Integration by Parts
Dr. Philippe B. Laval
Kennesaw State University
August 21, 2008
Abstract
This handout describes an integration method called Integration by
Parts. Integration by parts is to integrals what the product rule is to
derivatives.
1
1.1
Integration by Parts (5.6)
General Concept
If f and g are two di¤erentiable functions, then the product rule gives us:
0
(f (x) g (x)) = f (x) g 0 (x) + f 0 (x) g (x)
If we integrate both sides, we get
Z
Z
Z
0
(f (x) g (x)) dx = f (x) g 0 (x) dx + f 0 (x) g (x) dx
From the fundamental theorem of calculus, we know that the integral of the
derivative of a function is the function itself, therefore we have:
Z
Z
f (x) g (x) = f (x) g 0 (x) dx + f 0 (x) g (x) dx
Or, solving for the …rst integral
Z
f (x) g 0 (x) dx = f (x) g (x)
Z
f 0 (x) g (x) dx
This is the integration by parts formula. However, we usually do not remember
it this way. If we let u = f (x), then du = f 0 (x) dx. Similarly, if v = g (x), then
dv = g 0 (x) dx. Doing the two substitutions gives us:
Z
Z
udv = uv
vdu
(1)
1
The goal when
R using this formula is to pretend that the integral we are given is
of the form udv. We …nd u and dv that accomplish this. Once we have Ru and
dv, we …nd du and v. Then, we can rewrite the given integral as uv
vdu.
This will work if the new integral we obtained can be evaluated or is easier to
evaluate than the one we started with. Also, once we have selected dv, v is an
antiderivative of dv. therefore, we must be able to …nd an antiderivative for dv.
For de…nite integrals, the integration by parts formula becomes
Z
a
b
Z
b
f (x) g 0 (x) dx = f (x) g (x)ja
b
f 0 (x) g (x) dx
a
We illustrate this technique with several examples. For clarity, the quantities
we select from the integral (u and dv) will be in bold characters. The quantities
we deduce from our selection (du and v) will be in normal characters.
R
Example 1 Find x sin xdx
If we select
u=x
du = dx
v = cos x dv = sin xdx
Then, applying formula 1 gives us:
Z
x sin xdx = x cos x
Z
Z
cos xdx
=
x cos x +
cos xdx
=
x cos x + sin x + C
Remark 2 There is usually more than one way to select u and dv. However,
we should do it so that when we apply the integration by parts formula, we obtain
an integral we can solve, or at least one that does not seem more complicated.
In the above example, if we had selected
u = sin x du = cos xdx
x2
v=
dv = xdx
2
then the integration by parts formula would have given us
Z
Z
x2
1
sin x
x2 cos xdx
x sin xdx =
2
2
which is more complicated than the integral we started with.
R
Example 3 Find xex dx
Following the previous example, we select
u=x
v = ex
du = dx
dv = ex dx
2
Then, applying formula 1 gives
Z
xex dx = xex
Z
= xex
Example 4 Find
If we select
Re
1
ex dx
ex + C
ln xdx
1
dx
x
dv = dx
u = ln x du =
v=x
Then, applying formula 1 gives
Z e
e
ln xdx = x ln xj1
Z
1
e
dx
1
e
e
= x ln xj1 xj1
= (e ln e ln 1)
=1
(e
1)
Remark 5 You will note that in the problem we just did, we found that
Z
ln xdx = x ln x x + C
Example 6 Find
If we select
R
tan
1
xdx
u = tan
1
1
dx
1 + x2
dv = dx
x du =
v=x
Then, applying formula 1 gives
Z
tan 1 xdx = x tan
R
1
x
Z
x
dx
1 + x2
(2)
x
dx separately, using the substitution u = 1 + x2 . Then,
1 + x2
du = 2xdx therefore
Z
Z
x
1
du
dx
=
2
1+x
2
u
1
= ln juj + C
2
1
= ln 1 + x2 + C
2
1
= ln 1 + x2 + C
2
Using what we just found in equation 2, gives us
Z
1
tan 1 xdx = x tan 1 x
ln 1 + x2 + C
2
We will do
3
1.2
Repeated Integration by Parts
In some cases, applying the integration by parts formula one time will not be
enough. You may need to apply it twice, or more. We look at some example to
illustrate the various cases which can occur.
R
Example 7 Find x2 ex dx
If we select
u = x2 du = 2xdx
v = ex dv = ex dx
Then, applying formula 1 gives
Z
x2 ex dx = x2 ex
2
Z
xex dx
At this point, we are left with an integral that we still cannot do. However, we
see that it looks simpler than the one we started with. So, we try the integration
by parts formula again. In fact, we already did this integral in an example above,
so we will simply use the result.
Z
x2 ex dx = x2 ex 2 (xex ex + C)
= x2 ex
2xex + 2ex + C
Note that we replaced 2C by C. We only wish to denote that there is some
constant.
R
Example 8 Find x2 sin xdx
If we select
u = x2
du = 2xdx
v = cos x dv = sin xdx
Then, applying formula 1 gives
Z
x2 sin xdx =
x2 cos x + 2
We apply the integration by parts formula to
R
Z
x cos xdx
x cos xdx. If we select
u=x
du = dx
v = sin x dv = cos xdx
Then, applying formula 1 gives
Z
x cos xdx = x sin x
Z
sin xdx
= x sin x + cos x + C
4
(3)
If we use what we just found in equation 3, we obtain
Z
x2 sin xdx = x2 cos x + 2 (x sin x + cos x + C)
=
x2 cos x + 2x sin x + 2 cos x + C
Here again, we replaced 2C by C.
In these two examples, we applied the integration by parts formula several
time. We noticed that every time, the integral was getting easier. So, our hope
was that we would eventually be able to …nd it, which we did. In some cases, a
di¤erent situation will present itself, as illustrated in the next example.
R
Example 9 Find ex sin xdx
If we select
u = ex
du = ex dx
v = cos x dv = sin xdx
Then, applying formula 1 gives
Z
ex sin xdx =
ex cos x +
Z
ex cos xdx
(4)
The integral we obtain is not simpler, but it is not more di¢ cult either. Furthermore, it looks like the original, sin x having been replaced by cos x. We apply
the integration by parts formula to it. If we select
u = ex
du = ex dx
v = sin x dv = cos xdx
Then, applying formula 1 gives
Z
ex cos xdx = ex sin x
Z
ex sin xdx
If we replace what we just found in equation 4, we obtain
Z
Z
x
x
x
e sin xdx = e cos x + e sin x
ex sin xdx
R
Therefore, we can solve for ex sin xdx. We obtain
Z
Z
ex sin xdx + ex sin xdx = ex cos x + ex sin x
Z
2 ex sin xdx = ex (sin x cos x)
Z
ex
(sin x cos x) + C
ex sin xdx =
2
Note, we inserted the constant C in the last step.
5
Using integration by parts, we can prove important formulae, called reduction formulae. The most important ones are listed as a theorem.
Theorem 10 Let n be an integer such that n
1.
2.
3.
4.
R
R
R
R
1
cos x sinn
n
sinn xdx =
cosn xdx =
1
sin x cosn
n
n
n
(ln x) dx = x (ln x)
xn ex dx = xn ex
n
R
1
n
xn
1
x+
R
n
x+
n
n
n
1R
1R
n 1
(ln x)
2.
sinn
cosn
2
2
xdx
xdx
dx
1 x
e dx
Proof. We only prove part 1. The other parts are left as an exercise.
1. Prove that
select
R
sinn xdx =
u = sinn 1 x
v = cos x
1
cos x sinn
n
=
=
sinn xdx + (n
1)
n
Z
Z
Z
x+
n
n
1R
sinn
2
xdx. If we
du = (n 1) sinn 2 x cos xdx
dv = sin xdx
Then, applying formula 1 gives
Z
sinn xdx =
Z
1
n 1
sin
sinn
sinn
1
1
x cos x + (n
x cos x + (n
x cos x + (n
1)
1)
1)
sinn xdx =
sinn
1
x cos x + (n
1)
sinn xdx =
sinn
1
x cos x + (n
1)
sinn xdx =
1
cos x sinn
n
1
x+
n
1
n
Z
Z
Z
Z
Z
Z
2. see exercise 34
3. see exercise 37
4. do as an exercise
We illustrate with an example how these formulas might be used.
6
cos2 x sinn
1
2
xdx
sin2 x sinn
sinn
2
xdx
sinn
2
xdx
sinn
2
xdx
sinn
2
xdx
(n
2
xdx
Z
1) sinn xdx
Example 11 Find
R
3
(ln x) dx = 6x
1 3
ln x
6
Using part 3 of theorem 10, we obtain
Z
3
3
(ln x) dx = x (ln x)
To …nd
R
1 2
ln x + ln x
2
3
Z
1
2
(ln x) dx
2
(ln x) dx, we apply the same formula again.
Z
Z
2
2
(ln x) dx = x (ln x)
2 ln xdx
The last integral was found in an example above (otherwise, we would …nd it
using integration by parts). It was found to be
Z
ln xdx = x ln x x
Combining the three gives us
Z
3
3
(ln x) dx = x (ln x)
3
Z
2
(ln x) dx
3 x (ln x)
3
3x (ln x) + 6
3
3x (ln x) + 6 [x ln x
= x (ln x)
= x (ln x)
3
= x (ln x)
2
Z
3
= x (ln x)
2
2
2
2
Z
ln xdx
ln xdx
3x (ln x) + 6x ln x
x]
6x
Remark 12 As this example shows, the reduction formula may have to be applied several times before we …nd the answer.
1.3
Things to know
Know and be able to apply the integration by parts formula. Remember
that this formula may have to be applied more than once.
Using integration by parts, be able to prove and use the following reduction
formulas
Z
Z
1
n 1
sinn xdx =
cos x sinn 1 x +
sinn 2 xdx
n
n
Z
Z
1
n 1
cosn xdx = sin x cosn 1 x +
cosn 2 xdx
n
n
Z
Z
n
n
n 1
(ln x) dx = x (ln x)
n (ln x)
dx
Z
Z
xn ex dx = xn ex n xn 1 ex dx
7
Related homework assigned:
– # 1-19 odd #, 25, 33, 34, 35a, 35b, 37, 39, 45 on pages 398, 399.
R
– Find sin 1 xdx
– Finish proving theorem 10
8