More U-Substitution:The “Double-U”Substitution with ArcTan(u)
Chapter 5.5
February 13, 2007
Techniques of Integration so far…
1. Use Graph & Area ( )
2. Use Basic Integral Formulas
3. Simplify if possible (multiply out, separate fractions…)
4. Use U-Substitution…..
r2 −x2 , x , ...
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)dxa
b
∫ = f(u)dug(a)
g(b)
∫
Evaluate: sec2 (2t) tan2 (2t)dt∫
1
2tan2 (u)sec2 (u)du∫
1
2tan2 (2t)2sec2 (2t)dt∫
1
2sec2 (2t) tan2 (2t)2dt∫
u =2tdu=2dt
u =tan(2t)
du=2sec2 (2t)dt
w =tan(u)
dw=sec2 (u)du1
2w2dw∫
1
2u2du∫
Compare the two Integrals:
x 5 −xdx1
3
∫5 −xdx1
3
∫ u =5 −xdu=−dx
5 −xdx1
3
∫ x 5 −xdx1
3
∫
− 5 − x (−dx1
3
∫ )when x =1, u=5 −1=4when x=3, u=5 −3 =2
udu2
4
∫
2u3
2
32
4
=2(4)
3
2
3−
2(2)3
2
3=
16
3−
4 2
3
Extra “x”
Notice that the extra ‘x’ is the same power as in the substitution:
u =5 −xdu=−dx
x 5 −xdx1
3
∫
− x 5 − x (−dx1
3
∫ )
when x =1, u=5 −1=4when x=3, u=5 −3 =2 (5 −u) udu
2
4
∫
=10u
3
2
3−
2u5
2
52
4
=10(4)
3
2
3−
2(4)5
2
5
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
−10(2)
3
2
3−
2(2)5
2
5
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
Extra “x”
x =5 −u
(5u1
2 −u32 )du
2
4
∫
Compare:
x 3+ x2( )4dx∫ x2 3+ x2( )
4dx∫ x3 3+ x2( )
4dx∫
Let : u =3+ x2
du=2xdx
1
23+ x2( )
42xdx∫
1
2u( )4 du∫
1
2x 3+ x2( )
42xdx∫
1
2x2 3+ x2( )
42xdx∫
Since : u =3+ x2
We have: x2 =u−3
1
2(u−3) u( )4 du∫
Still have an extra “x” that can’t be related to the substitution.
U-substitution cannot be used for this integral
Evaluate:
tdt
1+ 9t4∫
Let : u =3t2
du=6tdt
1
6
6tdt
1+ 3t2( )2∫ =
1
6
du
1 + u( )2∫ =
1
6tan−1 u( ) +C
=1
6tan−1 3t 2( ) +CReturning to the original variable “t”:
Evaluate:
dt
1+ 2t−5( )2∫
Let : u =2t−5 du=2dt
1
2
2dt
1+ 2t−5( )2∫ =1
2
du
1 + u( )2∫ =
1
2tan−1 u( ) +C
=1
2tan−1 2t − 5( ) +CReturning to the original variable “t”:
Evaluate:
dt
9 + t2∫
Let : u =t3
du=13dt
=1
9
dt
1+t
3⎛⎝⎜⎞⎠⎟
2∫
=1
3
du
1 + u( )2∫ =
1
3tan−1 u( ) +C =
1
3tan−1 t
3⎛⎝⎜⎞⎠⎟
+C
We have the formula:dt
1+ t2∫ =tan−1 t( ) +C
Factor out the 9 in the expression 9 + t2:
dt
9 + t2∫ =dt
9 1+t2
9⎛⎝⎜
⎞⎠⎟
∫
=3
9
1
3dt
1+t
3⎛⎝⎜⎞⎠⎟
2∫
In general:
dt
a2 + t2∫
Let : u =ta
du=1adt
=1
a2
dt
1+t
a⎛⎝⎜
⎞⎠⎟
2∫
=1
a
du
1 + u( )2∫ =
1
atan−1 u( ) +C =
1
atan−1 t
a⎛⎝⎜
⎞⎠⎟
+C
We now have the formula:dt
a2 + t2∫ =1atan−1 t
a⎛⎝⎜
⎞⎠⎟+C
Factor out the a2 in the expression a2 + t2:
dt
a2 + t2∫ =dt
a2 1+t2
a2
⎛⎝⎜
⎞⎠⎟
∫
=a
a2
1
adt
1 +t
a⎛⎝⎜
⎞⎠⎟
2∫
Evaluate:
dt
25 + 3t+1( )2∫
Let : u =3t+1 du=3dt
1
3
3dt
52 + 3t+1( )2∫ =1
3
du
52 + u( )2∫ =
1
3⎛⎝⎜⎞⎠⎟
1
5tan−1 u
5⎛⎝⎜
⎞⎠⎟
+C
=1
15tan−1 3t +1
5⎛⎝⎜
⎞⎠⎟
+CReturning to the original variable “t”:
Use:
dt
25 + t−1( )2∫ =dt
t2 −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 +b)2 =a2 + 2ab+b2
=x2 + x +3
(2ab) 2xb =x⇒ b=12
a =x
(x +12)2 =x2 + 2x
12
+12
⎛⎝⎜
⎞⎠⎟
2
+1
4−
1
4
= x2 + x +1
4⎛⎝⎜
⎞⎠⎟
+ −1
4+
12
4⎛⎝⎜
⎞⎠⎟
=x2 + x +1
4
=(x +1
2)2 +
11
4
x2 + x+ 3
Use to solve:
How do you know WHEN to complete the square?
x2 + x+ 3 =(x+12)2 +
114
dx
x2 + x+ 3∫
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
3x2 + x+1∫
dx
2x2 −7x+ 5 ∫
Try these:
dx
x2 −x+ 2∫
dx
x2 + 7x−5 ∫
1+ x1+ x2∫ dx
In groups of two/three, use u-substitution to complete:
1. −4z−5( )∫32 +
17z+ 4
dz
2. 7e5−2rdr∫
3.x
x32 +1( )
2
+12
⎛
⎝
⎜⎜⎜
⎞
⎠
⎟⎟⎟
∫ dx
4.z
z2 −4⎛⎝⎜
⎞⎠⎟∫ dz