CK-12-Calculus

7.2. Integration By Parts http://www.ck12.org

Tabular Integration by Parts

Sometimes, we need to integrate by parts several times. This leads to cumbersome calculations. In situations like
these it is best to organize our calculations to save us a great deal of tedious work and to avoid making unpredictable
mistakes. The example below illustrates the method oftabular integration.
Example 5:
Evaluate∫x^2 sin 3xdx.
Solution:
Begin as usual by lettingu=x^2 anddv=sin 3xdx.Next, create a table that consists of three columns, as shown
below:

TABLE7.1:

Alternate signs uand its derivatives dvand its antiderivatives

+ x^2 ↘ sin 3x

− 2 x↘ − 31 cos 3x

+ 2 ↘ − 91 sin 3x

− 0 271 cos 3x

To find the solution for the integral, pick the sign from the first row(+),multiply it byuof the first row(x^2 )and
then multiply by thedvof the second row,− 1 /3 cos 3x(watch the direction of the arrows.) This is the first term in
the solution. Do the same thing to obtain the second term: Pick the sign from the second row, multiply it by theuof
the same row and then follow the arrow to multiply the product by thedvin the third row. Eventually we obtain the
solution

∫

x^2 sin 3xdx=− 31 x^2 cos 3x+^29 xsin 3x+ 272 cos 3x+C.

Solving for an Unknown Integral

There are some integrals that require us to evaluate two integrations by parts, followed by solving for the unknown
integral. These kinds of integrals crop up often in electrical engineering and other disciplines.
Example 6:
Evaluate∫excosxdx.
Solution:
Letu=ex,anddv=cosxdx.Thendu=exdx,v=sinx,and

∫

excosxdx=exsinx−

∫

exsinxdx.

Notice that the second integral looks the same as our original integral in form, except that it has a sinxinstead of
cosx.To evaluate it, we again apply integration by parts to the second term withu=ex,dv=sinxdx,v=−cosx,
anddu=exdx.
Then