SolitaryRoad.com

```Website owner:  James Miller
```

[ Home ] [ Up ] [ Info ] [ Mail ]

INTEGRATION BY PARTS

A study of integration is largely a study of methods of transforming various types of integrands into forms in which the fundamental integration formulas can be applied.

Integration by parts. One of the most useful aids to integration is the process known as integration by parts. The formula for the differential of a product

d(uv) = udv + vdu

may be written

udv = d(uv) - vdu .

Integrating both sides, we obtain

which is the formula for integrating by parts. In order to use this formula we must regard the given integrand as the product of a function u and the differential dv of another function v. There is no general rule that one can give for breaking the integrand into two parts. One usually takes as much of it as he can easily integrate as dv and calls the remainder u. Two rules can be stated, however:

(a)       the part selected as dv must be readily integrable.

(b)        v du should not be more complex than u dv

Example 1. Find

Solution. We can integrate ex dx, so let us choose u = x and dv = ex dx.

Then du = dx and

Then our integral becomes, integrating by parts,

and this last integral is one of the standard forms. Hence

Example 2. Find

Solution. Set u = x2 and dv = sin x dx.

Then du = 2x dx and v = -cos x.

Using the formula

Note that this integral is of the same type except that we have x instead of x2 as a multiplier of the trigonometric part. We therefore apply the rule once more to the integral

Set u = x and dv = cos dx. Then du = dx and v = sin x. Our integral then becomes

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

In some cases it may be possible to use the procedure illustrated by the following example.

Example 3. Find

Solution. Set u = ex and dv = sin x dx. Then du = exdx and v = -cos x.

For the last integral set u = ex, dv = cos x dx. Then du = ex and v = sin x. So

Transposing we have

or finally

Example 4. Find

Solution. Set u = x2 and

Then du = 2xdx and

by the formula for integrals of functions type un. Then

This last integral is again one containing a function of type un so

One gains proficiency in this method of integration only by solving many problems. Choosing the parts u and dv is not easy without wide experience.

Method for some special integrals.

● Integrals

can be evaluated by integration by parts. Set dv = xm dx.

● Integrals of the forms

can be evaluated by n successive integrations by parts. Set u = xn on the first integration.

Reduction formulas. The labor involved in successive applications of integration by parts to evaluate an integral may be materially reduced by the use of reduction formulas. In general, a reduction formula yields a new integral of the same form as the original but with an exponent increased or reduced. A reduction formula succeeds if ultimately it produces an integral which can be evaluated. Among the reduction formulas are:

More from SolitaryRoad.com:

[ Home ] [ Up ] [ Info ] [ Mail ]