calculus 2 Chapter 7 Test Practice 1) Identify the integration method you would use for each of the following problems. Do NOT integrate!

a.

€

1x 2 + 8x + 25

dx∫ b.

€

3x + 4x 2 + 2x −15

dx∫ c.

€

64 − x 2dx∫

tan-1(x) partial fractions trig substitution__

(x = 8sin(θ)) d.

€

x 2 cos(x)dx∫ e.

€

cos4 (x)sin3(x)dx∫ f.

€

5x 25 − x 2dx∫ integration by parts trigonometric integration u-substitution___

2) Find the limits using L’Hospital’s Rule, when necessary.

a.

€

limx→ 0

sin(3x)4x

= limx→ 0

3cos(3x)4

=34

€

limx→ 0

sin(3x)4x

=

€

34

b.

€

limx→ 0

sin−1(2x)x

= limx→ 0

21− 4x 21

= limx→ 0

21− 4x 2

=21

= 2

€

limx→ 0

sin−1(2x)x

= 2

c.

€

limx→ 0

xx( )

lny=

€

limx→ 0

x ln x( )=

€

limx→ 0

ln xx −1

$

% &

'

( ) =

€

limx→ 0

1x

−x −2$

%

& &

'

(

) ) = limx→ 0

1x

− 1x2

$

%

& &

'

(

) ) = limx→ 0

−x 2

x$

% &

'

( ) = lim

x→ 0−x( ) = 0

lny=0 → y = e0 = 1

€

limx→ 0

xx( ) = 1

d.

€

x→+∞lim 1− 2

x%

& '

(

) * 3x

lny =

€

x→+∞lim 3x ln 1− 2

x%

& '

(

) * =

€

3x→+∞lim

ln x − 2x

%

& '

(

) *

1x

=

€

3x→+∞lim

xx − 2%

& '

(

) * ⋅ 2x −2( )

−1x 2

=

€

= 3x→+∞lim

−2x 3

x 2 x − 2( )= 3

x→+∞lim

−2xx − 2( )

= 3⋅ −2 = −6

lny= -6 → y = e-6 ≈ 0.00248

€

x→+∞lim 1− 2

x%

& '

(

) * 3x

= e-6 ≈ 0.00248

e.

€

x→∞lim x e

1x −1( ) =

€

x→∞lim

e1x −1( )1x

=

€

x→∞lim

e1x ⋅ −x −2( )−x −2

=

€

x→∞lim

e1x( )1

=x→∞lim e

1x = e0 =1

€

x→∞lim x e

1x −1( ) = 1

Name _________________________ Period _____

3) Integrate each indefinite integral:

a.

€

x ln(x)dx∫ =x 2 ln x2

−x2dx∫ =

x 2 ln x2

−x 2

4+C

€

x 2 ln x2

−x 2

4+C

€

u = ln x v =x 2

2du =

1xdx dv = xdx

b.

€

x 2exdx∫ = x 2ex − 2xexdx∫ = x 2ex − 2xex − 2exdx∫[ ] = x 2ex − 2xex + 2ex +C=

€

x 2ex − 2xex + 2ex +C

€

u = x 2 v = ex

du = 2xdx dv = exdx

€

u = 2x v = ex

du = 2dx dv = exdx

c.

€

x 2 4 − x 3dx∫ =

€

−13

udu∫ =

€

−29u32

#

$ %

&

' ( +C =

€

−29(4 − x 3)

32

#

$ %

&

' ( +C

€

u = 4 − x 3,du = −3x 2dx

−13du = x 2dx

d.

€

x 3 4 − x 2dx∫

€

x = 2sin(θ),dx = 2cos(θ )dθ

=

€

8sin3(θ ) 4 − 4sin2(θ )2cos(θ )dθ∫ =

€

16 sin3(θ)cos(θ) 4cos2(θ )dθ∫

=

€

32 sin3(θ)cos2(θ)dθ∫ =

€

32 [1− cos2(θ )]cos2(θ)sin(θ)dθ∫

=

€

32 [cos2(θ ) − cos4 (θ )]sin(θ)dθ∫ =

€

32 −cos3(θ )3

+cos5(θ )5

$

% &

'

( ) +C

=

€

324 − x 2( )

532

5−

4 − x 2( )38

3

#

$

% % %

&

'

( ( (

+C=

€

4 − x 2( )5

5−4 4 − x 2( )

3

3+C

e.

€

x 5 4x 2 −16∫ dx

€

x = 2sec(θ ),dx = 2tan(θ )sec(θ )dθ

=

€

32sec5(θ ) 4(4 sec2(θ )) −16 ⋅ 2sec(θ )tan(θ )dθ∫ =

€

64sec6(θ ) 16(sec2(θ ) −1) ⋅ tan(θ)dθ∫

=

€

256 sec6(θ) tan2(θ) ⋅ tan(θ)dθ∫ =

€

256 sec6(θ)tan2(θ )dθ∫

=

€

256 1+ tan2(θ)( )2 tan2(θ)sec2(θ)dθ∫ =

€

256 1+ 2tan2(θ) + tan4 (θ )( ) tan2(θ )sec2(θ )dθ∫

=

€

256 tan2(θ) + 2tan4 (θ ) + tan6(θ)( )sec2(θ )dθ∫ =

€

256 u2 + 2u4 + u6( )du∫ =

€

256 u3

3+2u5

5+u7

7"

# $

%

& '

=

€

256 tan3θ3

+2tan5θ5

+tan7θ7

#

$ %

&

' ( =

€

256

x 2 − 42

#

$ %

&

' ( 3

3+2 x 2 − 4

2#

$ %

&

' ( 5

5+

x 2 − 42

#

$ %

&

' ( 7

7

#

$

% % % %

&

'

( ( ( (

=

€

256x 2 − 4( )

3

24+

x 2 − 4( )5

80+

x 2 − 4( )7

896

#

$

% % %

&

'

( ( ( =

€

32 x 2 − 4( )32

3+16 x 2 − 4( )

52

5+2 x 2 − 4( )

72

7

#

$

% % %

&

'

( ( (

+C

f.

€

e2x cos(3x)dx∫ =

€

cos(3x)e2x

2 –

€

−32e2x sin(3x)dx∫ =

€

cos(3x)e2x

2+

€

32

€

sin(3x)e2x

2−

32e2x cos(3x)dx∫

$

% &

'

( )

€

u = cos(3x) v =e2x

2du = −3sin(3x)dx dv = e2xdx

€

u = sin(3x) v =e2x

2du = 3cos(3x)dx dv = e2xdx

So,

€

e2x cos(3x)dx∫ =

€

cos(3x)e2x

2+

€

34sin(3x)

e2x

2−94

e2x cos(3x)dx∫ ,

€

134

€

e2x cos(3x)dx∫ =

€

cos(3x)e2x

2+

€

34sin(3x) e

2x

2,

€

e2x cos(3x)dx∫ =

€

213

€

cos(3x)e2x +

€

326sin(3x)e2x + C,

g.

€

5cos8(x)sin3(x)dx∫ =

€

5 cos8(x) 1− cos2(x)[ ]sin(x)dx∫ =

€

5 cos8(x) − cos10(x)( )sin(x)dx∫

=

€

5 cos8(x)sin(x)dx∫ − 5 cos10(x)sin(x)dx∫ =

€

5−cos9(x)

9− 5

−cos11(x)11

+C

=

€

5cos11(x)11

−5cos9(x)

9+C

h.

€

6sec4 (2x)tan3(2x)dx∫ =

€

6 sec3(2x)tan2(2x)sec(2x)tan(2x)dx∫

=

€

6 sec3(2x) sec2(2x) −1( )sec(2x)tan(2x)dx∫ =

€

6 sec5(2x) − sec3(2x)( )sec(2x)tan(2x)dx∫

=

€

6 u5 − u3( ) 12 du∫ =

€

3 u6

6−u4

4#

$ %

&

' ( +C =

€

u6

2−3u4

4#

$ %

&

' ( +C =

€

sec6(2x)2

−3sec4 (2x)

4#

$ %

&

' ( +C

OR

€

6sec4 (2x)tan3(2x)dx∫ =

€

6 sec2(2x)tan3(2x)⋅ sec2(2x)dx∫

=

€

6 tan3(2x) tan2(2x) +1( )⋅ sec2(2x)dx∫ =

€

6 tan5(2x) + tan3(2x)( )⋅ sec2(2x)dx∫

=

€

6 u5 + u3( ) 12 du∫ =

€

3 u6

6+u4

4"

# $

%

& ' +C =

€

u6

2+3u4

4"

# $

%

& ' +C =

€

tan6(2x)2

+3tan4 (2x)

4"

# $

%

& ' +C

i.

€

4(x − 3)(x +1)2

dx∫

€

4(x − 3)(x +1)2

=A

(x − 3)+

B(x +1)

+C

(x +1)2

€

4 = A(x +1)2 + B(x − 3)(x +1) +C(x − 3) A = 1/4, B = -1/4, C = -1

€

14(x − 3)

−1

4(x +1)−

1(x +1)2

#

$ %

&

' ( dx∫ =

€

ln x − 34

−ln x +14

+1

(x +1)+C