3
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
du =
1
dx
x
x4
v = ∫ x dx =
4
3
∫x
3
ln xdx = uv − ∫ vdu
= (ln x)
x4
−
4
= (ln x)
x4
1 3
−
x dx
4
4∫
=
x4 1
∫ 4 x dx
x4
1 x4
+ C
(ln x) −
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
3
15
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Answer
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
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Answer
Page 5 of 7
Practice Problems:
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
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Page 6 of 7
Solutions:
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.
x(ln x) 2 − 2 x ln x + 2 x + C
u = (ln x) 2
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