CALCULUS : INTEGRATION BY PARTS, CLASS NOTES AND PRACTICE QUESTIONS WITH SOLUTIONS

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 (L comes before A in LIATE)
3
dv = x dx
1
du = dx
x
x4
 x dx 
3
v=
4

x ln xdx  uv   vdu
3




x4 x4 1
 (ln x)
4
  4 x dx
x4 1 3
4
 (ln x)  x dx
4
x4 1 x4
 (ln x)   C
4 4 4
x4 x4
 (ln x)   C ANSWER
4 16




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, Example B:  sin x ln(cos x) dx

u = ln(cosx) (Logarithmic Function)
dv = sinx dx (Trig Function [L comes before T in LIATE])
1
du = ( sin x) dx   tan x dx
cos x

v=  sin x dx   cos x


 sin x ln(cos x) dx  uv   vdu
 (ln(cos x))( cos x)   ( cos x)( tan x)dx
sin x
  cos x ln(cos x)   (cos 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)

1
du  dx
1 x2
v   1dx  x


 sin  vdu
1
x dx  uv 

1
 (sin 1 x)( x)  x 1 x2
dx

 1
 x sin 1 x     (1  x 2 )  (2 x) dx
 2

1
 x sin 1 x  (1  x 2 ) (2)  C
2

 x sin 1 x  1 x2  C ANSWER



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