Techniques of Integration - Unit 8
Integration by Parts:
This
□ Sample Problem (by Becca Bassett and Theo Kulczycki): This is an intermediate problem designed to help
students do trig integrals and recognize when you cannot use natural logs to integrate.
Problem:
 3x
Integrate,
1
dx
2
2
Solution:
It is really easy to just write this off as an natural log integral and quickly write the solution as, ln 3x 2  2  C

BUT THIS IS WRONG! You cannot do a natural log integration with xn if n is greater than 1.
So you have to do a trig integral.
d 1 1x 
1
 tan   2
a  x  a 2
dx a

First, take out a 1/3 so that you have x2 with no coefficient. 

1
3

1
2
x 
3
dx
2
Now we have it in the form that we want and we can use solve the integral. It is set up so a 

 

 
1 1  1 x 
 
tan
C
3
2 
 2 
 
 3 
3 

2
3


□ Sample Problem (by Becca Bassett and Theo Kulczycki): Chap 8
This is an hard problem designed to help students practice doing integration by parts several times.

Problem:
Integrate

e4 x
dx
sec3x 
Solution:
First, you should notice you have to use integration by parts. Notice that 1/sec(3x) is just cos(3x).
 u dv = uv -  v du
For this problem, because we have a trig function and a exponential, it does not matter which one you pick to be u and
which you pick to be dv.
So, u  cos3x  and dv  e4 x dx

du  3sin 3x dx and v 
e4
4
now plug those in:
e4 x 
 cos3x   
4 x 


e4
 3sin 3x dx
4
here we have to do integration by parts again for



e4 x
 3sin 3x dx
4
e4 x
so, u  sin 3x  and dv 
4
e4 x

du  9cos3x dx and v 
16

e4 x  
e4 x  e4 x 
4 x
 e  cos3xdx  cos3x  4  3sin 3x 16    16  9cos3xdx


let I 
e
4 x
cos3xdx notice that you have a form of I on both sides of the equation and since I is what we are looking
for, we solve for I


e4 x  
e4 x  9
I  cos3x 

3sin
3x



 
 I (notice that we pulled the constant (9/16) out of the integral)
4  
16  16




Subract (9/16)I from both sides,
e4 x 
7
e4 x
I  cos3x    3sin 3x 
16
16
4 
Divide both sides by (7/16) to get : I 


4 
3
cos3x  e4 x  sin 3x e4 x 


7 
7
Chap 8

□ Sample Problem (by Becca Bassett and Theo Kulczycki): This is an intermediate problem designed to help
students identify trig identities and apply them in integration problems. It will also help students practice usubstitutions with sine and cosine
Problem:
Integrate,


tan 2 (x)
dx
cos4 (x)
Solution:
1
 sec 4 x  and separate that into sec 2 xsec 2 xso the entire integral becomes
4
cos (x)
2
2
 tan xsec xsec 2 xdx
First, notice that


 tan x 1 tan xsec xdx
 tan x  tan xsec xdx
2
Then change one of the sec 2 x  to 1 tan x  , which makes the integral,


Then distribute tan 2 x over (1 tan 2 x , which makes the integral,
Then let u  tan x,du  sec 2 xdx , so

Then integrate,


 u
2

 u
2
 u4 du
 u4 du =  1/3u3  1/5u5



2
2
2
4
2
2
Then substitute u  tan x back in,  1/3tan 3 x  1/5tan 5 x  C
Chap 8
 Problem (by 
□ Sample
Becca Bassett and Theo Kulczycki): This is an intermediate problem designed to help
students practice partial fractions

sec 2 x 
tan 3 x   tanx 
The hardest part about this problem is figuring out what to do first. Some students might try to re-write the secant and
tangents to simplify the equation. However, the most important thing to do is think about using old integration methods.
Like an easy u-substitution.

So, let u  tanx  and du  sec 2 x dx
Now rewrite the integral in terms of u,

u
du
u
3
Now the equation is much simpler and it is easier to see that you should use partial fractions to solve the integral.
bottom and put it into “partial fraction form”
Factor out the
du
2

 1
A Bu  C 
    2

u  1 
u
A(u2 1)  Bu  Cu 1
uu




now you have to choose values that will cancel out either A or B,C and D
the first choice to cancel out B and C is easy, let u=0
when u=0: A(1)  0 1 A 1
the next one is more difficult to figure out what will make A drop out, many students will notice quickly that -1 will not
work and are left to think of a number that will
the number that works this time is i. i 2  1 and will make A drop out

A0  Bi  C i  1
B  Ci  1
B  1

C0
now go back to the integral and plug in the values you just found

1
u 
    2 du
u u 1
separate the integral,


1
 u du   u
u
du
1
2
 ln u
for the second integral let v equal the bottom of the fraction, so dv  2udu
1 dv

2 v
1
 ln v
2



now plug back in what v and u are equal to, when you do this your final answer will be,

1
ln tanx   ln tan 2 x   1  C
2
1
 ln tanx   ln sec 2 x   C
2
Chap 8
□ Sample Problem (by Becca Bassett and Theo Kulczycki): This is a basic problem designed to help students practice
finding limits using L’Hopital’s rule.

Problem::
Solve,
2x  6 

2
 9 
limx
x 3
Solution:

If you just plug in 3 you will get 0/0 which is indeterminate. This is a situation where you can use L’Hopital’s rule.
So you take the derivative of the top, divided by the derivative of the bottom:
d
2x  6  2
dx
d 9
x  9 2x
dx
2

2x
now simplify and plug in 3 to get the limit,


1
x
1  1
lim x  3
x 3
Chap 8
□ Sample Problem (by Becca Bassett and Theo Kulczycki): This is a basic problem to help students practice solving
improper integrals

Problem:


1
1
dx
3x
To do this, we have to take the limit of this integral as the upper limit goes to infinity, so we write


1
3 lim
b 

b
1
b
dx 1
1
1
 lim ln x 1  lim ln b  ln 1  lim ln b  
x 3 b 
3 b 
3 b 
Chap 8: □ Sample Problem (by Marina Mendoza & Beni Atibalentja): INTEGRATION BY PARTS:
 udv  uv   vdu
PROBLEM: (Hard)
3
 sec xdx
SOLUTION:
 (sec
2
x)(sec x)dx
u= sec x
dv= sec 2 xdx
du= sec xtan x
v= tan x
= sec xtan x-  (tan x)(sec x tan x)dx
 (tan
= sec xtan x-  (sec
= sec xtan x-  (sec
= sec xtan x-  (sec
= sec xtan x-  (sec
= sec xtan x-
2
x)(sec x)dx
2
x  1)(sec x)dx
3
x  sec x)dx
3
3
xdx   sec xdx
xdx   sec xdx
sec x  tan x
dx
sec x  tan x
sec 2 x  sec x tan x
dx
2  sec 3 xdx = sec xtan x+ 
sec x  tan x
 sec
3
xdx = sec xtan x-  sec 3 xdx   sec x
Using a “u substitution”:
u=sec x + tan x, du= sec x tan x + sec2 x
2  sec 3 xdx = sec xtan x+

du
u
2  sec 3 xdx = sec xtan x+ ln |sec x + tan x|
 sec
3
xdx =
sex tan x  ln | sec x tan x |
+C
2
Chap 8: □ Sample Problem (by Marina Mendoza & Beni Atibalentja):
U-SUBSTITUTION AND INTEGRATION
PROBLEM: (hard)
Integrate :  cot x(ln(sin x)) dx
SOLUTION: let u= ln(sin x) , du=
With substitution,
=
1
cos xdx
sin x
 cot x(ln(sin x)dx)   udu =
u2
2
[ln(sin x)] 2
2
Chap 8: □ Sample Problem (by Marina Mendoza & Beni Atibalentja): (Easy)

cos 3 x
sin x
dx
SOLUTION:

cos 2 x
(sin x)
1
cos xdx
2

1  sin 2 x
cos xdx
1
(sin x) 2
u= sin x
du= cos xdx
2
1 u
=  1 du
u 2
3
3
1
1
=  1 du   u 2 du =  u 2 du   u 2 du
u 2
=
5
2u 2
= 2u 
5
5
1
2
= 2 sin 2 x  sin 2 x
5
1
2
Chap 8: □ Sample Problem (by Marina Mendoza & Beni Atibalentja): (Medium)

sec 4 (6 x) tan 3 (6 x)dx
SOLUTION:

sec 4 (6 x) tan 3 (6 x)dx
Pull out a sec2x
=  sec 2 (6 x) tan 3 (6 x)(sec 2 (6 x)) dx
=  [1  tan 2 (6 x)][tan 3 (6 x)](sec 2 (6 x)) dx
Let u= tan( 6 x) , du= sec 2 (6 x)dx
=  (1  u 2 ) (u 3 ) du
=  (u 3  u 5 )du
=  u 3 du   u 5 du
u4 u6

4
6
1
1
= tan 4 (6 x)  tan 6 (6 x)
4
6
=
Chap 8: □ Sample Problem (by Marina Mendoza & Beni Atibalentja): PARTIAL FRACTIONS:
PROBLEM: (Medium)
cos x
 sin x(sin x  1)dx
SOLUTION: Let u= sin x, du  cos xdx
du
=
 u(u  1)
=
1
A
B
 
u (u  1) u u  1
 1  A(u  1)  B(u )
u  1, B  1
u  0, A  1
du
du
=

u 1
u
= ln |u – 1| - ln |u|
u 1
u
sin x  1
= ln
sin x
= ln
Chap 8: □ Sample Problem (by Marina Mendoza & Beni Atibalentja):
INTEGRATION BY PARTS
PROBLEM: (Hard)
e
3x
1
cos xdx
3
SOLUTION:
ue
e
3x
1
cos xdx
3
1
dv  cos xdx
3
1
v  3 sin x
3
3x
du  3e 3 x
* uv   vdu
1
1
x   9e 3 x sin xdx
3
3
1
dv  sin xdx
3x
u  9e
3
3x
1
du  27e
v  3 cos x
3
= 3e 3 x sin
= 3e 3 x sin
e
3x
1
1
1
x  27e 3 x cos x  81 e 3 x cos xdx
3
3
3
1
1
1
1
cos xdx = 3e 3 x sin x  27e 3 x cos x  81 e 3 x cos xdx
3
3
3
3
 80  e 3 x cos xdx = 3e 3 x sin
1
3
1
1
x  27e 3 x cos x
3
3
3e 3 x sin
1
1
x  27e 3 x cos x
3
3
80
1
  e 3 x cos xdx = 
3
Chap 8: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Improper Integrals
Solve


1


dx
. Remember to use those limits of b!
x5
b dx
dx
 Lim 1 5
5
x
x
b 

1
 Lim  x 5 dx  Lim[
b
1
b 
b 
1 4 b
x ]
1
4
1 4 1 4 b
b  (1) ]
1
4
b  4
b
1 1
1
 Lim[ 4  ]  [0  ]
41
4
b  4b
1

4
 Lim[
Chap 8 □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): (arc length, parts): Deutsches
Bargeldwachstum
 Herr Geldkostenzähler, a former Reichsbank employee of the early 1920’s, graphs German inflation on the remains of his
tattered bed spread (which is his only possession) in blood. It is in the shape of ln(x) from x=1 meter to x=3 meters.
If rope to hang yourself with costs 27 billion marks per meter at 3:00 PM, how much would suicide cost Herr
Geldkostenzähler if he bought some hanging rope as long as his blood streak? You will have to use trigonometric
substitution (hint: let x=cot) and integration by parts.
First find the arc length of the blood streak (ln x):

s

3
1

3
1
(1 f '(x) 2 )1 2 dx
(1
1 12
) dx
x2
Let x  cot( )  dx  csc 2 ( )d


(1
1 12
) dx    [1 tan 2  ]1 2 csc 2 ( )d
2
x
   sec( ) csc 2 ( )d
Integration by parts :
 udv uv   vdu
Let u  sec   du  tan  sec d
Let dv  csc 2 d  v  cot 
   sec( )csc 2 ( )d  sec( )cot  
 sec  tan

sec d
The rest is common sense!


cot( )tan( )sec( )d
sec   tan 
 sec d   sec [ sec   tan  ]d


sec 2   sec  tan 
d
sec   tan 
Let u  sec   tan   du  (sec  tan   sec 2  )d
du
 
 ln( u)  ln(sec   tan  )
u
…Or at least sense as common as it could be under the circumstances…

 sec  tan  
 sec d
 sec  tan   ln(sec   tan  )
x2 1
x2 1 1 3
)x  ln(
 ) ]1
x
x
x
10 1
2
 [ 10  ln(
 ) ]  [ 2  ln(
 1) ]
3
3
1
 2.301987535
 [(
 2.3 meters
Since rope costs 27 billion marks per meter, Herr Geldkostenzähler will have to cough up about 62.1 billion marks to
leave this world behind, assuming that the price of rope hasn’t risen in the time it takes him to figure out how much he
 owes.
Chap 8: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Improper Integrals
Solve

1
0
x ln x dx
Use integration by parts:


1
0
x ln x dx  lim
b 0

1
b
x ln x dx
Let u  ln x and dv  xdx
du 
dx
x
v
1 2
x
2
1
1
1
(ln x)( x 2 )   ( x 2 )( )dx
2
2
x
1

1
x 2 
 lim (ln x)( x 2 )  
b 0
2
4 b
1
1
(1) 2 1 2
(b) 2 
 lim  (1) 2 ln(1) 
 (b) ln( b) 

b 02
4 2
4 b
1
 1 1
(b) 2 
 lim 0   (b) 2 ln( b) 

b 0
4 2
4 b
The lim (b2 ln b) is indeterminate in this form, so rewrite it as lim (
b 0


b 0
ln b
) and use L’Hôpital’s Rule.
b2
1
2
ln b
b )  lim ( b )  0
)

lim
(
b 0 b 2
b 0 2
b 0
2
b2
Therefore :
1
1
0  14  0  0 
 0 x ln x dx  lim
b 0
4
lim (



Chap 8 □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): (parts, surf area): Surface Area
The arc of the curve of y = ln x lying in the fourth quadrant is revolved around the y-axis. Find the surface area generated
by the created volume.
1
 dx 2 2
ds  1    dy

 dy  

1
 dx 2 2
dS  2xds  2x1    dy

 dy  

y  ln x
dy 1
dx


x
dx x
dy
dx
dy 
x

Upon substitution:

1
 dx 2 2
dS  2x1    dy

 dy  

1
dx 
 2x 1 x 2 2  
 x 
1
 2 1 x 2 2 dx
 S  2
1
2 2
 1 x  dx
1
0
Let x  tan  and dx  sec 2  d
 S  2
1
 1 tan 2  2 sec 2  d
1
2
 2  (sec  ) sec 2  d  2  sec 3  d
2
Integrate by parts:

Let
u = sec 
du = sec  d 

 
dv = sec2  d 
v = tan 
 

2  sec 3 d
 tan  sec  d 

 2 tan  sec    (sec  1)sec  d 
 2 tan  sec    (sec   sec  )d 
 2 tan  sec  
2
2
3
tan  sec    sec  d 

 2 
2




tan  sec   ln tan   sec  
 2 

2


  tan  sec   ln[tan   sec  ]



 S   x x 2  1  ln x  x 2  1
 S
1
0
 2  ln  2  1
 Chap 8: □ Sample problem (by Derek Caetano-Anolles & Ken Cebrian): Integral or Something
8sin 
2 

d

3
cos  cos 2  

Find the anti-derivative of 
. You will have to use partial fractions.
tan 2  1
First simplify the integral to make it less nasty, then use a substitution followed by partial fractions to seal the
deal.
8sin
2 

d

3
cos  cos2  


tan 2  1
8tan  sec 2   2sec 2 
 
d
tan 2  1
sec 2  [8tan   2]
 
d
tan 2  1
Substitution :
u  tan  ; du  sec 2 



sec 2  [8(tan  )  2]
d 
tan 2  1

8u  2
du
u 2 1

8u  2
du 
u 2 1
8u  2
 (u  1)(u 1) dv
Solve by partial fractions :
8u  2
A
B


(u  1)(u 1) u  1 u 1
 8u  2  A(u 1)  B(u  1)
 A3 ; B5
8u  2
3
5



(u  1)(u 1) u  1 u 1
Therefore :
 3
8u  2
5 
 (u  1)(u 1) du   u  1  u 1du 
3
5
 
du  
du
u 1
u 1
 3ln u  1  5ln u 1  C
Don’t forget to put add a +C and revert back to your theta variables.

3ln u  1  5ln u 1  C
 3ln tan   1  5ln tan  1  C
Chap 8: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an advanced-level problem.
Students need to be familiar with the concept of integration by parts, and will also be required to derive the integral of
 tanx, and use a u-substitution. This problem could be used in a review that covers several topics, after students have
learned integration by parts.
 x sec
2
xdx  ?
Solution: For this problem we will begin by using integration by parts. We will then have to derive the integral of tanx and
use a u-substitution in order to find our solution.
For integration by parts, remember:
 udv  uv   vdu
The goal is to ensure that vdu is simpler than udv. So, we set u = x and dv = sec2xdx. Then du=dx, and v=tanx. So,
 x sec
2
xdx   udv  uv   vdu
 x tan x   tan xdx
Now, we need to figure out what
 tan xdx is:
sin x
 tan xdx   cos x dx
let u = cosx, du = -sinxdx
 
du
1
1
  ln u  ln  ln
 ln sec x
u
u
cos x
Therefore,
 x sec
2
xdx  x tan x   tan xdx  x tan x  ln sec x  C
Chap 8: □ Sample problem (by Emily Selen, Alex Chew, & Brian Atchley): This is an advanced-level problem.
Students need to be able to do partial fractions, as well as u-substitutions. This could be used as a more advanced
problem after students are familiar with these concepts.
Integrate:
 2x
3x  2
dx
2
 3x  1
Solution:
To solve this problem, we will first use integration by parts, then use two u-substitutions.
First, you need to factor the denominator:
3x  2
 (2 x  1)( x  1) dx
We can now use partial fractions:
3x  1
A
B


(2 x  1)( x  1) 2 x  1 x  1
Eliminate fractions:
3x  2  A( x  1)  B(2 x  1)
 Ax  A  2Bx  B
 x( A  2 B)  ( A  B)
Therefore, A+2B = 3 and A+B = 2
A=B=1
3x  2

1
1 
1
1
 (2 x  1)( x  1) dx =   2 x  1  x  1 dx =  2 x  1 dx   x  1 dx
For the first, let u = 2x+1, du = 2dx:
1 du 1
1
= ln u  ln 2 x  1

2 u
2
2
For the second part let u = x+1, du = dx:
Putting it together, our final answer is:

du
 ln u  ln x  1
u
1
ln 2 x  1  ln x  1  C
2
1
□ Sample problem (by Fan Huang & Fernanda Mendez): Chap 8: Evaluate
Fix: Do as improper integral.
Check to see if function is continuous over the interval (0, 1).
Function is discontinuous when the denominator is equal to zero:
7x – 4 =0
2x 2
0 7 x  4 dx .
x=
4
7
at x =
Integral is undefined - Fundamental Theorem of Calculus doesn’t apply because of discontinuity
4
.
7
□ Sample problem (by Fan Huang & Fernanda Mendez): Chap 8 (Tab meth.):
Evaluate
3 x
 ( x 3 dx) .
Solution:
Tabular Method:
+
x3
3x
-
3x2
3x
(ln 3)
+
3x x3 3x x 2
3x 6x
3 x (6)
3 x
=



C
(
x
3
dx
)

ln 3 (ln 3) 2 (ln 3) 3 (ln 3) 4
+
6x
3x
(ln 3) 2
6
3x
(ln 3) 3
0
3x
(ln 3) 4
□ Sample problem (by Fan Huang & Fernanda Mendez): Chap 8: L’Hopital’s Rule, Evaluate lim x x .
x 0
Solution:
y = lim x
x
x 0
ln (y) = lim ln( x x )
x 0
ln (y) = lim [ x ln( x)]
x 0
ln (y) = lim
x 0
ln( x)

=
1

x
Put what’s inside the limit in indeterminate form so that we can use
L’Hopital.
d
1
ln( x)
L’Hopital: ln (y) = lim dx
 lim x 2  lim  x  0
x 0
x 0
d  1  x 0  x
 
dx  x 
y = e0
y=1
lim x x = 1
x0
7 □ Sample problem (by Fan Huang & Fernanda Mendez):
triangle trig Chap 8: Evaluate
7 x  4 dx
 x(3x  2)
2
Partial Fractions, u-substitution, trig-substitution, and basic
.
Solution: Split the integrand into partial fractions:
7x  4
A
B
C
 

2
x 3 x  2 (3 x  2) 2
x(3 x  2)
A(3x+2)2 + B(x)(3x+2) + C(x) = 7x + 4
(9x2 + 12x + 4)A + (3x2+2x)B + Cx = 7x + 4
9Ax2 + 3Bx2 + 12Ax + 2Bx + Cx + 4A = 7x + 4
(9A + 3B)x2 + (12A + 2B + C)x + 4A = 7x + 4
Coefficients on the left have to match up with the coefficients on the right, so:
4A = 4
A=1
9x2A + 3x2B = 0x2
9A + 3B = 0
9(1)+3B=0
B = -3
12xA + 2xB + Cx = 7x
12A + 2B + C = 7
12 – 6 + C =7
C=1
7x  4
1
3
1
 

2
x 3 x  2 (3 x  2) 2
x(3 x  2)
Substitute A, B, and C back into the partial fractions:
7 x  4 dx
 x(3x  2)
1
 x dx
2
=
3
1
1
 x dx   3x  2 dx   (3x  2)
2
dx
 ln x  C
3
U Substitution:
3
 3x  2 dx

 3x  2 dx

Trig Substitution:
 u  3x  2
du  3 dx
 du
  ln 3 x  2  C
u
1
 (3x  2)
2
dx  x 
1
 (3x  2) 2 dx 
4
tan  sec 2 
3
d 
2
 2

2
 (3) tan   2 
 3

V Substitution:

2
tan 2 
3
dx 
4
tan  sec 2  d
3
4
tan  sec 2 
1 tan  sec 2 
1
3
d


d   tan  cos 2  d
4
 2 2 tan 2   1 2

3
3
sec 


1
1
tan  cos 2  d   sin  cos  d  v  sin 

3
3
dv  cos  d

1
1 2
sin 2 
3x
x
v
dv

v

C

C 
C 
C

3
6
6
6(3x  2)
6x  4
2
tan 2 
3
x
tan
1
3x  2
3x
3

( x)
2
7 x  4 dx
 x(3x  2)
2
sin  
3x
3x  2

2
= ln x  ln 3 x  2 
x
x
x
 C  ln

C
6x  4
3x  2 6 x  4
Cauchy Mean Value Theorem: pg 615, Salas
L’Hospital’s Rule:
Chap 8: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu): L’Hospital’s
rule Find
7 x  5 x  3x  1
.
x 
3x 4  7 x
4
2
lim
Solution: Since both numerator and denominator tend toward infinity as x  , we have the indeterminate form  / .
L’Hospital’s rule does apply to this form. We begin by taking the limit as x   of the derivative of the top over the
derivative of the bottom. In doing so, we see that we will have to use L’Hospital’s rule repeatedly. Eventually, we arrive
at the limit of a constant:
28 x 3  10 x  3
7 x 4  5 x 2  3x  1
168 x
168 168
84 x 2  10
lim
lim
 lim
 lim

=
= 2.33.

lim
4
3
2
x 
x 
x   72 x
x   72
x 
72
3x  7 x
12 x  7
36 x
1. Clearly explaining the relationship between L’Hospital’s rule and horizontal asymptotes (if they exist) for
rational functions.
□ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu):
2  x2  4
that are in indeterminate form 0 0 .
x 0
x2
1
1
3



1 2
1 2
2
2
2
 ( x  4) (2 x)
 ( x  4)  ( x)(  )( x  4) 2
2  x2  4
2
2
 lim
 lim
Solution: lim
2
x

0
x

0
x 0
2
2x
x


1
1
x
  1 (  1  0)   1 .
 lim  

x 0 2 
4
x 2  4 2 ( x 2  4) 3  2 2

Using L’Hopital’s
Rule to evaluate limits of function lim
Chap 8: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu): Improper integral
Does the integral


2
1
dx converge?
x 1
3
(using the Limit Comparison Test to evaluate if a function converge or diverge)
Solution: By definition,


2
1
dx =
x3  1
lim

N
N  2
1
dx
x 1
3
We cannot evaluate the latter integral, there is no simple formula for the
Since
1
1
1
 3,
>0 on [ 2, ] and 3
3
x 1
x 1 x


2

1
.
x 1
3

1
1
dx   3 .
3
x 1
2 x
1
x
3
2
N
 lim
N 
1
x
3
2
N
 1

 lim  x  2 
N 
 2
2

1
1 
 lim 
 (
)
2
N 
2( 2 2 ) 
 2N
1
 0
8
1
 
8
Therefore, the given infinite integral


2
1
dx converges.
x 1
3
Chap 8 (volumes, parts): □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu):
Find the volume of the solid when the region bounded by f ( x)  x sin x  2 and g ( x)  x 2  1 is rotated around the
x-axis. (using Washer method to find the volume of a solid and use the short cut of Integration by Parts, Tabular Method)
Solution:
x sin x  2  x 2  1
intercept on x :  .7696426
 xsinx  2 2  ( x 2  1) 2 dx

.7696426 

V 

.7696426
.7696426
x sin x  x
4
.7696426

 2 x 2  1 dxV
   x sin xdx    ( x 4  2 x 2  1)dx
   x sin xdx   [ 15 x 5  23 x 3  x]
u
+ x
- 1
+ 0
dv / dx
sin x
-cos x
-sin x

Tabular method
 x sin xdx   x cos x  sin x

V   x cos x   sin x  15 x 5  23 x 3  x .7696426
.7696426
 1.7431  1.7431
 3.4862
□ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu):
1
3

Evaluate
0
9  x2
dx .
(The limit of an Improper Integral. “If the limit exists, thee integral converges and the value of the integral is the limit.
If the limit doesn’t exist, the integral diverges and the integral cannot be evaluated there.”

Solution: If any x value in [a, b] makes f (x) undefined, then
here, since the integrand is undefined at the upper limit, x =
b
a
f ( x)dx is an improper integral. That’s the situation
3 . So, we rewrite the integral with the limit as b
approaches 3 from the left. Then we transform the integrand into sine inverse form by factoring out a 9 inside the
radical and using a u-substitution, remembering to change our limits we do:

0
1
3
9  x2
 lim 
b 3

dx  lim 
b/3
0
b 3
(3 du )
3 1 u
2

1
b
dx  lim 
9  x2
0
b 3
b
1
0
3 1  ( 13 x) 2

dx



u  13 x
du  13 dx
 lim  [sin 1 u ]bo / 3  lim  [sin 1 ( 13 b)  sin 1 (0)]  sin 1
b 3
b 3
3
3
.
Chap 8: □ Sample problem (by Frances Ha, Lusiana Hadi, & Amity Xu): Partial fracs w/ diffeq
Find the solution of the differential equation
dy y 3  3 y 2  y  3  e x



2x
dx
4y  9
 1 e

 , with the initial condition: y(0) = -1.


(Use various kind of technique of integration to solve an initial value problem. The technique use here includes partial
fraction, u-substitution and trig-substitution).
Solution:
dy y 3  3 y 2  y  3  e x



2x
dx
4y  9
 1 e
 ex
4y  9

dy


2x
y3  3y 2  y  3
 1 e
To evaluate the left side of
y
3

dx 




 . Separating the variables and integrating,


 ex 
4y  9
dy

  1  e 2 x dx (*)
y3  3y 2  y  3


4y  9
dy , we use partial fractions:
 3y 2  y  3
4y  9
A
By  C

 2
2
( y  3)( y  1) ( y  3) ( y  1)
4 y  9  A( y 2  1)  ( By  C ) ( y  3)
4 x  9  Ay 2  A  By 2  3By  Cy  3C
4 x  9  ( A  B ) y 2  (3B  C ) y  ( A  3C )
( A  B)  0
(3B  C )  4
( A  3C )  9
-3
3
31
A
, B
,C 
10
10
10
So
y
3
4y  9
dy
 3y2  y  3


3
y  31
 3 10
dy   10 2 10 dy
y3
y 1
3
1
31
1
3
y
dy   2
dy   2
dy

10 y  3
10 y  1
10 y  1
u  y2 1
du  2 ydy

3
31
3
ln y  3  tan 1 y 
ln y 2  1 dy
10
10
20
Next, evaluate the right hand side of (*),.
 ex


  1  e 2 x dx


x

 ue


du  e x dx

du

1 u2
 sin 1 u  C
 sin 1 (e x )  C
Thus
3
31
3
ln y  3  tan 1 y  ln y 2  1  sin 1 (e x )  C .
10
10
20
Use the initial condition y(0)=-1,
3
31
3
ln 2  tan 1 (1) 
ln 2  sin 1 (e 0 )  C
10
10
20
C  1.6162
3
31
3
ln y  3  tan 1 y 
ln y 2  1  sin 1 (e x )  1.6162
10
10
20
Chap 8: □ Sample problem (by David Mesri & Jake Mathis): Using integration by parts, find
Solution: The formula that must be known in order to apply integration by parts is as follows:
 u dv
x

u
 uv 
sec 2  x  dx

v
 v du
x
secx 2  dx .
u  x
dv  sec 2 x  dx
du  dx
v  tan x 
x tan  x    tan  x  dx

 tan x  dx
*We must rewrite tan (x) x
sin x 
 cosx  dx

z  cosx 
*Perform a simple substitution
dz   sin x  dx

1
  z  dz
 ln z   C
ln cos x   C
*Take answer and replace
tan (x) x w/ it
x tan  x   ln cos x   C

*Don’t forget minus times minus
is positive
Chap. 8: □ Sample problem (by David Mesri & Jake Mathis): With your knowledge of partial fractions, solve this
x
integral:
15 x 4  2 x 2  6
4
 3x 3  7 x 2  21x
dx
Solution: The first step to solving this problem would be to rewrite it.
x
=
15 x 4  2 x 2  6
4
 3x 3  7 x 2  21x
15 x 4  2 x 2  6
 xx

 3 x
2
 7

dx =
15 x 4  2 x 2  6
 x
2

 3x x 2  7
dx *Factor out an x
=




 dx
*Factor the numerator
15 x 4  2 x 2  6


xx  3 x

2
 7



A

x
 
B
Cx  D

x  3
x2  7





15 x 4  2 x 2  6  A x  3 x 2  7  Bx x 2  7  Cx  D x  x  3
6  A3 7 
if x = 0:
if x = -3:
2

 A
7

1215  18  6  B 32
 200.5  B






15 x 4  2 x 2  6  A x 3  3x 2  7 x  21  B x 3  7 x  C x 3  3x 2  D x 2  3x

 2 x 2  3 Ax 2  3Cx 2  Dx 2
0 x 3  Ax 3  Bx 3  Cx 3
0  
 2
 2811 
 2  3    3
  D
 7
 14 
2
401

 C
7
2
2811
 C
14

1207
 D
2
So, rewriting the integral
 2 7
 401 2


x
x

3

 

  1207 2
 dx
x2  7



 2 7
2811 14x   1207 2 dx
 401 2




 x
x  3
x2  7
x 2  7 




2811 14x






2
401
2811
1207
1
 ln x 
ln x  3 
ln x 2  7 
dx

2
7
2
28
2
x  7

Chap 8: □ Sample problem (by Morgan Holbrook & Justin Parks): Integrate:
 sec
Solution:
(tan
=> 
2
4
( x) tan 4 ( x)dx
 (u
6
 sec
2
 sec
4
( x) tan 4 ( x)dx
( x) tan 4 ( x) sec 2 ( x)dx
( x)  1) tan 4 ( x) sec 2 ( x)dx
u= tan(x)
=>
=

du= sec2(x)
 u 4 )du
1 7 1 5
1
1
u  u
tan 7 ( x )  tan 5 ( x )  C
5 +C = 7
5
=> 7
Chap 8: □ Sample problem (by Morgan Holbrook & Justin Parks): Integrate:
Solution:
=
A
B
C


( x  4) ( x  2) ( x  3)
=> 1=A(x-2)(x-3)+B(x-4)(x-3)+C(x-2)(x-4)
=> x=4; A=1/2
x=2; B=1/2
x=3; C= -1
=
1
1
1
1
dx  
dx 

2 x 4
2 x 2
1
=
1
1
ln x  4  ln x  2  ln x  3  C
2
2
 x  3dx
Chap 8: □ Sample problem (by Morgan Holbrook & Justin Parks):
1
 ( x  4)( x  2)( x  3)dx
x
3
cos( x)dx
NO TABULAR METHOD!
Solution: u = x3
dv = cos(x)dx
du = 3x2 v = sin(x)
x 3 sin( x)  3 x 2 sin( x)dx
=
u = x2 dv = sin(x)dx
du = 2x v = -cos(x)
x 3 sin( x)  3x 2 cos( x)  6 x cos( x)dx
=
u=x
dv = cos(x)dx
du = dx v = sin(x)
=
x 3 sin( x)  3x 2 cos( x)  6x sin( x)  6 sin( x)dx
3
2
= x sin( x)  3x cos( x)  6x sin( x)  6 cos( x)  C
Chap 8: □ Sample problem (by Liz King & Katherine Wallig): This problem uses the method of integration by parts,
and just as a warning do not give up in the middle of the problem, you will have to use the method twice in order to get to
an answer.
x
2
cos xdx
Solution: Remember that integration by parts is  udv  uv   vdu , so first determine what u and dv should be.
Let u=x2 and dv=cosxdx
So du=2xdx and v=sinx
x
2
cos xdx  x 2 sin x   sin x 2 xdx
Can you spot the problem yet?...okay I’ll tell you, we need to use integration by parts again for the antiderivative of
sinx2x.
This time let u=2x and dv=sinxdx
Which means du=2dx and v=-cosx
 2 x sin xdx  2 x cos x    cos x2dx  2 x cos x  2 cos xdx  2 x cos x  2 sin x
Put both of the answers together to get the final answer of
x
2
cos xdx  x 2 sin x  2 x cos x  2 sin x
chap 8: □ Sample problem (by Merla Hübler & Lisa Portis): Basic hyperbolic inverse functions & trig substitution
Find

dx
9  9x 2
using trig substitution and then using hyperbolic functions.
Solution: Add trig sub here
Method #2:
d
sinh 1 x 
dx
1
1 x2
In this particular problem you can factor out the 9
to get it into the general form.
dx

9  9x 2
1
1
dx
= sinh 1 x  c

3
3 1 x2
=
chap 8: □ Sample problem (by Merla Hübler & Lisa Portis): integration by parts using the tabular method: Find
x
3
cos xdx .
Solution:
(+)
(-)
(+)
(-)
cos x
0
dv
dx
u
x3
3x2
6x
6
cos x
sin x
-cos x
-sin x
 x cosxdx  x sinx  3x cosx  6xsinx  6cosx  C
3
3
2
Check:
d 3
( x sin x  3x 2 cos x  6 x sin x  6 cos x  C )
dx
 (3x 2 sin x  x3 cos x)  (6 x cos x  3x 2 sin x)  (6 sin x  6 x cos x)  (6 sin x)
 x3 cos x
chap 8: □ Sample problem (by Merla Hübler & Lisa Portis): integration by parts. get integral in terms of itself

Find e3 x cos 7 xdx .
Solution: Because the problem involves two distinct parts that can easily be integrated separately, integration by parts
will be used. The tabular method could not be used in this case, since the derivatives of both e3x and cos7x never go to
zero. In this example, u was chosen to be e3x, but it could have also been cos7x.
u  e 3 x du  3e 3 x dx   udv  uv   vdu
Because the integration


3x
3x 1
3x
1
1
dv  cos 7 x v  7 sin 7 x    e cos 7 xdx  e ( 7 sin 7 x)   ( 7 sin 7 x)(3e )dx
(
1
7
sin 7 x)(3e3 x )dx still has two distinct parts, integration by parts should be used again. To avoid undoing the work
just done, u should be du from the first integration, and dv should be v.
u  3e 3 x du  9e 3 x dx


   ( 17 sin 7 x)(3e 3 x )dx = 3e3x ( 491 cos 7 x)  499  cos 7 xe3 x dx
1
1
dv  7 sin 7 xdx v   49 cos 7 x 
 I=  cos 7 xe3 x dx . Now, notice the integration that’s left is the same as the original, but multiplied by 9/49, so call it
“I” and solve for it.
I  e 3 x ( 17 sin 7 x)  3e 3 x ( 491 cos 7 x)  499 I
58
49
I  17 e 3 x sin 7 x  493 e 3 x cos 7 x  C
I
49
58
( 17 e 3 x sin 7 x 
I
7
58
e 3x sin7x  583 cos7x  C
  e 3 x cos 7 xdx 
3
49
7
58
e 3 x cos 7 x)  C
e 3x sin7x  583 cos7x  C
chap 8: □ Sample problem (by Merla Hübler & Lisa Portis): partial fractions. matrices Find
4 x 3  7 x 2  6 x  12
 x 2  3 x  12 dx .


Solution:
4 x 3  7 x 2  6 x  12
x
2

 3 x  1 ( x)
2

Ax  B
C
D
E
2


 . Multiply both sides by x 2  3 x  1 and
2
2
x
x  3 x  1 x  1


reduce to obtain:
4 x 3  7 x 2  6 x  12  ( Ax  B)( x)( x  1) 2  C ( x)( x  1)( x 2  3)  D( x)( x 2  3)  E ( x 2  3)( x  1) 2 . Plug in
various values for x that are useful in negating terms:

x  1 : 4  7  6  12  2 D
D  92

x  0 : 12  3E

x  3 : ( 3 A  B)( 3 )( 3  1) 2  12 3  21  6 3  12
E  4
3 A  3B  18( 331)332

x  1:
4( A  B)  8C  2D  8E  4  7  6  12
4 A  4 B  8C  2( 9 2 )  8(4)  29
A  B  2C  3 2

x  2 : (2 A  B)(9)( 2)  C (2)(3)(7)  D(2)(1)  E (1)(9)  4(2) 3  7(4)  12  12
36 A  18B  42C  2( 9 2 )  9(4)  84
36 A  18B  42C  111
Solving this set of equations could be potentially nerve-racking, therefore it is very useful to use your graphing
calculator and solve them with a matrix:
Equations:
A  B  2C  3 2
36 A  18B  42C  111
3 A  3B  8.6
3
1

1
2
2


Matrix form: 36 18 42 111 
3
3 0  8.6

To use the calculator, put the matrix in and then use “rref” to solve for it in the form of an identity matrix, the solutions
to A, B, and C consecutively will be on the right.
A  4.67377  4.7
B  3.1311  3.1
C  0.02132  0.02
D  92
E  4


Ax  B
1
1
1
dx  (C ) 
dx  ( D) 
dx  ( E )  dx
2
2
1 x
x
x 3
( x  1)
 ( A) 
x
1
1
1
1
dx  ( B )  2
dx  (C ) 
dx  ( D) 
dx  ( E )  dx
2
1 x
x
x 3
x 3
( x  1)
2

 u  x 2  3, du  2 x


 tanh 1  ax  
1
a

1
a2  x2
dx
  v  x  1,
dv  dx

1
1
1
1
1
  A2  du  ( B ) 
dx

(
C
)
dv

(
D
)
dv

(
E
)
dx
2



u
v
x
v
( 3) 2  x 2

4.7
2
ln( x 2  3)  ( 33.1) tanh 1
   (0.02) ln( x  1)  3(
x
3
9
2
) x 3  (4) ln( x)  c
 x 
27
 2.35  ln(x 2  3)  1.79  tanh 1 
  0.02  ln(x  1)  3  4  ln(x)  c
2x
 3
chap 8: □ Sample problem (by Merla Hübler & Lisa Portis): L’Hopital’s Theorem. using natural log Find
lim x 2 / x
x
2
.
Solution: The limit is currently in the indeterminate form of  0 , which is a form that L’Hopital’s theorem can’t be
used on. However, by calling the limit “f(x)” and taking it’s natural log, it could be transformed into a form that
L’Hopital’s theorem can be used on. This can only be done because the natural log function is a one-to-one function.
2
2
f ( x)  lim x x
x 
2
2
ln f ( x)  ln( lim x x )
x 
* f (lim g ( x))  lim f ( g ( x))
x c
2
x2
 lim ln( x )
x c
x 
 lim
2
2
x  x
ln x  2 lim
ln x
2
x  x
Now that the limit is in the form
lim
1
x
x  2 x
 2 lim
1
2
x  2 x


, L’Hopital’s theorem can be used:
 2(0)  0
So, ln f(x)=0 and f(x)=1.
chap 8: Kevin Stanford & Mike Mitchner integration by parts Evaluate
 xe
9x
dx .
Solution: To find the solution to this problem integration by parts and the formula,
 udv  uv   vdv , needs to be
used. In order to do so we will first substitute a value for u, in this case x, then the value for dv will be the rest of the
initial integral, e9x. Then, the values of du and v can then be derived and the equation solved using the before
mentioned formula.
 udv  uv   vdv
Let u  x & dv  e
9x
du  dx & v 
Then,
1 9x
e
9
1
1

dx  x e 9 x    e 9 x
9
9

x 9x 1 9x
9x
xe
dx

e  e C

9
81
 xe
9x
The most important thing to be able to do in these problems is to designate good values for u and dv. A nice rule of
thumb when dealing with problems involving e multiplied by another variable, such as we have here, is to make the
other variable u, and designate the portion of the integral with e in it as dv.
sin x
.
x 0
x
chap 8: basic Patrick McCall & Nathan Dornfeld: Using L’Hopital’s Rule, find lim
Solution: This is a 0 / 0 indeterminant form, so L’Hospital’s rule applies:
d
(sin x)
sin x
cos x
dx
lim
 lim
 lim
1
x 0
x 0
x 0
d
x
1
( x)
dx
chap 8: Derive the integration by parts formula.
Let u and v each be a function of x. Using the product rule,
d
uv   u dv  v du
dx
dx
dx
dv d
uv   v du . Multiplying through by dx, we have u dv  d uv   v du . Finally, integrating both sides

dx dx
dx
of the equation yields  u dv   d uv    v du   u dv  uv   v du .
 u
chap 8: intermediate Patrick McCall & Nathan Dornfeld: integration by parts and the tabular method. Find
x 3 cos 2 xdx .

Solution: Since the integrand is comprised of a polynomial and a trig function, we let u
= the polynomial and dv = the trig function times dx. Since repeated differentiation of u
will eventually yield zero, this problem can be handled quickly with the tabular method.
(Without it we would have to use integration by parts three times.) Thus,
x
3
+
u
x3
-
3x2
+
6x
-
6
cos 2 xdx
1
1
1
1
cos 2 x  6 x  sin 2 x  6  cos 2 x  C
= x  sin 2 x  3 x 2 
2
4
8
16
3
2
x sin 2 x 3x cos 2 x 6 x sin 2 x 3 cos 2 x



C.
=
2
4
8
8
3
chap 8: intermediate Patrick McCall & Nathan Dornfeld: trig integral Find
 5 sin
0
5
dv / dx
cos 2x
1
sin 2 x
2
1
 cos 2 x
4
1
 sin 2 x
8
1
cos 2 x
16
x dx .
Solution: If we had the same integral with a factor of cos x, we would have a simple u-substitution problem by letting
u = sin x, yielding du = cos x dx. However, with this factor of cos x, it is a more difficult problem. Since we have an
odd power of sin x, we break one off to become part of du. Then we let u  cos x , so du   sin x dx :
 5 sin

5




x dx = 5 1  cos 2 x sin x dx =  5 1  u 2 du =  5 1  2u 2  u 4  du

2

=  5 du  10 u 2 du  5 u 4 du =  5u 
2
10 3
10
u  u 5  C =  5 cos x  cos 3 x  cos 5 x  C
3
3
e12 x  1
.
x 0
2x
chap 8 intermediate Patrick McCall & Nathan Dornfeld: (L’ Hop): Evaluate the limit: lim
Solution: Both the numerator and denominator approach zero as x does, so we have a 0 / 0 indeterminate situation for
d 12 x
(e  1)
e 1
12e12 x
dx
which L’Hospital’s rule applies. So, lim
= lim
= lim
= 6e0 = 6.
x 0
x 0
x 0
d
2
2x
(2 x)
dx
12 x
chap 8: hard Patrick McCall & Nathan Dornfeld: partial fractions Find
x 2  22 x  121
 x 6  24 x 5  163x 4  192 x 3  619 x 2  742 x  363 dx .
Solution: We begin by factoring the numerator and denominator. For the latter we note that 363 = 3(11)2. By the
rational root theorem, if there are any rational roots to the denominator, they must be among {1, 3, 11}, since the
leading coefficient is 1. (Note that there could be other irrational or nonreal roots.) By Descartes’ rule of signs we can
exclude any positive roots and conclude that there are 0, 2, 4, or 6 negative roots. Using synthetic division we find that
-1 and -11 are double roots. Thus the integrand factors into
use partial fractions we rewrite the integrand:
x  1
1
2
x
2
x  112
1
=
, x  -11. To
2
2
2
x  11 x  1 x 2  3 x  1 x 2  3
 3

A
B
Cx  D

 2
. Since the denominator
2
x  1  x  1
x 3
has a linear factor squared, we must list it twice, once to the first power, and once to the second. Each linear factor has
an unknown constant above it. The second degree factor requires an known linear function above it. So, we have four


unknown for which we need to solve. Multiplying through by  x  1 x 2  3 , we obtain
2


 
 


1  Ax  1 x 2  3  B x 2  3  Cx  D x  1 . When x = -1, we have 1 = 4B, or B = ¼. Substituting,
1
2
1  A x  1 x 2  3  x 2  3  Cx  D  x  1 . To solve for the remaining three unknowns, we multiply through
4
2
by 4, expand, and combine like terms: 4  4 A x  1x 2  3  x 2  3  4Cx  D  x  1

 
2
 
 12 A  12 Ax  4 Ax 2  4 Ax 3  x 2  3  4D  4Cx  8Dx  8Cx 2  4Dx 2  4Cx 3

 3  12 A  4D   (12 A  4C  8D) x  (1  4 A  8C  4D) x  (4 A  4C ) x . For two polynomials to be equal, all
coefficients of all powers of x must be equal. Since this entire polynomial is equal to 4, its constant term must equal 4,
while all other coefficients must equal zero. This results in the following system of linear equations:
2
3
3  12 A  4D  4, 12 A  4C  8D  0, 1  4 A  8C  4 D  0, 4 A  4C  0
We have four independent equations relating three unknowns, A, C, and D (since we’ve already found B), so we can use
any three of the equations to find a solution to the system. Creating an augmented matrix with the first, second, and
fourth equations, and using row operations to simplify it, we have:
12 0 4 1  12 0 4 1  1
12 4 8 0 ~  3 1 2 0  ~  3

 
 
 4 4 0 0   1 1 0 0  12
1 0  1 0 1
0 
1 0



~ 0 1 1 0 ~ 0 1 1
0 

 
0 0  8 1  0 0 1  1 / 8
1 0
1 0 0  1


1 2 0  ~ 0  2 2
0 4 1 0  12 4
0  1
1 0 1

~ 0 1 1
0  ~ 0

0 0 1  1 / 8 0
0
0
1
0
1
0
1 0 0
1

~ 0
1  1 0

0  12 4 1 
0 1/ 8 
0  1 / 8
1  1 / 8
The final row equivalent matrix is in reduced row echelon form, and clearly shows that A = 1/8, C = -1/8, and D = -1/8.
We had already determined that B = 1/4. Substituting for these values we obtain
1/ 8
1/ 4
 x / 8  1/ 8


. Integrating term by term yields
2
x2  3
x  1 x  3 x  1 x  1
1
1 1
1
1
1 x 1
 x  12 x 2  3 dx  8  x  1 dx  4  x  12 dx  8  x 2  3 dx . The first integral on the right is a simple natural
log. For the second we let u = x + 1, implying du = dx. We split the third integral into two. Thus,
1
2

2


1
1
1
1
x
1
1
dx  ln | x  1 |   u  2 du   2
dx   2
dx . For the second integral we have
8
4
8 x 3
8 x 3
3
1 2
1
1
x
1
u du   u 1 
dx
. For the third integral we let v = x2 + 3, implying dv = 2x dx. So,   2

4
4
8 x 3
4( x  1)
1
2x
1 dv
1
1
  2
dx   
  ln | v |   ln x 2  3 . (Absolute values are unneeded here since x2 + 3
16 x  3
16 v
16
16
dx
1
 x
 tan 1   
is always positive.) Finally, the last integral is a tangent inverse:  2
2
a
a x
a
 x 
1
dx
1
  2

tan 1 
 . Putting it all together:
8 x 3
8 3
 3
 x  1 x
2
2

 x 
1
1
1
1
dx  ln | x  1 | 
 ln x 2  3 
tan 1 
  C . Since
8
4( x  1) 16
3
8 3
 3
1
1 1
1
ln x 2  3   ln x 2  3  ln x 2  3 , we can combine the two logarithms to rewrite our answer as
16
8 2
8
1
1
1 | x 1|
1  x 
 x  12 x 2  3 dx  8 ln x 2  3  8 3 tan  3   C .
1
 x  1 x
2
2



What is the rational root theorem?
Explain how Descartes’ rule of signs was used to exclude positive roots.
How does the fundamental theorem of algebra apply to this factoring the denominator.
Do the synthetic division not shown in the problem above in order to factor the denominator.
Apart Function
chap 8: hard Patrick McCall & Nathan Dornfeld: integration by parts, trig identities Find
1
e
2
x
csc x
1  sin
2


x  cos 2 x 1  cos 2 x  sin 2 x dx
Solution: Let’s begin by simplifying the integrand. Using the double angle formula for cosine, 1  sin 2 x  cos 2 x
= 1  sin 2 x  (cos 2 x  sin 2 x) = 1  cos 2 x  sin 2 x . For the second factor of the integrand we use the fact that
1  sin 2 x  cos 2 x to obtain 2 cos 2 x . Our integral now becomes

1
e
2
x
1
e
2
x
csc x
sin x 2 cos x dx
2
2
csc x  2 sin x cos x dx   e x cos x dx . Now it’s time for integration by parts:
 u  ex
dv  cos x

 
x
d u  e dx v  sin x 
e
x
cos x dx  e x sin x   e x sin x dx . To find the integral on the right we use parts
 u  ex


x
x
x
x
d v  sin x 
e
cos
x
dx

e
sin
x


e
cos
x

e
cos x dx


x
d
u

e
dx
v


cos
x


x
x
x
x
 e cos x dx  e sin x  e cos x   e cos x dx . We now have the integral we’re seeking in terms of itself, so now
a second time: 



it’s just a matter of algebraically solving for the integral: 2 e x cos x dx  e x sin x  e x cos x

e
x
cos x dx 
x
e
sin x  cos x   C .
2
chap 8 Medium (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): integration by parts Find
 (1  x)e
x
dx .
Solution: Using integration by parts,
dv= e dx
so, u= 1-x
 dv   e
du= -dx
x
x
x
dx = (1 x)e

 (1 x)e x 
v  ex
(dx)
Plug it in 
 (1  x)e
 udv  uv   vdu
x
 e dx
 e (dx)
x
x
 e x  xex  e x  c

 2e x  xex  c
 e x (2  x)  c
chap 8: (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): integration by parts Find the antiderivative of
e x cos x .

Solution: Step 1: Use u-substitution
e

x
cos xdx
 udv  uv   vdu
Use
cos xdx  e
 e 
x



e
x
sin xdx



e
Step3: Set original equation equal to e x cos x  e x sin x 
x
cos xdx
 e cos xdx = e cos x  e sin x   e
x
x
e
x
cos xdx


x
cos xdx using algebra
e x sin x  C
2  e cos xdx = e cos x 

e x cos x  e x sin x
x
C
=
e
cos
xdx
 
2

Step 4: Solve for
x

cos x 
e x cos x  ex sin x   e x cos xdx
x

x
du  sin xdx v  e x dv  e x dx
Step 2: Do substitution again with new integral: u  sin x du=cosxdx v  e x dv  e x dx


u  cosx
x
chap 8: trig, partial fractions, special factoring, Mathematica Find the antiderivative of

tan x .
First have students expand (u 2  u a  1)(u 2  u a  1) and show that if this expansion equal u4 + 1,
then a = 2.

Factor 1 u4, Extension
Sqrt 2
2
1
2 u u
1
2 u
2
u
2
2
Solution: Step1 U-substitution; Step2 sec x  tan x  1 &substitute; Step3 Plug in U for
tan x ; Step4: Solve
2u
for dx; Step6; Factor u 4  1; Step7 Getting partial fractions, Multiplying u 4  1by
u 1
each side, Then foil out the rest,
Solve for A,B,C and D; Step8
 Put like terms together, Set coefficients equal, 
Substitute A,B,C and D into original equation; Step9 Split into two integrals; Step10 Perfect square


 dx

polynomials;
 arctan x  c ; Step12 Substitution U and du;
Step11; Setting up for arctan integral  2
 x 1

Step13 Break up into 4 integrals; Step14 Solving each of the 4 integrals; Step15 Combine ln’s;
for dx; Step5 Plug in
4

Step16 Back substituting, Plug in expressions for V&W that are in terms of U, Square out
 2u 1 and  2u  1 , Simplify 1+1=2, Cancel the 2’s, Plug in expression for u
2
u=


2
du 
tan x

Thus,

u4 1
sec 2 x
tan 2 x 1
2u
dx  dx  4
dx =
dx =
du
2u
u  1
2 tan x
2 tan x
tan x dx   u dx =


2u 2
 u 4 1 du

Step6: u 4  1  (u 2  u 2  1)(u 2  u 2  1)
2u 2
Au  B
Cu  D
 2
 2
4
u  1 (u  u 2  1) (u  u 2  1)
Step 7:
2u 2  ( Au  B)(u 2  u 2  1)  (Cu  D)(u 2  u 2  1)
2u 2 = Au3  2Au2  Au  Bu 2  2Bu  B  Cu3  2Cu2  Cu  Du 2  D 2u  D
2u 2 =(A+C) u 3  ( 2A  B  2C  D)u 2  (A  2B  C  2D)u  B  D


A+C=0  2A  B  2C  D=2

A  2B  C  2D =0
B+D=0
 for A, B, C and D : *=original equation
Solve
*A+C=0 * A  2B  C  2D =0 A  C  2B  2D  0  2B  2D  0
B  D  0 B=D done.

*B+D=0 B+B=0 2B=0 B=0 D=0 done.
 2
*
 2A  B  2C  D=2  2A  2C  2 -A+C=

2
2
*A+C=0 2C= 2 C 


A
 2
2

 1 

2u 2
1 
u
u

 2
+
 2

 u 4  1




2
u

2u

1
2
u

2u

1


Step 8:

Step 9:




tan xdx 
Step 10:

tan xdx =
1
2
1
2

tan xdx =

tan xdx =


u

u
1
2

1
2

2
u
1
du +
 2u  1
2
2
u
du
 2u  1
u
1
u
dudu

2
 2u  1
2 u  2u  1

u
1
u
(du)
1 1
2 u 2  2u  1  1
u 2  2u  
2 2
2 2
2
u
1
du 1 2 1
2
(u 2 
) 
2
2


u


u
du
1 2 1
2
(u 
) 
2
2
u  tan x , Simplify.
Step 11:

tan xdx =
1
2

2u
1
du 1 2
2
2(u 
) 1
2
dv= 2du w= 2u +1 w-1= 2u dw= 2du
 1
v 1
1
dv 
=

2
2 v 1
2




Step 15:
Step 16:
 





2u
du
1 2
2(u 
) 1
2
2
Make the equation look like x  1
2u
2u
- 2
du
du
 tan xdx = 2 
1 2
1 2
2(u 
) 1
2(u 
) 1
2
2

u
u
du - 2 
du
Simplify:  tan xdx = 2 
2
( 2u 1)  1
( 2u  1) 2  1
v 1
Step12: v + 1= 2u , v = 2u -1
u
2
1
v
1
dv
1
w
1
dw

 xdx 
Step13:  tan
dv   2

dw 



2
2
2 v 1
2 v 1
2 w 1
2 w2  1


------------------- ---------------

1
2
ln u  1


2 2
1
arctan u
2
Step14:

s  v2 1
ds  2vdv

1
ds
1
1

ln s 
ln v 2  1
.

2 2 s 2 2
2 2
1
dv
1

arctan v

2
2 v 1
2
1
1
1
1
ln v 2  1
arctan v 
ln w 2  1
arctan w  c
2 2
2
2 2
2





w 1
u
2
w 1
dw
2
1

v 2  1
 1
1
1
ln  2 
arctan v 
arctan w  C
tan xdx =
2 2 w  1 2
2
w
( 2u 1) 2  1 1
1
1
arctan( 2u 1) 
arctan( 2u  1)  C
ln 
tan xdx =
+
2
2
2
2
2
(
2u

1)

1



(2u 2  2 2u  1 1 1
1
1
arctan( 2u 1) 
arctan( 2u  1)  C
ln  2
tan xdx =
+
2
2 2 (2u  2 2u  1 1 2
(2u 2  

1
2 2u  2  1
1
arctan( 2u 1) 
arctan( 2u  1)  C
ln  2
tan xdx =
+
2
2 2 (2u  2 2u  2  2


2
u  2u  1 1
1
1
arctan( 2u 1) 
arctan( 2u  1)  C
ln  2
tan xdx =
+
2
2
2
2
u

2u

1




( tan x ) 2  2tan x  1 1
1
ln 
arctan( 2 tan x  1) 
arctan( 2 tan x  1)  C
tan xdx =
+
2
2 2 ( tan x )  2tan x  1 2
2
1



tan x  2tan x  1 1
1
1
ln 
arctan( 2 tan x  1) 
arctan( 2 tan x  1)  C
tan xdx =
+
2 2 tan x  2tan x  1 2
2

Integrate


2 2
ArcTan
Tan x
Tan x
^ 1 2 ,x
2 2
ArcTan
Tan x
2
2
2
Log
1
2
2
Tan x
Tan x
Log 1
2
Tan x
2 2
Have students show the answers are equivalent.
2
Tan x
2
chap 8: (by Raquel Roney, Nayeon Kang, & Ayush Dulguun): integration by parts Find the average value of ln x
on the interval [1, e].
Solution: Using
 udv  uv   vdu
e
1
ln xdx
e  1 1
u=lnX
e

e
1

1 
=
since   x 1
x ln x    dx 
x

e 1
1

1

1
1

(e  0  e  1) 
(plug in e&1)
.
e 1
e 1





x
2
5
dx , factor →
 25x  24
dv=dx
e 
1 
eln e  ln1 x 
e 1 
1 

Chap 8: (by Shaofeng Sun & Artem Rogachev): basic partial fractions. Find
Solution:
1
du= dx
x
v=x
x
2
5dx
.
 25x  24
5
A
B


( x  24)( x  1) x  24 x  1
5dx
 ( x  24)( x  1)
→ 5=(x+1)A+(x+24)B. since this equation must be true for any x pick one that will make one of the terms drop out x
5
5
x = -24, A= →
23
23
5
x 1
) + C.
ln (
x  24
23
= -1, B=
5
5
  23( x  24) dx   23( x  1) dx
= -
5
5
ln(x+24) +
ln(x+1) + C =
23
23
Chap 8: (by Shaofeng Sun & Artem Rogachev): advanced u substitution and partial fractions. Find
 sin
2
cos x
dx .
x  7 sin x  10
Solution: the solution is straightforward just follow the steps.
u
2
du
du

(u  2)(u  5)
 7u  10
 sin
let u=sinx then du=cosx
2
cos x
dx , the integral simplifies to
x  7 sin x  10
Partial fractions:
1 = A(u+5)+B(u+2) Now plug
1
1
u   2, A 
3
3
1
1
1
1
1 sin x  2
(
du  
du)  (ln u  2  ln u  5 ) = ln
C
3 u2
u5
3
3 sin x  5
.
in u values that will make one of the terms drop out. u  5, B  
Chap 8: (by Shaofeng Sun & Artem Rogachev): intermediate two u-substitutions and integration by parts. Evaluate:
 ( x  5)
7
sin( x  5) 4 dx .
Solution: Well the solution is straightforward so just follow the steps
use u=x-5, so it’ simply :
u
7
 ( x  5)
7
sin( x  5) 4 dx , for this problem we
sin u 4 du . now we use a v substitution, let v=u4 then dv=4u3du, now it’s
And now we use integration by parts, let s=
v
dv
, and df=sinvdv, ds=
, and f=-cosv
4
4
 cosv
v
sin v
dv = - cosv +
+ C plug back the beginning values for u and v
4
4
4
( x  5) 4
sin( x  5) 4
=
cos( x  5) 4 
C.
4
4


v sin v
dv .
4
v sin v
v
dv = - cosv 4
4

Chap 8: (by Shaofeng Sun & Artem Rogachev): basic L’Hopital’s rule Use L’Hopital’s rule to find the limit:
Lim
x 
2x
x2
2x
Lim 2
Solution: x  x
since plugging in the limit gives us indeterminate form we can take
x
Lim 2 ln 2
= x 
the derivative of numerator and denominator and take the limit.
2x
Lim 2 x (ln 2) 2
= x 
=  repeating l’hopital’s rule and evaluate plugging in infinity.
2