Integration by Parts - Math and Science with Dr. Taylor



Integration by Parts

I. Basic Technique

By the Product Rule for Derivatives, [pic]. Thus,

[pic]

[pic]. This formula for integration

by parts often makes it possible to reduce a complicated integral involving a product to

a simpler integral. By letting [pic]

[pic]

we get the more common formula for integration by parts:

[pic]

Example 1: Find [pic].

Let [pic] and [pic] and [pic]. Thus,

[pic]

[pic]

[pic].

It is possible that when you set up an integral using integration by parts, the resulting

integral will be more complicated than the original integral. In this case, change your

substitutions for u and dv.

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Example 2: Find [pic].

Let [pic]and [pic] and [pic]. Thus,

[pic]

[pic]. Notice that this resulting integral is

more complicated that the original one. Therefore, let [pic] and [pic]

[pic] and [pic]. Thus, [pic]=

[pic]

[pic].

Example 3: Find [pic].

Let [pic] and [pic] and [pic]. Thus,

[pic]

[pic].

Example 4: Find [pic].

As in the previous example, let [pic] and [pic]

and [pic]. Thus, [pic][pic]

[pic]

[pic]= [pic].

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Example 5: Find [pic].

Let [pic]and [pic]and [pic]

[pic]. Thus, [pic]

[pic]

[pic].

Sometimes, it is necessary to use integration by parts more than once.

Example 6: Find [pic].

Let [pic] and [pic] and [pic]. Thus,

[pic]

[pic]. Notice that the resulting integral [pic]is less complicated

than the original one, but integration by parts is needed to evaluate it.

Let [pic] and [pic] and [pic]. Thus,

[pic]

[pic]. Finally, we get [pic]

[pic]. [Note: We will be

able to integrate [pic] more easily with tabular integration; this

technique will be described later.]

The next illustration of repeated integration by parts deserves special attention.

Example 7: Find [pic].

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Let [pic] and [pic] and [pic]. Thus,

[pic]

[pic]. Notice that integration by parts is now needed to

evaluate [pic]. Let [pic]and [pic] and

[pic]. Thus, [pic]

[pic]. Returning to the original problem,

[pic] = [pic].

Thus, [pic][pic] = [pic]

[pic] and this does check.

The next example illustrates an interesting type of integral that surprisingly requires integration by parts.

Example 8: [pic].

Let [pic][pic]

[pic]. In example 2, we got the following using integration by parts:

[pic] = [pic] or [pic] = [pic].

Thus, [pic]= [pic]= [pic]

[pic] and this does check.

In general, to evaluate [pic], let [pic]

[pic]= [pic].

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II. Tabular Integration

Integrals of the form [pic], in which f can be differentiated repeatedly

to become zero and g can be integrated repeatedly without difficulty, can be

evaluated using tabular integration.

Example: Find [pic].

[pic] and its derivatives [pic] and its antiderivatives

[pic] (+) [pic]

[pic] (–) [pic]

6x (+) [pic]

6 (–) [pic]

0 [pic]

[pic] = [pic]

= [pic]

III. Reduction Formulas

Integration by parts can be used to derive reduction formulas for integrals. These

are formulas that express an integral involving a power of a function in terms of an

integral that involves a lower power of that function

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Example 1: Prove the reduction formula [pic] and

use the result to find [pic].

Let [pic] and [pic] and [pic]. Thus,

[pic]

[pic]. Thus, [pic]

[pic] and this checks.

Example 2: Prove the reduction formula [pic]

[pic].

Let [pic] and [pic] and v =

[pic]. Thus, [pic]

[pic]

[pic]

[pic]

[pic]

[pic].

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Practice Sheet for Integration by Parts

(1) [pic] dx =

2) Prove the following reduction formula: [pic]

[pic] for [pic].

(3) [pic] dx =

(4) [pic] dx =

(5) [pic] dx =

(6) [pic] arcsec x dx =

(7) [pic] dx =

(8) [pic] dx =

(9) [pic] dx =

(10) [pic] dx =

(11) [pic] dx =

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(12) [pic]

(13) [pic]

(14) [pic]

Solution Key for Integrations by Parts

(1) Let [pic] and [pic] [pic] and [pic]

[pic] dx = [pic]

[pic].

2) Let [pic] and [pic] and [pic]

[pic][pic]

[pic].

(3) Let [pic] and [pic] and [pic]

[pic] dx = [pic]

[pic]

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(4) Let [pic] [pic] dx = [pic]. Let

[pic] and [pic] and [pic][pic] dx =

[pic] = [pic]

[pic].

5) Let [pic] and [pic] and [pic]

[pic] dx = [pic]

[pic].

(6) Let [pic] and [pic] and [pic]

[pic] arcsec x dx = [pic].

(7) Let [pic] and [pic]and[pic][pic]

[pic]. Use integration by parts to evaluate [pic]. Let

[pic] and [pic] and [pic][pic] =

[pic]. Thus, [pic] dx = [pic] =

[pic]

[pic][pic].

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(8) Let [pic]

[pic]. Let [pic] and [pic] and [pic]

[pic]

[pic].

(9) Let [pic][pic] dx = [pic]. Let [pic] and

[pic] and [pic][pic]

[pic][pic] dx = [pic].

(10) Use tabular integration:

f (x) and its derivatives g (x) and its antiderivatives

[pic] [pic]

(+)

[pic] [pic]

(–)

2 [pic]

(+)

0

[pic]

Thus, [pic] dx = [pic].

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(11) Use tabular integration:

f (x) and its derivatives g (x) and its antiderivatives

[pic] [pic]

(+)

[pic] [pic]

(–)

2 [pic]

(+)

0 [pic]

Thus, [pic] dx = [pic].

(12) Let [pic][pic][pic]. [pic] can

integrated using tabular integration:

f (x) and its derivatives g (x) and its antiderivatives

[pic] (+) [pic]

[pic] (–) [pic]

2 (+) [pic]

0 [pic]

[pic] = [pic][pic]

[pic]+ C.

(13) Let [pic][pic][pic].

In solution to problem #9 previously, [pic][pic]

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[pic].

(14) Let [pic]sin(ln x) and [pic]and[pic]

[pic] [pic]. Use integration by parts to evaluate

[pic]. Let [pic] and [pic] and

[pic][pic]= [pic]. Thus, [pic]

[pic]= [pic]

[pic]

[pic][pic].

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