Integration By Parts
Z
There is NO formula for
f (x)g(x)dx.
Z
It almost never happens that
Z
f (x)g(x)dx =
Z
f (x)dx
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Notice that
g(x)dx
Z
df = f (x) + C. We often shorten this to
df = f to indicate that the integral and
differential operators “cancel” each other.
d
d
d
(uv) = u
(v) +
(u)v
dx
dx
dx
The Product Rule for derivatives,
has no simple counterpart for antiderivatives. It can be restated in terms of differentials as
d(uv) = udv + vdu, and if we apply indefinite integral signs, we get
Z
Z
d(uv) =
Z
udv +
Z
vdu, or uv =
Z
udv +
vdu.
Z
We usually use the equivalent formula
Z
udv = uv −
1
vdu .
Z
Example: Evaluate
x sin xdx
Solution: Use integration by parts, with u = x, and dv = sin xdx.
Then du = dx, and v = − cos x, so
Z
Z
Z
x sin xdx =
udv = uv −
Z
vdu = x(− cos x) − (− cos x)dx =
Z
−x cos x +
cos xdx = −x cos x + sin x + C
2
We can also use this technique with definite integrals:
Evaluate
Solution:
Zπ
0
x sin xdx
Zπ
0
π
x(− cos x)|0
−x cos x|π
0 +
Z
Z
x sin xdx =
−
Zπ
Zπ
0
0
udv = uv −
vdu =
(− cos x)dx =
π
cos xdx = −x cos x|π
0 + sin x|0 =
−π cos π − (−0 cos 0) + sin π − sin 0 = −π (−1) = π
3
Question: How do I know what u and dv should be?
Students first encountering the technique of using the equation
knowing what to take for u and what to take for dv.
R
R
udv = uv − vdu have trouble
Answer: Get lots of experience. This is an area where we learn a lot from experience. The Integration
by Parts technique is characterized by the need to select u from a number of possibilities. Once u
has been chosen, dv is determined, and we hope for the best.
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The basic idea underlying Integration by Parts is that we hope that in going from
Z
udv to
vdu we
will end up with a simpler integral to work with. In the example we have just seen, we were lucky.
Let’s try it again, the unlucky way:
4
Z
Example: Evaluate
(sin x)xdx
Solution: Use integration by parts, with u = sin x, and dv = xdx.
x2
Then du = cos xdx, and v =
, so
2
Z
Z
Z
(sin x)xdx =
x2
1
sin x −
2
2
Z
udv = uv −
vdu = sin x
x2
−
2
Z
x2
cos xdx =
2
x 2 cos xdx
which involves a tougher looking integral than we started with.
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As a rule of thumb, one-third of the possible choices will lead to an easier integral, one-third will
lead to a harder one, and one-third will lead to one of equal difficulty.
It is also possible to spin your wheels, and go around in circles, as we shall soon see.
Let’s try a strategic approach to our example: u has to be selected so that udv = x sin xdx, so we
x sin xdx
,
look at the possible choices for u: x, sin x, or x sin x. Once u is selected, we have dv =
u
R
and all we have to is find du(easy) Zand v = dv (possibly very hard or impossible). Then we try to
decide if we can get anywhere with
vdu. Thus, we can let u be any factor of f (t), including 1, and
the corresponding dv is determined. (Of course, if we let u = 1, the problem of finding v is just
our original integration problem, so we will omit it.) If we cannot then find v we know we have a
non-viable selection of the pair u and dv.
We shall illustrate the rather inefficient technique of examining all the possibilities and discarding
the non-viable ones in the following examples. Organizing our information in a table is helpful:
u
x
sin x
R
dv =
v = dv
x sin xdx
u
x sin xdx
x
=
x sin xdx
sin x
=
x sin xdx
x sin x
=
sin xdx
xdx
x sin x
dx
du
vdu
R
sin xdx = dx
− cos x
R
xdx =
x2
R2
dx =
x
− cos xdx
x2
2
cos xdx
cos xdx
sin x
x sin x
+x cos x +x 2 cos x
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R
vdu
R
− cos xdx =
− sin x
R
R
x2
2
cos xdx
Better?
YES!
NO!
x(sin x + x cos x)dx NO!
There are some simple integrals where little choice is available: knowing which of a large number
techniques to use is crucial.
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Example:
ln xdx obviously requires
u = ln x, dv = dx, so that v = x and du =
Z
Z
Z
ln xdx =
udv = uv −
dx
.
x
Z
vdu = (ln x)x −
dx
= x ln x −
x
x
Z
dx =
x ln x − x + C
Z
Example:
arctan xdx also obviously requires
u = arctan x, dv = dx, so that v = x and du =
Z
Z
Z
arctan xdx =
x arctan x −
udv = uv −
dx
1 + x2
Z
vdu
= (arctan x)x −
1
ln(1 + x 2 ) + C =
2
p
x arctan x − ln 1 + x 2 + C
7
x
dx
1
= x arctan x −
1 + x2
2
Z
2xdx
=
1 + x2
Indefinite Integration of ekt times another factor
We now look at a family of integrals which show up in LSD, (Linear Systems Design), particularly in
the calculation of Laplace Transforms.
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We define G(G) = ekt G(t)dt, where G is any function. The types of continuous function G that
arise in practical situations are often sums and products of polynomials, exponential functions, and
sinusoidal functions. Fortunately, we know how to evaluate these using the technique of integration
by parts.
Examples:
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(1) G(t) = c, a constant. Integration by parts is not needed here. Then G(c) =
ekt G(t)dt =
R
(2) G(t) = t. Then G(t) = ekt tdt.
Here f (t) = ekt t has four different possible factorizations:
u
t
ekt
ekt t
dv =
ekt dt
tdt
dt
f (t)dt
u
v
1 kt
e
k
2
t
t
du
vdu
1 kt
dt
e dt
k
kt
ke dt
kt 2 ekt dt
kt
(t + k)e dt t(t + k)ekt dt
Viable?
Yes!
No!
No!
We
use the one
viable factorization,
u = t, v = ekt dt:
R
R
R
t(ekt dt)
= udv = uv − vdu =
R 1 kt
1 kt
t k e − e dt = kt ekt − k12 ekt + C
k
=
t
k
−
1
k2
ekt + C
R
(3) G(t) = t 2 . Then G(t 2 ) = ekt t 2 dt, so f (t) = ekt t 2 . This has four possible factorizations:
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c kt
e +C
k
u
dv =
kt
t
f (t)dt
u
e tdt
2
t
ekt
ekt t
ekt t 2
kt
e dt
t 2 dt
tdt
dt
v
t
k
−
1 kt
e
k
1 3
t
3
1 2
t
2
1
k2
du
e
t
kt
vdu
t
k
dt
−
1
k2
Viable?
kt
e dt
2
tekt dt
k
k 3 kt
t e dt
3
1 2
t (t + k)ekt dt
2
2
kt
2tdt
kekt dt
(t + k)ekt dt
(kt 2 + 2t)ekt dt t(kt + 2t)e dt
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Yes, but messy
Yes!
No!
No!
No!
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R
We see that if we use the factorization u = t 2 , dv = ekt dt, we get t 2 ekt dt = udv = uv − vdu =
t 2 kt
2 R
e − k tekt dt =
k
t 2 kt
e − k2
k 2
t
2t
− k2 +
k
t2
−2
k k
2
ekt
k3
ekt =
+C
R
R
(4) We notice how the problem of evaluating t 2 ekt was reduced to the evaluation
of ekt tdt, which
R kt n
had just been done. We suspect the existence of a reduction formula for e t dt. Having been
successful in taking dv = ekt dt in the two preceding examples, we decide to do this again, and we
kt n
have u = e ektt dtdt = t n . We calculate v = k1 ekt and du = nt n−1 dt, so that:
R kt n
R
R
1
n−1 R
e t dt = udv = uv − vdu = k t n ekt − k ekt t n−1 dt.
Thus we have G(t n ) = k1 t n ekt −
n−1
G(t n−1 ).
k
(5) G(t) = sin at. We have f (t) = ekt sin at, so there are just three choices for u:
u
sin at
ekt dt
ekt sin at
f (t)dt
dv = u
ekt dt
sin atdt
dt
v
1 kt
e
k
− a1 cos at
t
du
vdu
a kt
a cos atdt
e cos atdt
k
kt
ke dt
− ak ekt cos atdt
(k sin at + a cos at)ekt dt t(k sin at + a cos at)ekt dt
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Viable?
Yes
Yes
No!
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The first two choices are both viable, and we see that they both lead to the evaluation of ekt cos atdt.
We will examine both cases closely:
First Choice: u1 = sin
at, dv1 = ekt dt
R kt
R
R
R
We get: G(sin at) = e sin atdt = u1 dv1 = u1 v1 − v1 du1 = sin at k1 ekt −
1
aR
sin at k ekt − k ekt cos atdt =
sin at k1 ekt − ak G(cos at)
or
a
1
G(sin at) = sin at ekt − G(cos at)
k
k
a kt
e
k
cos atdt =
R
In evaluating ekt cos atdt we again have three choices, two of which are viable:
u
cos at
ekt dt
ekt cos at
f (t)dt
dv = u
ekt dt
cos atdt
dt
v
1 kt
e
k
1
sin at
a
t
du
vdu
a
−a sin atdt
− k ekt sin atdt
k kt
kekt dt
e sin atdt
a
kt
(k cos at − a sin at)e dt t(k sin at + a cos at)ekt dt
Again we have two choices. First and Best Choice:
We will let UR1 = cos at, dV1 =
ekt dt, and we get:
R
R
G(cos at) = R cos atekt dt = U1 dV1 = U1 V1 − V1 dU1 =
cos at k1 ekt − k1 ekt (−a sin atdt) =
1
a R kt
kt
cos
ate
+
e sin atdt =
k
k
1
a
kt
cos ate + k G(sin at),
k
or
1
a
G(cos at) = cos atekt + G(sin at)
k
k
so we are back where we started! However, if we substitute this into the equation
1
a
G(sin at) = sin at ekt − G(cos at)
k
k
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Viable?
Yes
Yes
No!
we get
a 1
a
1
a
a2
1
G(sin at) = sin at ekt −
cos atekt + G(sin at) = sin at ekt − 2 cos atekt − 2 G(sin at)
k
k k
k
k
k
k
which can be solved for G(sin at):
(1+
a2
a 2 + k2
1 kt a
k a
ekt
kt
kt
)G(sin
at)
=
G(sin
at)
=
sin
at
−
cos
ate
=
e
(sin
at
−
cos
at)
=
(k sin at−a c
e
k2
k2
k
k2
k2 k2
k2
so
G(sin at) =
k sin at − a cos at kt
e
a2 + k 2
Last and Worst Choice: We will now let U2 = ekt , dV2 = cos at, and we get:
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1
1
1
k
sin at)−
sin at(ekt kdt) = ekt sin at−
a
a
a
a
This time, however, if we substitute this into the equation
G(cos at) = ekt (
Z
ekt sin atdt =
1 kt
k
e sin at− G(sin at)
a
a
1
a
G(sin at) = sin at ekt − G(cos at)
k
k
we get
1
a
G(sin at) = sin at ekt −
k
k
!
1 kt
k
e sin at − G(sin at) = G(sin at)
a
a
so there is no information gained.
ekt , dv2 = sin atdt
Second Choice: u2 =
R kt
R
R
R
We get: G(sin at) = e sin atdt = u2 dv2 = u2 v2 − v2 du2 = ekt (− a1 cos at) −
− a1 ekt
cos at −
k
G(cos at)
a
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k kt
e
a
cos atdt =
or
1
k
G(sin at) = − ekt cos at − G(cos at)
a
a
Using, from above,
G(cos at) =
we get
1
a
cos atekt + G(sin at)
k
k
1
k 1
a
G(sin at) = − ekt cos at −
cos atekt + G(sin at) = G(sin at)
a
a k
k
so there is no new information. On the other hand, if we use
G(cos at) =
1 kt
k
e sin at − G(sin at)
a
a
we get the same value as before.
(6)G(cos at). This is easily derived from the calculations of the previous example.
1
a
a
1
G(cos at) = cos atekt + G(sin at) = cos atekt +
k
k
k
k
so
G(cos at) =
k sin at − a cos at kt
e
a2 + k2
!
=
a sin at + k cos at kt
e
a2 + k 2
a sin at + k cos at kt
e
a2 + k 2
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(7)G(t n sin at) = t n sin atekt dt
1
dv = ekt dt, so Rthat du = (nt n−1
sin at + at n cos at)dt, and v = k ekt . Then
Let u = t n sin at,
R
R
G(t n sin at) = tRn sin atekt dt = udv = uv − vdu =
t n sin at( k1 ekt ) − k1 ekt ((nt n−1 sin at + at n cos at)dt) =
1 n
t sin atekt − nk G(t n−1 sin at) − ak G(t n cos at)
k
or
1
n
a
G(t n sin at) = t n sin atekt − G(t n−1 sin at) − G(t n cos at)
k
k
k
R
(8)G(t n cos at) = t n cos atekt dt. Let U = t n cos at and dV = ekt dt,
so that dU = (nt n−1 cos at − at n sin at)dt and V = k1 ekt . Then
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R
R
R
G(t n cos at) = R t n cos atekt dt = UdV = UV − V du =
t n cos at k1 ekt − ekt ((nt n−1 cos at − at n sin at)dt) =
1 n
t cos atekt − nk G(t n−1 cos at) + ak G(t n sin at).
k
We can now solve for G(t n sin at):
G(t n sin at) =
h
i
1
n kt
n−1
(k
sin
at
−
a
cos
at)t
e
−
n(G(t
(k
sin
at
−
a
cos
at)))
a2 + k2
and hence
G(t n cos at) =
h
i
1
n kt
n−1
(k
cos
at
−
a
sin
at)t
e
+
n(G(t
(k
cos
at
−
a
sin
at)))
a2 + k2
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