8.2 Integration by Parts
Summary of Common Integrals Using
Integration by Parts
1.
For integrals of the form

n
ax
x e dx ,

n
x sin axdx ,

n
x cos axdx
Let u = xn and let dv = eax dx, sin ax dx, cos ax dx
2.

For integrals of the form
n

x
 ln xdx ,

x 
arcsin axdx ,
n

n
x arccos axdx
Let u = lnx, arcsin ax, or arctan ax and let dv = xn dx
3.

For integrals of the form

ax

e sin bxdx ,
or


e
ax
cos bxdx ,
Let u = sin bx or cos bx and let dv = eax dx
Integration by Parts
If u and v are functions of x and have
continuous derivatives, then
u
dv

uv

v
du


Guidelines for Integration by Parts
1. Try letting dv be the most complicated
portion of the integrand that fits a basic
integration formula. 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 simpler
function than u. Then dv will be the
remaining factor(s) of the integrand.
Evaluate
 xe
x
dx
To apply integration by parts, we want
to write the integral in the form u dv .
There are several ways to do
this.
  x e
x
dx
u dv
  e  xdx   1xe
x
u
dv
u
x
dx
dv

   xe dx 
x
u dv
Following our guidelines, we choose the first option
because the derivative of u = x is the simplest and
dv = ex dx is the most complicated.
  x e
u
x
dx
dv

u=x
du = dx
v = ex
dv = ex dx
u
dv

uv

v
du


 xe 
x
e
dx

x
 xe  e  C
x
x
x
2
Since x2 integrates easier than
ln x, let u = ln x and dv = x2
ln x dx
u = ln x
1
du 
v
x
3
u
dv

uv

v
du


3
dx dv = x2 dx
x

x
3
3
ln x 
3

x
3
3
ln x 

x
2
3
dx

x
3
3
ln x 
x
3
9
C

x 1
3 x
dx
Repeated application of integration by parts

2
u = x2
x sin x dx
v = -cos x
du = 2x dx
 u dv
 uv   v du
  x cos x 
2
u = 2x
 2 x cos
xdx
du = 2 dx
  x cos x  2 x sin x 
2
Apply integration by
parts again.
dv = cos x dx v = sin x
 2 sin
xdx
  x cos x  2 x sin x  2 cos x  C
2
dv = sin x dx
Repeated application of integration by parts
e
x
Neither of these choices for u
differentiate down to nothing,
so we can let u = ex or sin x.
sin xdx
Let’s let u = sin x
v = ex
du = cos x dx dv = ex dx
x
du = -sinx dx dv = ex dx
e sin x  e cos x   e sin xdx 
x
x
v = ex
u = cos x
e sin x   e cos xdx
x
x
e
x
sin xdx
e sin x  e cos x  2  e sin xdx
x
x
e sin x  e cos x
x
x
x
2
C 
e
x
sin xdx