Calc 2 Lecture Notes
Section 6.2
Page 1 of 5
Section 6.2: Integration by Parts
Big idea: Integration by parts is an integration technique that allows you to integrate the product
of two functions (under certain circumstances). Integration by parts is really the product rule in
reverse. However, it does not work in every case where the integrand is a product of functions.
Big skill: You should be able to identify appropriate choices for u(x) and v(x), apply those
choices to the integration by parts formula, and know if and when you’ve made a bad choice for
u(x) and v(x).
Proof that the antiderivative of the product function f  x   x sin  x  is  x cos  x   sin  x 
d
  x cos  x   sin  x    1 cos  x     x    sin  x    cos  x 
dx 
  cos  x   x sin  x   cos  x 
 x sin  x 
Thus,  x cos  x   sin  x  is an antiderivative of x sin  x  .
i.e.,
 x sin  x  dx   x cos  x   sin  x   c
Notice that we can write this result as:
 x sin  x  dx   x cos  x   sin  x 
 x  dx  cos  x  dx  x    cos  x       cos  x    dx  x  dx
d
d
So now let’s look at re-writing the general product rule in reverse:
d
du
dv
v  u 
u  x  v  x   
dx
dx
dx
d
du
dv
 vx  u  x
u  x  v  x   x 

dx
dx
dx
 d u  x  v  x    v  du  u  dv
  d u  x  v  x     vdu   udv
  udv   d u  x  v  x     vdu

 udv  u  x  v  x    vdu
(This is the integration by parts “formula”)
(Recall that the differential df of a function f(x) is defined as df  f   x  x (p. 243).)
Calc 2 Lecture Notes
Section 6.2
Page 2 of 5
So, if you have a product of functions, think of one of the functions in terms of its antiderivative.
The integral then equals the product of the first function times the antiderivative plus the integral
of the product of the derivative of the first function and the antiderivative.
Note that the “under certain circumstances” part of this technique is that you can compute the
antiderivative for  du  v .
So, to compute
let u  x   x
 x sin  x  dx
using integration by substitution,
and dv  sin  x  dx .
To find v(x), we have to evaluate v  x    dv   sin  x  dx   cos  x  .
Also,
du  x 
dx
 1  du = dx.
So:
 x sin  x  dx  u  x  v  x    du  v
 x   cos  x     dx    cos  x  
  x cos  x    cos  x  dx
  x cos  x   sin  x   c
Practice:
1. Evaluate
2. Evaluate
 xe dx .
x
 x ln  x  dx .
Calc 2 Lecture Notes
3. Evaluate

ln  x 
x
Section 6.2
Page 3 of 5
dx .
4. Evaluate  ln  x  dx .
5. Evaluate
 x cos  x  dx either by yourself or with a neighbor.
Don’t let your instructor
tell you the answer until you’ve had a chance to work it out for yourself!
6. Evaluate
 x e dx .
2 x
Calc 2 Lecture Notes
Section 6.2
7. Evaluate  e x sin  x  dx .
8. Derive a power reduction formula for
 x e dx .
9. Use the reduction formula to evaluate
 x e dx .
10. Evaluate  sin  ln  x   dx .
n x
4 x
Page 4 of 5
Calc 2 Lecture Notes
Section 6.2
11. Evaluate  e x dx . Hint: Use a substitution first.
3
Evaluating Definite Integrals Using Integration by Parts:
x b
 udv  uv
xa
x b
xa
x b

 vdu
xa
Practice:
2
12. Evaluate
 x ln  x  dx
5
1
Page 5 of 5