Integration by Parts
Main Objectives
• Find an antiderivative using integration by parts
What is Integration by Parts?
Theorem 1 If u and v are functions of x that have continuous derivatives,
then
Z
Z
udv = uv − vdu
Let’s look at an example:
Example 1
R
x sin(x)dx
Solution.
Notice that the techniques we have learned so far (the basic
rules and substitution) will not help us with this integral. Maybe our new
technique will work.
Before we continue we must choose u and v.
Let u = x (I will give you a rule to help you in choosing your u later)
Now, dv must be everything else inside of the integral. So, dv = sin(x)dx.
In order to use integration by parts we need four things, u, v, du, dv.
u=x
u=x
du = dx
u=x
du = dx
dv = sin(x)dx
R
R
dv = sin(x)dx
dv = sin(x)dx
v = − cos(x)
dv = sin(x)dx
Now that we have u, v, du, and dv, we can use the integration by parts formula:
Z
Z
x sin(x)dx = −x cos(x) − (− cos(x))dx
1
= −x cos(x) + sin(x) + C
Note 1 It is considered a rule of thumb to remember the acronym LIPET
when performing integration by parts. This acronym will help you to determine what to use as u. L-logarithms, I-inverse trigonometric functions,
P-polynomials, E-exponentials, T-trigonometric functions.
More Examples
Example 2 Find
R
x2 ln(x)dx
Solution. By LIPET, we should let u = ln(x), so we get the following:
v = 13 x3
dv = x2 dx
u = ln(x)
du = x1 dx
So by Integration by Parts, we find
Z
Z
1 3
1 3 1
2
x · dx
x ln(x)dx = x ln(x) −
3
3
x
Z
x3 ln(x) 1
=
−
x2 dx
3
3
=
Example 3 Find
R
x3 ln(x) x3
−
+C
3
9
arcsin(x)dx
2
Solution. Here we only have one term in the integrand. Substitution won’t
work, and this doesn’t follow one of the general integration rules. Can Integration by Parts help?
u = arcsin(x)
1
du = √1−x
2 dx
v=x
dv = dx
Z
Z
x
dx
1 − x2
Now, the integral which has been left over can be done with a simple substitution:
Let w = 1 − x2
Then dw = −2xdx
Z
x −1
dw
= x arcsin(x) − √ ·
w 2x
Z
1
w−1/2 dw
= x arcsin(x) +
2
√
= x arcsin(x) + w + C
√
= x arcsin(x) + 1 − x2 + C
arcsin(x)dx = x arcsin(x) −
√
Example 4 Find
R
x2 sin(x)dx
Solution. By LIPET we should choose the polynomial portion of the integrand for u.
u = x2
du = 2xdx
Z
=
2
v = − cos(x)
dv = sin(x)dx
2
x sin(x)dx = −x cos(x) + 2
Z
x cos(x)dx
Now, the integral left over still requires integration by parts:
3
u=x
du = dx
v = sin(x)
dv = cos(x)dx
Z
= −x cos(x) + 2 x sin(x) − sin(x)dx
2
= x2 cos(x) + 2x sin(x) + 2 cos(x) + C
Example 5 Find
R
ex cos(x)dx
Solution. By LIPET, we choose the exponential as our u. So, we obtain
u = ex
du = ex dx
v = sin(x)
dv = cos(x)dx
So by Integration by Parts
Z
Z
x
x
e cos(x)dx = e sin(x) − ex sin(x)dx
Again, we use Integration by Parts
u = ex
du = ex dx
v = − cos(x)
dv = sin(x)dx
Z
x
x
= e sin(x) − −e cos(x) + e cos(x)dx
x
x
Z
x
= e sin(x) + e cos(x) −
ex cos(x)dx
The integrand left over is the same as the one we started with. That is:
Z
Z
x
x
x
e cos(x)dx = e sin(x) + e cos(x) − ex cos(x)dx
4
Z
=⇒ 2
ex cos(x)dx = ex sin(x) + ex cos(x)
=
ex (sin(x) + cos(x))
+C
2
5