Merced College; Don Power

MATH 4B -- LAB 3 -- TABULAR INTEGRATION BY PARTS NAME __________________

Example 1. Integrate:

Step 1: Since , we must first identify u and dv, as follows:

u dv The original integrand is the product of u and dv; the product is positive.

x² sin x

Step 2: Differentiate u and integrate dv, writing the results (du and v) on the next line

u dv The product on the diagonal is uv, and the product on the bottom line is v du

x² sin x The extra negative on the bottom line is to give us the negative of ∫ v du.

2x -cos x

Step 3: The factors on the last row become a new u and dv; repeat step 2:

u dv The "-" on the bottom line in step 2 is carried forward to the second product uv.

x² sin x The same "-" also applies to the second "-∫ v du," which is therefore positive.

2x -cos x Notice that we will have a negative product on every other line.

2 -sin x

Step 4: Repeat the process again.

u dv Since the final integral (on the bottom line) has a factor of 0, we are done.

x² sin x (When we get a factor of 0, we routinely omit the last line)

2x -cos x

2 -sin x

0 cos x I = -x² cos x + 2x sin x + 2cos x + C

Example 2: Whenever possible, if u is a polynomial, continue until you get a 0.

u dv

x³ e^-2x

3x²

6x

6

0

Example 3:

As soon as we get an integral we can evaluate, we should do it.

Also, we may need to simplify before the integration:

u dv

ln x 1

x

Example 4:

Integrals involving sines, cosines, and possibly exponential functions can "cycle" after two stages of integration by parts.That is, the later integral is a multiple of the original integral.

u dv

sin 2x

2 cos 2x

-4sin2x

or

Example 5:

Most textbooks tell you to use product-to-sum identities on this problem.

Integration by parts is easier (you don't have to remember the trig formula).

In addition, you can check more easily by differentiating.

(The product-to-sum identities change the angles, so you'd have to convert the angles to check)

u dv

sin 3x cos 5x

3 cos 3x

-9sin 3x

Example 6:

Integrals should be simplified if possible at each step before continuing.

In this case, we get 1/x in the u column and x in the dv column.

It also helps avoid extra calculations if we move negative signs into the integrand.

First stage:

u dv

sin(ln x) 1

x

Second stage:

u dv

+

-cos(ln x) 1

-

x

Assignment: Use the tabular procedure to integrate:

1. This should work like examples 1 and 2

2. The inverse trig functions should all work like example 3

3. This will cycle like examples 4 and 5

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