AP CALCULUS AB
CHAPTER 6:
DIFFERENTIAL EQUATIONS AND
MATHEMATICAL MODELING
SECTION 6.3:
ANTIDIFFERENTIATION BY PARTS
What you’ll learn about
 Product Rule in Integral Form
 Solving for the Unknown Integral
 Tabular Integration
 Inverse Trigonometric and Logarithmic Functions
… and why
The Product Rule relates to derivatives as the
technique of parts relates to antiderivatives.
Section 6.3 – Antidifferentiation by Parts
 Remember the Product Rule for Derivatives:
d
dv
du
 uv   u  v
dx
dx
dx
 By reversing this derivative formula, we obtain the
integral formula
 dv 
d
  du 
  u dx  dx    dx  uv      v dx  dx
 du 
 uv    v  dx
 dx 
Section 6.3 – Antidifferentiation by Parts
 Integration by Parts Formula
udv

uv

vdu


Section 6.3 – Antidifferentiation by Parts
 Example:
  x cos x dx
Let u  x and dv  cos xdx
then du  dx and v  sin x
So,
  x cos x  dx  x sin x   sin xdx
 x sin x    cos x   C
 x sin x  cos x  C
Example Using Integration by Parts
Evaluate  x sin xdx.
Use  udv  uv -  vdu with
ux
dv  sin xdx
du  dx v  -cos x
 x sin xdx  - x cos x   cos xdx
 - x cos x  sin x  C
Section 6.3 – Antidifferentiation by Parts
 Try to pick for your u something that will get smaller,
or more simple when you take the derivative.
 Sometimes, you may have to integrate by parts more
than one time to get to something you can integrate.
 Sometimes, integration by parts does not work.
Section 6.3 – Antidifferentiation by Parts
 Repeated use
2
x
 cos xdx
Let u  x 2 and
then du  2 xdx
dv  cos xdx
v  sin x
So,  x 2 cos xdx  x 2 sin x   2 x sin xdx
Repeating the process
let u  2 x and
then du  2dx
dv  sin xdx
v   cos x

And  x 2 cos xdx  x 2 sin x  2 x cos x   2 cos xdx
 x 2 sin x  2 x cos x  2  cos xdx
 x 2 sin x  2 x cos x  2sin x  C

Example Repeated Use of Integration by
Parts
Evaluate  2 x e dx.
2
Let u  2 x
x
dv  e dx
2
x
du  4 xdx v  e
 2 x e dx  2 x e  4  xe dx
Now let u  x dv  e dx
du  dx v  e
x
2
x
2
x
x
x
x
 2 x e dx  2 x e  4  xe   e dx 
 2 x e  4 xe  4e  C
2
x
2
x
2
x
x
x
x
x
Example Antidifferentiating ln x
Find  ln xdx.
Let u  ln x dv  dx
1
du  dx v  x
x
1
 ln xdx  x ln x -   x    dx
 x
 x ln x -  dx
 x ln x - x  C
Section 6.3 – Antidifferentiation by Parts
 Solving for the Unknown Integral
Sometimes, when performing this process, we seem
to get a circular argument going. When this occurs,
collect your like terms, and you can solve for the
unknown integral.
 Example:
e
x
cos xdx
Let u  e x
and
dv  cos xdx
then du  e x dx
v  sin x
So,  e x cos xdx  e x sin x   e x sin xdx
Repeating the process
let u  e x
and
then du  e x dx
And
dv  sin xdx
v   cos x

x
x
x
x
e
cos
x
dx

e
sin
x


e
cos
x


e

 cos xdx
 e x sin x  e x cos x   e x cos dx
Adding the like integrals together we get
2 e x cos xdx  e x sin x  e x cos x  e x  sin x  cos x 
ex
So,  e cos dx   sin x  cos x   C
2
x

Example Solving for the Unknown Integral
Evaluate  e sin xdx.
x
Let u  e dv  sin xdx
du  e dx v  -cos x
 e sin xdx  e cos x   e cos xdx
Now let u  e dv  cos xdx
du  e dx v  sin x
 e sin xdx  e cos x  e sin x   e sin xdx
2  e sin xdx  e cos x  e sin x
e cos x  e sin x
C
 e sin xdx 
2
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
Section 6.3 – Antidifferentiation by Parts
 Tabular Integration
When many repetitions of integration by parts is
needed, there is a way to organize the calculations
that saves a great deal of work and confusion. It is
called tabular integration.
Section 6.3 – Antidifferentiation by Parts
 x e dx
 Example:
3 x
u and its derivatives
dv and its integrals
x3
(+)
ex
3x 2
(-)
ex
(+)
ex
(-)
ex
6x
6
ex
0
So  x e dx  x e  3x e  6 xe  6e  C
3 x
3 x
2 x
x
x