Transcript More U-Substitution: The “Double

More U-Substitution:
The “Double-U”
Substitution with ArcTan(u)
Chapter 5.5
February 13, 2007
Techniques of Integration so far…
1.
2.
3.
4.
Use Graph & Area ( r 2  x 2 , x , ... )
Use Basic Integral Formulas
Simplify if possible (multiply out, separate
fractions…)
Use U-Substitution…..
Substitution Rule for Indefinite Integrals

If u = g(x) is a differentiable function whose range is an interval I
and f is continuous on I, then
 f g(x) g'(x)dx   f (u)du
Substitution Rule for Definite Integrals

If g’(x) is continuous on [a,b] and f is continuous on the
range of u = g(x), then
 f g(x) g'(x)dx
g(b)
b
a


g(a)
f (u)du
Evaluate:
2
2
sec
(2t)
tan
(2t)dt

u  2t
du  2dt
1
2
2
sec (2t) tan (2t)2dt

2
1
2
2
tan
(u)sec
(u)du

2
w  tan(u)
dw  sec 2 (u)du
1
2
w dw

2
u  tan(2t)
du  2sec2 (2t)dt
1
2
2
tan (2t)2 sec (2t)dt

2
1 2
u du

2
Compare the two Integrals:
3

u5x
du  dx
5  xdx
1
3
  5  x(dx )
when x  1, u  5  1  4
when x  3, u  5  3  2
1
4

udu
2
2u
3
3 4
2
2
3
2
2(4)
2(2)


3
3
3
2
16 4 2


3
3
3
x
5  xdx
1
Extra “x”
Notice that the extra ‘x’ is the same power
as in the substitution:
Extra “x”
3
u5x
du  dx
x 5u
x
5  xdx
1
3
  x 5  x(dx )
1
4
when x  1, u  5  1  4
when x  3, u  5  3  2
 (5  u)
udu
2
4
1
2
3
2
 (5u  u )du 
2
3
2
10u
2u

3
5
5
2
4
2
3
5
3
5




2
2
2
2
10(4)
2(4)   10(2)
2(2) 




 3
5   3
5 

 

Let : u  3  x 2
Compare:
 x 3  x  dx
2 4

1
2
3

x
2
 2xdx
4
1
4
u  du

2
du  2xdx
 x 3  x  dx
2

1
2
x
3

x
2
2 4
 2xdx
4
Still have an extra “x”
that can’t be related to
the substitution.
U-substitution cannot be
used for this integral
 x 3  x  dx
2 4
3

1 2
2
x
3

x
2
Since :
 2xdx
4
u  3  x2
We have : x 2  u  3
1
4
(u

3)
u
du



2
tdt
 1  9t 4
Evaluate:
Let :
u  3t 2
du  6tdt
1
6tdt
6  1  3t 2
 
2
1 1
1
du
 tan u   C
 
2
6
6 1  u 
Returning to the original variable “t”:
 
1 1 2
 tan 3t  C
6
dt
 1  2t  5 
2
Evaluate:
Let :
u  2t  5
du  2dt
1 1
1
du
1
2dt
 tan u   C
 
2
2

2
2 1  u 
2 1  2t  5 
Returning to the original variable “t”:
1
 tan 1 2t  5   C
2
dt
 9  t2
Evaluate:
dt
1

tan
t   C
 1 t2
We have the formula:
Factor out the 9 in the expression 9 + t2:
dt
 9  t2 

dt

t2 
9 1  
9

1
 
9
Let :
dt
t
1  
 3
2
t
u
3
1
du  dt
3
1
dt
1
du
3
1 1
1 1  t 
3
 
tan u   C  tan    C
2 
2 
 3
9
3
3 1  u 
3
t
1  
 3

In general:
dt
 a2  t 2
Factor out the a2 in the expression a2 + t2:
dt
 a2  t 2 

dt
2


t
2
a 1  2 
a 

1
dt
a
 2 a
2
a
t
 
1  
 a
1
 2
a

Let :
dt
t
1  
 a
2
t
u
a
1
du  dt
a
1 1
1
du
1 1  t 
 tan u   C  tan    C
 
2
 a
a
a
a 1  u 
We now have the formula:
dt
1 1  t 
 a2  t 2  a tan  a   C
dt
 25  3t  1
2
Evaluate:
Let :
u  3t  1
du  3dt
1
3dt
1
du
1  1 1  u 

tan    C
2 
2  


2
2
 3 5
 5
3 5  3t  1 3 5  u 
Returning to the original variable “t”:
1
1  3t  1 
 tan 
C

 5 
15
dt
Use:
 25  t  1
2
dt
 2
t  2t  26
It’s necessary to know both forms:
t2 - 2t +26 and 25 + (t-1)2
t2 - 2t +26 = (t2 - 2t + 1) + 25 = (t-1)2 + 25
Completing the Square:

Comes from
ax
(2ab)
(a  b)2  a2  2ab  b2
x x3  x x
2
2xb  x  b  1 2
1 2
1  1
2
(x  )  x  2x   
2
2  2
1
2
 x x
4
2
2
1 1
 
4 4
3
1   1 12 
 2
 x  x      

4  4 4 
1 2 11
 (x  ) 
2
4
Use
1 2 11
x  x  3  (x  ) 
2
4
2
to solve:
dx
 x2  x  3

How do you know WHEN to complete the
square?
Ans: The equation x2 + x + 3 has NO REAL ROOTS
(Check b2 - 4ac)
If the equation has real roots, it can be factored and
later we will use Partial Fractions to integrate.
Evaluate:
dx
 3x 2  x  1
dx
 2x 2  7x  5
Try these:
1 x
 1  x 2 dx

dx
x 2  7x  5
dx
 x2  x  2
In groups of two/three, use u-substitution to complete:
3
1.
2.
 4z  5 
2
1

dz
7z  4
5  2r
7e
dr


x

3.  
2
3
 x 2 1 



 z 
4.   2
dz

 z  4


dx

1

2