8.2 Integration by Parts
Question: How do we integrate expressions such as
! x ln x
dx
or
!x e
2 x
dx
or
!e
x
sin x dx ???
We will use a new method, integration by parts, to integrate these complicated looking integrals. The product
rule is the basis for deriving the formula for integration by parts.
Given u(x) and v(x), the derivative of the product u( x ) ! v ( x ) is given by:
d
u( x ) ! v ( x ) = u( x ) ! v '( x ) + v ( x ) ! u'( x )
dx
Integrating both sides of the equation gives:
d
" dx [u(x) ! v(x)]dx = " [u(x) ! v'(x) + v(x) ! u '(x)]dx
u(x) ! v(x) = " u(x) ! v'(x)dx + " v(x) ! u '(x)dx
Rearrange the equation to get the formula for Integration by Parts:
# u(x) ! v'(x)dx = u(x) ! v(x) " # v(x) ! u '(x)dx
Using a simpler notation we get:
"u
dv = uv ! " v du
The new formula expresses the original integral in terms of another integral that should be easier to evaluate.
To apply the Integration by Parts formula:
1.
2.
3.
4.
Determine the appropriate substitution for u and dv.
Integrate dv to solve for v, and differentiate u to solve for du.
Substitute the values for u, v, du, and dv into the formula.
Solve the new integral.
Example 1
! x sin x
dx
Example 2
! xe
x
dx
Example 3
!e
x
cos x dx
Example 4
! ln x
dx
The Tabular Method
This method is useful when you must apply integration by parts multiple times. For instance, when integrating
any of the forms below, the tabular method is helpful.
!x
n
sin ax dx or ! x n cos ax dx or
!x e
n ax
dx
Example 5
!x
2
sin 4x dx
The sign applies to
u and u'.
Alternating Sign +/-
Let u = x2 and dv = sin 4x dx
Differentiate until you get to 0.
u and du
Integrate.
dv and v
The solution is obtained by adding the signed product of the diagonal entries.