Integration By Parts
Formula
 udv  uv   vdu
I. Guidelines for Selecting u and dv:
(There are always exceptions, but these are generally helpful.)
“L-I-A-T-E” Choose ‘u’ to be the function that comes first in this list:
L: Logrithmic Function
I: Inverse Trig Function
A: Algebraic Function
T: Trig Function
E: Exponential Function
Example A:  x 3 ln x dx
*Since lnx is a logarithmic function and x 3 is an algebraic
function, let:
u = lnx
dv =
(L comes before A in LIATE)
3
x dx
1
du =
dx
x
v=
x
3
3
 x dx 
x4
4
ln xdx  uv   vdu
 (ln x)
x4

4
x4 1
 4 x dx
 (ln x)
x4
1 3

x dx
4
4

x4
1 x4
(ln x) 
 C
4
4 4

x4
x4
(ln x) 
 C
4
16
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ANSWER
Page 1 of 7
Example B:  sin x ln(cos x) dx
u = ln(cosx) (Logarithmic Function)
dv = sinx dx (Trig Function [L comes before T in LIATE])
du =
v=
 sin x
1
( sin x) dx   tan x dx
cos x
 sin x
dx   cos x
ln(cos x) dx  uv 
 vdu
 (ln(cos x))( cos x)   ( cos x)( tan x)dx
  cos x ln(cos x)   (cos x)
sin x
dx
cos x
  cos x ln(cos x)   sin x dx
  cos x ln(cos x)  cos x  C
ANSWER
Example C:  sin 1 x dx
*At first it appears that integration by parts does not apply, but let:
u  sin 1 x (Inverse Trig Function)
dv  1 dx (Algebraic Function)
du 
1
1 x2
dx
v   1dx  x
 sin
1
x dx  uv 
 vdu
 (sin 1 x)( x) 
x
1
1 x2
dx
 1
 x sin 1 x     (1  x 2 ) 1 / 2 (2 x) dx
 2
 x sin 1 x 
 x sin 1 x 
1
(1  x 2 )1 / 2 (2)  C
2
1 x2  C
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ANSWER
Page 2 of 7
II. Alternative General Guidelines for Choosing u and dv:
A.
Let dv be the most complicated portion of the integrand
that can be “easily’ integrated.
B.
Let u be that portion of the integrand whose derivative du is a
“simpler” function than u itself.
Example:
x
3
4  x 2 dx
*Since both of these are algebraic functions, the LIATE Rule of
Thumb is not helpful. Applying Part (A) of the alternative
guidelines above, we see that x 4  x 2 is the “most complicated part
of the integrand that can easily be integrated.” Therefore:
dv  x 4  x 2 dx
u  x 2 (remaining factor in integrand)
du  2 x dx
1
(2 x)(4  x 2 )1 / 2 dx

2
1
 1  2 
     (4  x 2 ) 3 / 2   (4  x 2 ) 3 / 2
3
 2  3 
v
x
3
x
4  x 2 dx  
4  x 3 dx  uv   vdu
1
 1

 ( x 2 )  (4  x 2 ) 3 / 2     (4  x 2 ) 3 / 2 (2 x) dx
3
 3


 x2
1
(4  x 2 ) 3 / 2   (4  x 2 ) 3 / 2 (2 x) dx
3
3
 x2
1
2

(4  x 2 ) 3 / 2  (4  x 2 ) 5 / 2    C
3
3
5

 x2
2
(4  x 2 ) 3 / 2  (4  x 2 ) 5 / 2  C Answer
3
15
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Page 3 of 7
III. Using repeated Applications of Integration by Parts:
Sometimes integration by parts must be repeated to obtain an answer.
Note: DO NOT switch choices for u and dv in successive applications.
Example:
x
2
sin x dx
u  x 2 (Algebraic Function)
dv  sin x dx (Trig Function)
du  2 x dx
v   sin x dx   cos x
x
2
sin x dx  uv   vdu
 x 2 ( cos x) 
  cos x
  x 2 cos x  2
x
2 x dx
cos x dx
Second application of integration by parts:
u  x (Algebraic function) (Making “same” choices for u and dv)
dv  cos x (Trig function)
du  dx
v   cos x dx  sin x
x
2
sin x dx   x 2 cos x  2 [uv   vdu]
  x 2 cos x  2 [ x sin x   sin x dx]
  x 2 cos x  2 [ x sin x  cos x  c]
  x 2 cos x  2 x sin x  2 cos x  c
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Answer
Page 4 of 7
Note:
Example:
After each application of integration by parts, watch for the
appearance of a constant multiple of the original integral.
e
x
cos x dx
u  cos x (Trig function)
dv  e x dx (Exponential function)
du   sin x dx
v   e x dx  e x
e
x
cos x dx  uv   vdu
 cos x e x   e x ( sin x) dx
 cos x e x   e x sin x dx
Second application of integration by parts:
u  sin x (Trig function) (Making “same” choices for u and dv)
dv  e x dx (Exponential function)
du  cos x dx
v   e x dx  e x
e
e
x
x
cos x dx  e x cos x  (uv   vdu)
cos x dx  e x cos x  sin x e x   e x cos x dx
Note appearance of original integral on right side of equation. Move to left side
and solve for integral as follows:
2 e x cos x dx  e x cos x  e x sin x  C
e
x
cos x dx 
1 x
(e cos x  e x sin x)  C
2
Answer
Practice Problems:
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Page 5 of 7
1.
 3x
2.

3.
x
4.
x
5.
 cos
6.
 (ln x)
7.
x
8.
e
9.
x
10.
 x(ln x)
e  x dx
ln x
dx
x2
2
cos x dx
sin x cos x dx
1
x dx
2
dx
9  x 2 dx
3
2x
sin x dx
x  1 dx
2
1
3
dx
Solutions:
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Page 6 of 7
1.
 3xe  x  3e  x  C
u  3x
dv  e  x dx
2.

ln x 1
 C
x
x
u  ln x
dv 
3.
x 2 sin x  2 x cos x  2 sin x  C
1
dx
x2
u  x2
dv  cos x dx
4.

x cos 2 x sin 2 x

C
4
8
note:
sin 2 x
 sin x cos x
2
ux
dv  sin 2 x cos x dx
5.
x cos 1 x 
u  cos 1 x
1 x2  C
dv  dx
6.
u  (ln x) 2
x(ln x) 2  2 x ln x  2 x  C
dv  dx
7.

x2
2
(9  x 2 ) 3 / 2  (9  x 2 ) 5 / 2  C
3
15
u  x2
dv  (4  x 2 )1 / 2 x dx
8.
2e 2 x sin x e 2 x cos x

C
5
5
u  sin x
dv  e 2 x dx
9.
2 x 2 ( x  1) 3 / 2 8 x( x  1) 5 / 2 16( x  1) 7 / 2


C
3
15
105
u  x2
dv  ( x  1)1 / 2 dx
10.
1
C
2(ln x) 2
u
1
 (ln x) 3
3
(ln x)
dv 
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1
dx
x
Page 7 of 7