4.2. INTEGRATION BY PARTS 131
Problem 4.26 Perform the following definite integrals.
1.
Z
4
0
.2x 3/
8
dx
2.
Z
p
0
x sin.x
2
/ dx
3.
Z
1
0
.3x C1/e
9x
2
C6xC1
dx
4.
Z
2
1
ln.x/
x
dx
5.
Z
2
2
x
5
x
2
C 1
dx
6.
Z
4
p
3
0
2x
x
4
C 1
dx
4.2 INTEGRATION BY PARTS
Probably the easiest of the complex derivative rules is the product rule. e reversed product
rule is a technique called integration by parts. Let’s derive it.
.U V /
0
D V dU C U dV is is just the product rule
U dV D .U V /
0
V dU Rearrange
Z
U dV D
Z
.U V /
0
Z
V dU Integrate both sides
Z
U dV D U V
Z
V dU is is the formula
Knowledge Box 4.3
Integration by parts
Z
U dV D U V
Z
V dU
e technique for using this rule is not obvious from its statement. e “parts” are U and V
together with their derivatives. Examples are needed.
132 4. METHODS OF INTEGRATION I
Example 4.27 Compute
Z
x cos.x/ dx:
Solution:
Choose U D x and dV D cos.x/ dx.
en dU D dx and V D
Z
cos.x/ dx D sin.x/.
When doing integration by parts we add the CC as the last step of the integration pro-
cess.
Now that we have the parts and their derivatives, we apply the integration by parts for-
mula.
Z
U dV D U V
Z
V dU
Z
x cos.x/ dx D x sin.x/
Z
sin.x/ dx
D x sin.x/ .cos.x// CC
D x sin.x/ Ccos.x/ CC
Let’s check this result by taking the derivative:
.
x sin.x/ Ccos.x/ CC
/
0
D sin.x/ C x cos.x/ sin.x/ D x cos.x/
So the method worked.
˙
ere is a substantial strategic component to choosing the parts U and dV when doing
integration by parts. You take the derivative of U , and you must integrate dV , and when you’re
done it would be lovely if the result could be integrated without too much difficulty. In general
you choose U so that differentiation will make a problem go away.
Let’s do another simple example.
4.2. INTEGRATION BY PARTS 133
Example 4.28 Find
Z
x e
x
dx:
Solution:
We can integrate e
x
, and x goes away if we take its derivative.
So, choose U D x, dV D e
x
dx. is means d U D dx and V D e
x
.
Apply the formula:
Z
U dV D U V
Z
V dU
Z
xe
x
dx D xe
x
Z
e
x
dx
D xe
x
e
x
C C
˙
Sometimes the choice of parts is not obvious. Let’s pick up another basic integral using integra-
tion by parts.
Example 4.29 Compute
Z
tan
1
.x/ dx
Solution:
At first it might look like there are no parts, but there are.
Choose U D tan
1
.x/ and dV D dx. en d U D
dx
x
2
C 1
and V D x.
Apply the formula:
Z
U dV D U V
Z
V dU
134 4. METHODS OF INTEGRATION I
Z
tan
1
.x/ dx D x tan
1
.x/
Z
x
x
2
C 1
dx is is a substitution integral.
Let r D x
2
C 1. en
1
2
dr D x dx.
D x tan
1
.x/
1
2
Z
dr
r
D x tan
1
.x/
1
2
ln.r/ C C
D x tan
1
.x/
1
2
ln.x
2
C 1/ C C
Notice that, since U and d U are used in integration by parts, we used “r-substitution” instead
of u-substitution for the part of the integral that needed substitution.
˙
Knowledge Box 4.4
Integral of the arctangent function
Z
tan
1
.x/ dx D x tan
1
.x/
1
2
ln.x
2
C 1/ C C
Example 4.30 Compute
Z
x
2
ln.x/ dx:
Solution:
Tactically, we know that ln.x/ turns into a (negative) power of x when we take its derivative.
So a natural choice to try is U D ln.x/, dV D x
2
dx.
is makes d U D
dx
x
and V D
1
3
x
3
.
Apply the formula.
Z
U dV D U V
Z
V dU
4.2. INTEGRATION BY PARTS 135
Z
x
2
ln.x/ dx D
1
3
x
3
ln.x/
Z
1
3
x
3
dx
x
D
1
3
x
3
ln.x/
1
3
Z
x
2
dx
D
1
3
x
3
ln.x/
1
9
x
3
C C
˙
Sometimes integration by parts must be applied more than once to finish a problem.
Example 4.31 Compute
Z
x
2
e
x
dx:
Solution:
Since taking the derivative of a power of x at least makes it a smaller power of x, choose
u D x
2
, dV D e
x
dx.
en dU D 2x dx and V D e
x
.
Integrate by parts:
Z
x
2
e
x
dx D x
2
e
x
Z
2x e
x
dx
D x
2
e
x
2
Z
x e
x
dx Integrate by parts again.
U D xI dV D e
x
dx
d U D dxI V D e
x
D x
2
e
x
2
x e
x
Z
e
x
dx
D x
2
e
x
2x e
x
C 2e
x
C C
˙
..................Content has been hidden....................
You can't read the all page of ebook, please click here login for view all page.