198
4.6
4.6.1
CHAPTER 4. INTEGRALS
Integration by parts
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
0
f (x) g (x) = f (x) g (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
(4.4)
The goal whenR 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 u
and dv, we …nd du by di¤erentiating
R u and v by integrating dv. 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 b
Z b
b
f (x) g 0 (x) dx = f (x) g (x)ja
f 0 (x) g (x) dx
a
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 293 Find x sin xdx
If we select
u=x
du = dx
v = cos x dv = sin xdx
4.6. INTEGRATION BY PARTS
199
Then, applying formula 4.4 gives us:
Z
x sin xdx = x cos x
Z
cos xdx
Z
=
x cos x +
cos xdx
=
x cos x + sin x + C
Remark 294 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
x2
v=
2
du = cos xdx
dv = xdx
then the integration by parts formula would have given us
Z
x sin xdx =
x2
sin x
2
1
2
Z
x2 cos xdx
which is more complicated than the integral we started with.
R
Example 295 Find xex dx
Following the previous example, we select
u=x
v = ex
du = dx
dv = ex dx
Then, applying formula 4.4 gives
Z
xex dx = xex
= xex
Z
ex dx
ex + C
R
Remark 296 In the past two examples, we had an integral of the form xf (x) dx
where we knew an antiderivative of f (x). We did integration
R by parts with
u = x. This will be the case for most integrals of this form. x ln xdx is an
exception.
Example 297 Find
If we select
Re
1
ln xdx
u = ln x
v=x
1
dx
x
dv = dx
du =
200
CHAPTER 4. INTEGRALS
Then, applying formula 4.4 gives
Z e
e
ln xdx = x ln xj1
Z
1
e
dx
1
e
= x ln xj1
e
xj1
= (e ln e
ln 1)
(e
1)
=1
Remark 298 You will note that in the problem we just did, we found that
Z
ln xdx = x ln x x + C
Example 299 Find
If we select
R
tan
1
xdx
u = tan
1
x
v=x
Then, applying formula 4.4 gives
Z
tan 1 xdx = x tan
R
1
dx
1 + x2
dv = dx
du =
1
x
Z
x
dx
1 + x2
(4.5)
x
dx separately, using the substitution u = 1 + x2 . Then,
1 + x2
du = 2xdx therefore
Z
Z
x
1
du
dx =
1 + x2
2
u
1
= ln juj + C
2
1
= ln 1 + x2 + C
2
1
= ln 1 + x2 + C
2
We will do
Using what we just found in equation 4.5, gives us
Z
1
tan 1 xdx = x tan 1 x
ln 1 + x2 + C
2
Remark 300 In the previous two examples, the integrand consisted of just one
function. We used integration by parts with u set to that function. This technique should be remembered.
4.6.2
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.
4.6. INTEGRATION BY PARTS
Example 301 Find
If we select
R
201
x2 ex dx
u = x2
v = ex
du = 2xdx
dv = ex dx
Then, applying formula 4.4 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 302 Find x2 sin xdx
If we select
u = x2
du = 2xdx
v = cos x dv = sin xdx
Then, applying formula 4.4 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 4.4 gives
Z
x cos xdx = x sin x
Z
sin xdx
= x sin x + cos x + C
If we use what we just found in equation 4.6, 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.
(4.6)
202
CHAPTER 4. INTEGRALS
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 303 Find ex sin xdx
If we select
u = ex
du = ex dx
v = cos x dv = sin xdx
Then, applying formula 4.4 gives
Z
ex sin xdx =
ex cos x +
Z
ex cos xdx
(4.7)
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 4.4 gives
Z
ex cos xdx = ex sin x
Z
ex sin xdx
If we replace what we just found in equation 4.7, we obtain
Z
Z
ex sin xdx = ex cos x + ex 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.
Using integration by parts, we can prove important formulae, called reduction formulae. The most important ones are listed as a theorem.
Theorem 304 Let n be an integer such that n
1.
R
sinn xdx =
1
cos x sinn
n
1
x+
n
n
1R
2.
sinn
2
xdx
4.6. INTEGRATION BY PARTS
2.
3.
4.
R
R
R
cosn xdx =
1
sin x cosn
n
n
x+
n
n
R
n
n (ln x)
n
(ln x) dx = x (ln x)
xn ex dx = xn ex
1
203
R
n xn
1R
1
cosn
2
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 4.4 gives
Z
sinn xdx =
Z
1
n 1
sin
x cos x + (n
1)
=
sinn
1
x cos x + (n
1)
=
sinn
1
x cos x + (n
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
Z
Z
Z
Z
Z
n
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.
Example 305 Find
R
3
(ln x) dx = 6x
1 3
ln x
6
Using part 3 of theorem 304, we obtain
Z
3
3
(ln x) dx = x (ln x)
To …nd
R
2
3
1 2
ln x + ln x
2
Z
2
(ln x) dx
(ln x) dx, we apply the same formula again.
Z
Z
2
2
(ln x) dx = x (ln x)
2 ln xdx
1
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
204
CHAPTER 4. INTEGRALS
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
Combining the three gives us
Z
3
3
(ln x) dx = x (ln x)
3
Z
x
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
2
2
2
Z
3
= x (ln x)
2
Z
ln xdx
ln xdx
3x (ln x) + 6x ln x
x]
6x
Remark 306 As this example shows, the reduction formula may have to be
applied several times before we …nd the answer.
4.6.3
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. In particular,
remember the following techniques we learned in the examples:
R
– To integrate xn f (x) dx where f is a function for which we know
an antiderivative (such as sin x, cos x, ex , :::) we let u = xn . When
n > 1, we will have to do repeated integration by parts (n times).
R
– When the integrand contains only one function as in f (x) dx, one
thing to try is to let u = f (x).
Using integration by parts, be able to prove and use the following reduction
formulas
Z
Z
1
n 1
n
n 1
sin xdx =
cos x sin
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
4.6. INTEGRATION BY PARTS
4.6.4
205
Problems
1. Using integration by parts, evaluate
R
2. Evaluate x cos 2xdx
R t
3. Evaluate te 3 dt
R
4. Evaluate x2 sin xdx
R
5. Evaluate ln (5x + 4) dx
R
6. Evaluate arctan 3tdt
R
7. Evaluate e2x sin 5xdx
R
8. Evaluate 0 x sin 3xdx
R
x ln xdx with u = ln x and dv = xdx
R 2 ln x
dx
1 x2
R1 x
10. Evaluate 0 2x dx
e
R2
2
11. Evaluate 1 (ln x) dx
R
12. Evaluate sin 1 xdx
9. Evaluate
13. Use a substitution then integration by parts to evaluate
14. Using the formulas in theorem 304, prove that
15. Using the formulas in theorem 304, prove that
16. Using the formulas in theorem 304, …nd
17. Finish proving theorem 304
R
R
R
R
sin 2x
+C
4
n 1R2
sinn xdx =
sinn
0
n
sin2 xdx =
2
0
x
2
2
xdx
3
(ln x) dx
18. If f (0) = g (0) = 0 and f 00 and g 00 are continuous, show that
Ra
f (a) g 0 (a) f 0 (a) g (a) + 0 f 00 (x) g (x) dx
4.6.5
p
sin xdx
Ra
0
f (x) g 00 (x) dx =
Answers
R
1. Using integration by parts, evaluate x ln xdx with u = ln x and dv = xdx
R
x2 ln x x2
x ln xdx =
+C
2
4
R
1
1
2. Evaluate x cos 2xdx = cos 2x + x sin 2x + C
4
2
R t
1
3. Evaluate te 3 dt = 3e 3 t (t 3) + C
206
4. Evaluate
5. Evaluate
6. Evaluate
7. Evaluate
8. Evaluate
CHAPTER 4. INTEGRALS
R
R
R
R
R
x2 sin xdx = 2 cos x
ln (5x + 4) dx =
1
5
x2 cos x + 2x sin x + C
(5x + 4) ln (5x + 4)
arctan 3tdt = t arctan 3t
e2x sin 5xdx =
0
x sin 3xdx =
1
6
x+C
ln 9t2 + 1 + C
2 2x
e
sin 5x
29
5
cos 5x + C
2
1
3
R 2 ln x
dx = 12 12 ln 2
1 x2
R1 x
10. Evaluate 0 2x dx = 41 34 e 2
e
R2
2
2
11. Evaluate 1 (ln x) dx = 2 (ln 2 1)
p
R
12. Evaluate sin 1 xdx = 1 x2 + x arcsin x
9. Evaluate
13. Usepa substitution
then
p
p integration by parts to evaluate
2 x cos x + 2 sin x + C
14. Using the formulas in theorem 304, prove that
15. Using the formulas in theorem 304, prove that
16. Using the formulas in theorem 304, …nd
17. Finish proving theorem 304
R
R
R
2
3
sin
p
xdx =
sin 2x
+C
4
n 1R2
sinn xdx =
sinn
0
n
sin2 xdx =
0
R
(ln x) dx = 6x
1
6
18. If f (0) = g (0) = 0 and f 00 and g 00 are continuous, show that
Ra
f (a) g 0 (a) f 0 (a) g (a) + 0 f 00 (x) g (x) dx
x
2
ln3 x
Ra
0
1
2
2
xdx
ln2 x + ln x
f (x) g 00 (x) dx =
1