Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 2
1
Integration by Parts
Integration by Parts is a technique of integration that is useful when the integrand
involves a product of an algebraic with a transcendental expression, such as …
 x cos( x) dx
See Exercise 1.
x
3
e
ln( x) dx
See Exercise 2.
x
sin( x) dx
See Exercise 5.
Integration by Parts is based on the Product Rule for taking derivatives.
Product Rule:
(u  v)  =
where u and v are functions of x
If u  and v  are continuous, then you can integrate both sides of the Product Rule equation.
 (u  v) dx
This produces:
=
=
 u vdx
+
 v u dx
where dv =
and du =
Subtracting  v du from both sides of the previous equation, we obtain the following:
***
***
This last equation that we just created is the Integration by Parts formula.
 u dv
is the
“original integral,” the one that we are trying to integrate. Notice that it is written in
terms of another integral,  v du . Choosing u and dv wisely will make it easier to
determine the integral  v du than it would be to determine the original integral,  u dv .
Let’s look at some guidelines for choosing u and dv well. (“You have chosen…wisely.”
This quote is from what movie?)
Guidelines for Choosing u and dv
1. Try letting dv be the most “complicated” portion of the integrand that fits an
elementary integration rule. Then u will be the remaining factor(s) of the integrand.
2. Try letting u be the portion of the integrand whose derivative is a function that is
“simpler” than u. The dv will be the remaining factor(s) of the integrand.
Exercise 1: Determine  x cos( x) dx .
Let u =
Then du =
Thus,
 x cos( x) dx =
and let dv =
and v =  dv =
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 2
2
Exercise 2: Determine  x 3 ln( x) dx .
Fruitless First Attempt at Exercise 2
Hmmm. If we let u  x 3 , then we are forced to let dv  ln( x) dx .
Okay, du will be easy to compute. But then v =  dv =  ln( x) dx = ???
I hope and pray…that you did not say…that  ln( x) dx equals
1
is the derivative of ln(x).
x
1
… for you’d be astray!
x
1
is NOT an anti-derivative of ln(x).
x
Fruitful Second Attempt at Exercise 2
Let u =
Then du =
Thus,
x
3
and let dv =
and v =  dv =
ln( x) dx =
An Integrand with a Single Factor
A surprising use of Integration by Parts involves some problems in which the integrand consists
of a single factor. Bonusly (a new adverb…and a near-antonym of “bogusly”), we’ll also solve
the problem raised in the Fruitless First Attempt at Exercise 2.
Exercise 3: Determine  ln( x) dx .
In these sorts of cases, let dv =
Let u =
and let dv =
Then du =
and v =  dv =
Thus,  ln( x) dx =
Check this answer to see that you are right by derivating our anti-derivative answer.
x ln( x)  x = x ln( x) - x
=
.
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 2
3
Repeated Use of Integration by Parts
Some integrals require repeated use of the Integration by Parts formula, such as…
Exercise 4: Determine  x 2 cos( x) dx .
becomes “simpler,” whereas the derivative of
The derivative of
Therefore, let u =
and let dv =
Then du =
and v =
Thus,
x
2
(Compare with Exercise 1.)
cos( x) dx = x 2 sin( x) -
does not.
 sin( x) 2 xdx .
We now have to integrate  sin( x) 2 xdx by parts, with a new u and a new dv.
Let new u =
and let new dv =
Then new du =
and new v =
Thus,  sin( x) 2 xdx =
Finally,  x 2 cos( x) dx = x 2 sin( x) -
=
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 2
A Clever Trick
Exercise 5: Determine  e x sin( x) dx .
4
Let u =
 du =
Let dv =
 v=
Thus,  e x sin( x) dx =
Then,  e x sin( x) dx =
e
x
sin( x) dx =
Let new u =
 new du =
Let new dv =
 new v =
Bob Brown
CCBC Dundalk
Math 252 Calculus 2 Chapter 8, Section 2
5
Tabular Method
For problems involving repeated applications of Integration by Parts, a tabular method
can help to organize the work. This method works well for integrands with a factor that
is a power of x, such as…
x
n
cos( ax) dx
x
n
sin( ax) dx
x
n
e ax dx
Exercise 6: Determine  x 5 cos( 2 x) dx . Use the tabular method.
Let u =
and let dv =
Alternating Signs
u and its derivatives
dv = v  and its antiderivatives
+
x5
cos(2x)
-
+
-
Thus,
x
5
cos( 2 x) dx =