136 4. METHODS OF INTEGRATION I
Notice that the second integration by parts in this example was very similar to the one in Example
4.28. We could have simply plugged that result in. e example was worked in full to show how
two integrations by parts are needed.
Let’s consider for a minute what happened in the last example. To deal with the x
2
part of the
integral we integrated by parts twice – reducing the power by one in each step. at means that,
for example,
Z
x
5
e
x
dx
would require integration by parts five times.
Fortunately, there is a shortcut. What happens if we integrate
Z
f .x/ e
x
dx
Choose U D f .x/ and d V D e
x
dx.
en dU D f
0
.x/ and V D e
x
.
Integrate by parts and we get:
Z
f .x/ e
x
dx D f .x/e
x
Z
f
0
.x/e
x
dx
Which, used correctly, is a remarkable shortcut.
Knowledge Box 4.5
Shortcut for
Z
p.x/e
x
dx
Suppose that p.x/ is a polynomial. en
Z
p.x/e
x
dx D
p.x/ p
0
.x/ Cp
00
.x/ p
000
.x/ C
e
x
C C
e formula in Knowledge Box 4.5 comes from applying the formula for
Z
p.x/e
x
dx many
times. Let’s use the shortcut in an example.
4.2. INTEGRATION BY PARTS 137
Example 4.32 Find
Z
x
5
e
x
dx:
Solution:
Applying the shortcut we get that
Z
x
5
e
x
dx D
x
5
5x
4
C 20x
3
60x
2
C 120x 120
e
x
C C
All we need to do is take alternating signs of x
5
and its derivatives. If the polynomial is not just
a power of x this is a little more complicated.
˙
Example 4.33 Find
Z
.x
2
C 2x C2/e
x
dx:
Solution:
p.x/ D x
2
C 2x C2
p.x/
0
D 2x C2
p.x/
00
D 2
So
Z
.x
2
C 2x C2/e
x
D .x
2
C 2x C2 .2x C2/ C 2/e
x
C C
D .x
2
C 2/e
x
C C
e key fact is that, if you keep taking derivatives of a polynomial, then at some point you
get to zero. With a non-polynomial function the shortcut can produce an infinite object.
˙
Now we look at another technique, circularintegrationbyparts. You need this technique when,
in the course of integrating by parts, you arrive back at some version of the integral you started
with. is sounds a bit awful, but it is actually good news. When this happens the integral may
be finished using only algebra.
138 4. METHODS OF INTEGRATION I
Example 4.34 Compute
Z
sin.x/e
x
dx:
Solution:
ere is no natural choice of parts. So pick anything and plunge in.
Let U D sin.x/ and dV D e
x
dx. en dU D cos.x/ dx and V D e
x
:
Z
sin.x/e
x
dx D sin.x/e
x
Z
cos.x/e
x
dx
Let: U D cos.x/ and dV D e
x
dx (Second integration by parts.)
en: dU D sin.x/ dx and V D e
x
D sin.x/e
x
cos.x/e
x
Z
sin.x/e
x
dx
Z
sin.x/e
x
dx D .sin.x/ cos.x//e
x
Z
sin.x/e
x
dx
Notice that the remaining integral equals the left-hand side.
is is the circular part.
So, 2
Z
sin.x/e
x
dx D .sin.x/ cos.x//e
x
Divide by two and we get:
Z
sin.x/e
x
dx D
1
2
.sin.x/ cos.x//e
x
C C
Notice that we had to separately remember to put in the C C , because we finessed the final
integral where we would normally have generated it.
˙
e power of these methods of integration are expanded in the homework problems. We will
also see a horrific example of circular integration by parts in Section 4.3.
4.2. INTEGRATION BY PARTS 139
PROBLEMS
Problem 4.35 Do each of the following integrals using integration by parts.
1.
Z
x sin.2x/ dx
2.
Z
x e
5x
dx
3.
Z
ln.x/ dx
4.
Z
x ln.x/ dx
5.
Z
x sin.x/ cos.x/dx
6.
Z
ln.x
2
C 1/ dx
Problem4.36 For Examples 4.28-4.31, verify the formulas by taking an appropriate derivative.
Problem 4.37 Compute
Z
x
n
ln.x/ dx for n 1
Problem 4.38 In this section we develop a shortcut for
Z
p.x/e
x
dx where p.x/ is a polyno-
mial. Find the corresponding shortcut for
Z
p.x/e
x
dx
Problem 4.39 In this section we develop a shortcut for
Z
p.x/e
x
dx where p.x/ is a polyno-
mial. Find the corresponding shortcut for
Z
p.x/ sin.x/ dx
Problem 4.40 In this section we develop a shortcut for
Z
p.x/e
x
dx where p.x/ is a polyno-
mial. Find the corresponding shortcut for
Z
p.x/ cos.x/ dx
140 4. METHODS OF INTEGRATION I
Problem 4.41 Compute the following integrals using integration by parts.
1.
Z
x
4
e
x
dx
2.
Z
.x
3
C 2x
2
C 3x C4/e
x
dx
3.
Z
x
3
e
x
dx
4.
Z
x
2
C 1
e
x
dx
5.
Z
x
2
sin.x/ dx
6.
Z
x
5
cos.x/ dx
Problem 4.42 Compute the following integrals using integration by parts.
1.
Z
cos.x/e
x
dx
2.
Z
sin.2x/e
x
dx
3.
Z
cos.x/e
3x
dx
4.
Z
sin.x/e
x
dx
5.
Z
.
cos.x/ C sin.x/
/
e
x
dx
6.
Z
cos.2x/e
5x
dx
Problem 4.43 Compute
Z
x tan
1
.x/ dx
Problem 4.44 Compute
Z
x
3
.x
2
C 1/
10
dx
Problem 4.45 Compute
Z
x
.
cos.ax/ Csin.bx/
/
dx
Problem 4.46 Compute
Z
sin.ax/e
bx
dx
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