Tabular integration by parts Kowalski

Tabular integration by parts. The integration by parts formula can be difficult to apply repeatedly: ittakes a lot of space to write down and its easy to make a distribution error. Fortunately, there is a purelymechanical procedure for performing integration by parts without writing down so much called tabularintegration by parts. It is based on the following theorem and table.

High-order integration by parts. Suppose that u0(x) and v0(x) are given functions, and define a sequenceof derivatives of u0 by

u1(x) = u0(x), u2(x) = u1(x), . . . un(x) = un1(x)

and define a sequence of antiderivatives of v0 by

v1(x) =v0(x) dx, v2(x) =

v1(x) dx, . . . vn(x) =

vn1(x) dx.

Then the n-th order integration by parts formula isu0v

0 dx = u0v1 u1v2 + u2v3 u3v4 + + (1)n1un1vn + (1)nunv

n dx

for any integer n > 0.

The table. This theorem can be visualized in a table as

u

u0

dv

v0

u1 v1

u2 v2

u3 v3

... v4

un2 ...

un1 vn1

un vn

@@@R

+

@@@@R

@@@@R

+

@@@@R

@@@R

@@@@R

- R

In it, each diagonal arrow means to multiply the two functions with the appropriate sign (+ or ), whilethe lone horizontal arrow at the bottom means to integrate the product of the last two functions (with theappropriate sign).

Tabular IBP algorithm. This technique is often used when the integrand is a product of two differentclasses of functions. Remember that there are five classes of elementary functions, recalled by the mnemonicLIATE: Logarithms, Inverse trig, Algebraic, Trigonometric, and Exponential.) To use the IBP formula,first choose u and dv. Two helpful (though not mutually exclusive) rules of thumb are:

u: This should be an expression that simplifies upon differentiation. Often, the best choice is to setu equal to the first expression present from the term LIATE.

dv: This should be the most complicated expression you can integrate, even if you must use asubstitution to do so. If there is absolutely nothing you can integrate, set dv = 1 dx.

Place u and dv in the top of the appropriate columns and then write each new row in the table by differen-tiating the u column and integrating the dv column. There are really three main cases to worry about.

Case 1. The u column eventually goes to 0, i.e. u = A.If the u column reaches 0, then the antiderivative can be read off the table.

Best bet: This case is well suited for integrals of the form [polynomial

u

] [ something easy to integrate dv

] dx.

Example:x5 sinx dx, with u = x5 and dv = sin(x) dx:

u

x5dv

sin(x)

5x4 cos(x)

20x3 sin(x)

60x2 cos(x)

120x sin(x)

120 cos(x)

0 sin(x)

QQQs+

QQQs

QQQs+

QQQs

QQQs+

QQQs

-+R

Hence, the answer isx5 sin(x) dx = (x5)( cos(x)) (5x4)( sin(x)) + (20x3)(cos(x)) (60x2)(sin(x))

+ (120x)( cos(x)) (120)( sin(x)) +

0 dx

= x5 cos(x) + 5x4 sin(x) + 20x3 cos(x) 60x2 sin(x) 120x cos(x) + 120 sin(x) + C= (5x4 60x2 + 120) sin(x) + (x5 + 20x3 120x) cos(x) + C

Case 2. The u column eventually becomes algebraic, i.e. u = L or u = I.Recall that the algebraic functions are those without trigonometric, exponential, logarithmic, or inversetrigonometric terms, so essentially theyre combinations of power functions. In this case:

In the u column, cross out the last entry and write 1 beneath it. In the dv column, underneath the last entry write the product of the last u and dv terms.

Simply do one round of IBP, which should eliminate the logarithm or inverse trig term, and focus on thenew integral (which frequently will require a substitution to evaluate).

Best bet: This technique works very well on integrals of the form [logarithm

u

] [ algebraic dv

] dx or [

inverse trig u

] [ algebraic dv

] dx.

Example:x2 lnx dx, with u = ln(x) and dv = x2 dx:

u

ln(x)dv

x2

1x

13x3

@@R+

- R

Decoding the table gives x2 ln(x) dx = (ln(x))

(13x3) (

13x3)(

1x

)dx

=13x3 ln(x) +

13

x2 dx

=13x3 ln(x) +

13 13x3 + C

=19x3(3 ln(x) + 1) + C

This is more or less equivalent to doing standard integration by parts, but it omits all the u and dvnotation.

Case 3. Both the u and dv columns are periodic (ever repeating), i.e. u = T or u = E.In this case,

1. Apply two rounds of IBP and stop; the entry in the u column should match your original choice of uup to a multiple.

2. Extract the integral equation from the table.

3. Use basic algebra to solve for the unknown integral: move it to the LHS and divide out the multiplein front of the integral.

4. If its an indefinite integral, dont forget the +C.

This is the most complicated application of IBP, but its easy to get the hang of with a little practice.

Best bet: This technique works very well on integrals of the form [sine/cosine

u

] [ exponential dv

] dx or [

sine/cosine u

] [ sine/cosine dv

] dx.

Example:e3x cos 4x dx, with u = cos(4x) and dv = e3x dx:

u

cos(4x)dv

e3x

4 sin(4x) 13e3x

16 cos(4x) 19e3x

QQQQs

+

QQQQQs

-+R

Hence, the equation we obtain upon the first repeat ise3x cos(4x) dx = (cos(4x))

(13e3x) (4 sin(4x))

(19e3x)+(16 cos(4x))

(19e3x)dx

=13e3x cos(4x) +

49e3x sin(4x) 16

9

e3x cos(4x) dx.

Now, adding 169e3x cos(4x) dx to both sides, we find

259

x3x cos(4x) dx =

13e3x cos(4x) +

49e3x sin(4x)

whence e3x cos(4x) dx =

259

(13e3x cos(4x) +

29e3x sin(4x)

)=

325e3x cos(4x) +

425e3x sin(4x) + C

=125e3x(3 cos(4x) + 4 sin(4x)

)+ C.

where the final +C is added since this is a general antiderivative.