25Integration by Parts - University of California, Berkeley

Integration By Parts

3

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: x3 ln x dx

*Since lnx is a logarithmic function and x3 is an algebraic function, let:

u = lnx

(L comes before A in LIATE)

dv = x3 dx

1

du = dx

x

v = x3dx = x 4

4

x3 ln xdx = uv - vdu

= (ln x) x4 - x4 1 dx

4

4x

=

x4 (ln x)

-

1

x 3 dx

44

= x 4 (ln x) - 1 x 4 + C

4

44

=

x4 (ln x)

-

x4

+C

4

16

ANSWER

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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])

du = 1 (- 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

= - cos x

ln(cos

x)

-

(cos

x)

sin cos

x x

dx

= - 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 dx 1- x2

v = 1dx = x

sin -1 x dx = uv - vdu = (sin -1 x)(x) - x 1 dx

1- x2

= x sin -1 x - - 1 (1 - x2 )-1/ 2 (-2x) dx 2

= x sin -1 x + 1 (1 - x 2 )1/ 2 (2) + C 2

= x sin -1 x + 1 - x 2 + C ANSWER

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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:

x3 4 - x2 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 - x2 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 = 2x dx

v = x 4 - x2 dx = - 1 (-2x)(4 - x2 )1/ 2 dx 2

= - 1 2 (4 - x 2 )3/ 2 = - 1 (4 - x 2 )3/ 2

2 3

3

x3 4 - x3 dx = uv - vdu

= (x 2 ) - 1 (4 - x 2 )3/ 2 - - 1 (4 - x 2 )3/ 2 (2x) dx

3

3

= - x 2 (4 - x 2 )3/ 2 - 1 (4 - x 2 )3/ 2 (-2x) dx

3

3

= - x 2 (4 - x 2 )3/ 2 - 1 (4 - x 2 )5 / 2 2 + C

3

3

5

= - x 2 (4 - x 2 )3/ 2 - 2 (4 - x 2 )5/ 2 + C Answer

3

15

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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: x2 sin x dx

u = x 2 (Algebraic Function) dv = sin x dx (Trig Function)

du = 2x dx

v = sin x dx = - cos x x2 sin x dx = uv - vdu

= x2 (- cos x) - - cos x 2x dx = -x2 cos x + 2 x 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 x2 sin x dx = -x2 cos x + 2 [uv - vdu]

= -x2 cos x + 2 [x sin x - sin x dx]

= -x2 cos x + 2 [x sin x + cos x + c]

= -x2 cos x + 2x sin x + 2 cos x + c Answer

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Note: After each application of integration by parts, watch for the appearance of a constant multiple of the original integral.

Example:

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 x cos x dx = e x cos x + (uv - vdu)

e x 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 (e x cos x + e x sin x) + C Answer

2

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