Arkansas Tech University
MATH 2924: Calculus II
Dr. Marcel B. Finan
6.1
The Method of Integration by Parts
The integration by parts formula is an antidifferentiation method which
reverses the product rule of differentiation. To see this, let u and v be two
differentiable functions of x. Then the product rule asserts that the product
function uv is also differentiable and its derivative is given by
(uv)0 = uv 0 + u0 v.
This says that an antiderivative of uv 0 + u0 v is the function uv. In terms of
indefinite integrals we have
Z
Z
Z
0
0
(uv) dx = uv dx + u0 vdx
(6.1.1)
or equivalently
Z
uv 0 dx = uv −
Z
u0 vdx.
(6.1.2)
Remark 6.1.1
Since the differential of u is du = u0 dx and that of v is dv = v 0 dx, (6.1.2)
becomes
Z
Z
udv = uv − vdu.
(6.1.3)
Either formula (6.1.2) or (6.1.3) is known as the integration by parts
formula.
The corresponding integration by parts formula for definite integrals is given
by
Z
b
0
uv dx =
a
uv|ba
Z
−
Example
R x 6.1.1
Find xe dx.
1
a
b
u0 vdx.
(6.1.4)
Solution.
R
Let u = x and v 0 = ex . Then u0 = 1 and v = ex dx = ex . Note that in
finding v we did not includeR the constant of integration. We will write the
constant C in the answer of uv 0 dx. Now, substituting in formula (6.1.2) to
obtain
Z
xex dx =xex −
Z
ex dx
=xex − ex + C
Remark 6.1.2
If we chose u = ex and v 0 = x then we would have u0 = ex and v =
this case, formula (6.1.2) yields
Z
Z 2
x2 x
x x
x
xe dx =
e −
e dx
2
2
x2
2 .
In
and the second integral is definitely worse than
R xthe one to the left of the
formula we had in the previous example, i.e. xe dx. It follows that for the
method to be useful it is important to choose u and dv in such a way to
make the integral on the right easier to find than the integral on the left.
Example
6.1.2
R
Find x sin xdx.
Solution.
Let u = x and v 0 = sin x. Then u0 = 1 and v = − cos x. Substituting in
formula (6.1.2) to obtain
Z
Z
x sin xdx = − x cos x −
(− cos x)dx
Z
= − x cos x +
cos xdx
= − x cos x + sin x + C
There are examples which don’t look like good candidates for integration
by parts because they don’t appear to involve products, but for which the
method works well. Such examples often involves ln x or the inverse trigonometric functions.
2
Example R6.1.3
5
Calculate 1 ln xdx.
Solution.
Let u = ln x and v 0 = 1. Then u0 =
(6.1.4) we obtain
1
x
and v = x. Substituting in formula
5
Z
ln xdx
=x ln x|51
Z
5
−
1
1
=x ln x|51 −
=[x ln x −
Z
1
x dx
x
5
dx
1
x]51
=5 ln 5 − 5 − [ln 1 − 1]
=5 ln 5 − 4
Example
6.1.4
R
Find x3 ln xdx.
Solution.
Let u = ln x and v 0 = x3 . Then u0 =
(6.1.2) to obtain
Z
x4
x ln xdx =
4
x4
=
4
x4
=
4
x4
=
4
3
1
x
and v =
x4
4 .
Substituting in formula
x4 1
dx
4 x
Z
1
ln x −
x3 dx
4
1 x4
ln x − ·
+C
4 4
1
ln x − x4 + C
16
Z
ln x −
Sometimes evaluating an integral might require integration by parts more
than once as the following example shows.
Example
6.1.5
R
Find x2 cos xdx.
Solution.
Let u = x2 and v 0 = cos x. Then u0 = 2x and v = sin x. Substituting in
formula (6.1.2) to obtain
3
Z
x2 cos xdx =x2 sin x −
Z
2x sin xdx
Z
=x2 sin x − 2 x sin xdx
=x2 sin x − 2(−x cos x + sin x) + C
=x2 sin x + 2x cos x − 2 sin x + C
R
where we have used Example 6.1.2 to evaluate x sin xdx
The following example illustrates a very useful technique: Use integration by
parts to transform the integral into an expression containing another copy
of the same integral, possibly multiplied by a constant, then solve for the
original integral.
Example
6.1.6
R
Find sin2 xdx.
Solution.
Method I:
Using a half-angle formula to write sin2 x =
2x
lem reduces to integrating 1−cos
. That is,
2
Z
1
sin xdx =
2
Z
1
=
2
Z
2
1−cos 2x
.
2
In this case, the prob-
(1 − cos 2x)dx
1
dx −
2
1
1
= x−
2
4
Z
cos 2xdx
Z
cos udu
1
1
= x − sin u + C
2
4
1
1
= x − sin 2x + C
2
4
4
where the substitution u = 2x is used to evaluate the integral
R
cos 2xdx.
Method II:
We now use integration by parts formula to evaluate the given integral. Let
u = sin x and v 0 = sin x. Then u0 = cos x and v = − cos x. Substituting in
formula (6.1.2) to obtain
Z
2
Z
− cos2 xdx
Z
cos2 xdx.
sin xdx = − sin x cos x −
= − sin x cos x +
Using the trigonometric identities sin 2x = 2 sin x cos x and cos2 x + sin2 x =
1 we can rewrite the right-hand side of the above integral as
Z
1
sin xdx = − sin 2x +
2
Z
1
= − sin 2x +
2
Z
2
(1 − sin2 x)dx
Z
dx −
1
= − sin 2x + x −
2
Z
sin2 xdx
sin2 xdx.
Moving the right integral to the left side to obtain
Z
1
2 sin2 xdx = − sin 2x + x + C
2
and finally dividing both sides by 2 to obtain
Z
1
x
sin2 xdx = − sin 2x + + C
4
2
Integration by parts can be used to derive reduction formulas, i.e., formulas that reduce the integral of a power of a function to an integral with
lesser power progressively simplifying the integral until it can be evaluated.
We illustrate this method in the next example.
Example 6.1.7
R n x
R
n ex − n xn−1 ex dx.
(a) Derive the reduction
formula:
x
e
dx
=
x
R
(b) Use (a) to find x3 ex dx.
5
Solution.
(a) Letting u(x) = xn , v 0 (x) = ex , we find u0 (x) = nxn−1 and v(x) = ex .
Hence,
Z
Z
n x
n x
x e dx = x e − n xn−1 ex dx.
(b) We have
Z
3 x
3 x
x e dx =x e − 3
Z
x2 ex dx
Z
2 x
x
=x e − 3 x e − 2 xe dx
Z
=x3 ex − 3x2 ex + 6 xex − ex dx
3 x
=x3 ex − 3x2 ex + 6xex − 6ex + C
6