8/13/2019 Integration by Parts Shortcut DI METHOD

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INTEGRATION BY PARTS THE DI METHOD

This is a short cut to integration by parts and is especially useful when one has to integrate by parts

several times. It is a schematic method of the traditional udv method that usually is written as

in calculus books.udv uv vdu= We will illustrate and demonstrate by using examples. Remember that integration by parts is usually

indicated when you have to integrate a product of two functions. This method works exceedingly well when

one of these functions is a polynomial and the other is successively integrate. In this case it is possible tosuccessively differentiate until the last derivative is zero.

Example 1. Find xxe dx . The format is as follows:

D I

x ex

+

1 e

x

-

0 ex+

Dmeans to differentiate the functions in that column. Imeans to integrate the functions in that column.

Form the diagonal products as indicated by the arrows in the above table alternating algebraic signs as you

move down the table. Finally, you integrate the product of the last entry in the Dcolumn and the last entry in

the Icolumn, with the appropriate sign, of course.

It follows that the solution to the above problem is: 1 0x x x xxe dx x e e e dx= + + i i i which is: x x xxe dx xe e C= + .

Example 2. Find 2 xx e dx . The format is as follows:D I

x2

e

x

+

2x

e

x

-

2 ex

+

0 e

x

-

and therefore the answer to the problem is: 2 2 2 2 0x x x x xx e dx x e x e e e dx= + + i i i i or 2 2 2 2x x x xx e dx x e xe xe C= + + .

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Example 3. Find sinx xdx .D I

x sinx+

1 -cosx

-

0 -sinx+

and therefore the solution to the problem is: sin cos sinx xdx x x x C= + + .Here we have not written the last term as ( )0 sin x dx i but rather just as an arbitrary constant of integration C.

Example 4. Find .( )2 cos 3xe x dxNotice that here neither function is a polynomial. Since we have been picking the polynomial, xs to powers,

and differentiated until we reached zero we were able to discard the last integral of a product which was zeroand merely write C. We can use the DI method to work out the above integral in the following fashion:

D I

e-2x

cos3x

+

-2e

-2x

sin3x/3

-

4e-2x

-cos3x/9

+

Notice we stopped the process when the product of the functions in the Dand Icolumns in the third row wasthe same as the product of the functions in the Dand Icolumns in the first row, except for constants. That is

we have in the first row and we have2 cos3xe x 24

cos39

xe

x in the third row. They are the same except for

the constant multiple4

9 .

We have: ( ) ( )2 2 2 21 2 4

cos 3 sin 3 cos3 cos 33 9 9

x x x xe x dx e x e x e x dx

= + .

Solving for the integral we wish to evaluate, and adding an arbitrary constant of integration, we have:

( )2 2 23 2

cos 3 sin 3 cos313 13

x x xe x dx e x e x

C= +

.