Integration by Parts
Philippe B. Laval
KSU
Today
Philippe B. Laval (KSU)
Integration by Parts
Today
1 / 10
Introduction
We present a new integration technique called integration by parts. It is
to integrals what the product rule is to derivatives. The idea is to rewrite
an integral into one we can compute or at least a simpler one.
Philippe B. Laval (KSU)
Integration by Parts
Today
2 / 10
Integration by Parts: Formula
If f and g are two di¤erentiable functions, then the product rule gives us:
(f (x) g (x))0 = f (x) g 0 (x) + f 0 (x) g (x)
If we integrate both sides, we get
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 do not usually
remember it this way. If we let u = f (x) and v = g (x), then
du = f 0 (x) dx and dv = g 0 (x) dx. Replacing in the above formula and
rearranging gives:
Philippe B. Laval (KSU)
Integration by Parts
Today
3 / 10
Integration by Parts: Formula
Z
udv = uv
Z
vdu
(x)jba
Z
This is the formula we will remember and use. If we have limits of
integration, the formula becomes
Z
b
0
f (x) g (x) dx = f (x) g
a
Philippe B. Laval (KSU)
Integration by Parts
b
f 0 (x) g (x) dx
a
Today
4 / 10
Integration by Parts: Formula
Again, the formula we have is
Z
udv = uv
1
Z
vdu
The goal when using this
R formula is to pretend that the integral we
are given is of the form udv . We …nd u and dv that accomplish this.
Philippe B. Laval (KSU)
Integration by Parts
Today
5 / 10
Integration by Parts: Formula
Again, the formula we have is
Z
udv = uv
1
2
Z
vdu
The goal when using this
R 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 u and v by
integrating dv .
Philippe B. Laval (KSU)
Integration by Parts
Today
5 / 10
Integration by Parts: Formula
Again, the formula we have is
Z
udv = uv
1
2
3
Z
vdu
The goal when using this
R 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 u and v by
integrating dv .
R
Then, we can rewrite the given integral as uv
vdu.
Philippe B. Laval (KSU)
Integration by Parts
Today
5 / 10
Integration by Parts: Formula
Again, the formula we have is
Z
udv = uv
1
2
3
4
Z
vdu
The goal when using this
R 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 u and v by
integrating dv .
R
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 .
Philippe B. Laval (KSU)
Integration by Parts
Today
5 / 10
Integration by Parts: Examples
Example
Find
R
x sin xdx
Example
Find
R
xe x dx
Example
Find
Re
1
ln xdx
Example
Find
R
tan
1
xdx
Philippe B. Laval (KSU)
Integration by Parts
Today
6 / 10
Repeated Integration by Parts
We may need to apply the integration by parts formula twice, or more. We
look at some example to illustrate the various cases which can occur.
Example
Find
R
x 2 e x dx
Example
Find
R
x 2 sin xdx
In these two examples, we applied the integration by parts formula several
time. 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.
Example
Find
R
e x sin xdx
Philippe B. Laval (KSU)
Integration by Parts
Today
7 / 10
Integration by Parts: Reduction Formulas
Using integration by parts, we can prove important formulae, called
reduction formulae. The most important ones are listed as a theorem.
Theorem
Let n be an integer such that n 2.
R n
1
n 1R n 2
1
sin xdx =
cos x sinn 1 x +
sin
xdx
n
n
R
1
n 1R
2
cosn xdx = sin x cosn 1 x +
cosn 2 xdx
n
n
R
R
3
(ln x)n dx = x (ln x)n n (ln x)n 1 dx
R n x
R
4
x e dx = x n e x n x n 1 e x dx
Example
Find
R
(ln x)3 dx
Philippe B. Laval (KSU)
Integration by Parts
Today
8 / 10
Summary
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 x n f (x) dx where f is a function for which we know an
antiderivative (such as sin x, cos x, e x , :::) we let u = x n . 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 reduction
formulas listed above
Philippe B. Laval (KSU)
Integration by Parts
Today
9 / 10
Exercises
See the problems at the end of my notes on integration by parts.
Philippe B. Laval (KSU)
Integration by Parts
Today
10 / 10