274 integration by substitution

du

––

dx

du

––

dx

√x3–1

du

––

dx

du

––

dx

Then:

The integration-by-parts formula can be applied more

than once to complete an integration problem. For

example, to evaluate ∫excos(x)dx, one application of

the technique yields:

∫excos(x)dx = exsin(x) – ∫exsin(x)dx

Applying the integration-by-parts technique to the sec-

ond integral yields:

∫excos(x)dx = exsin(x) – ∫exsin(x)dx

= exsin(x) – [–excos(x) – ∫ex(–cos(x))dx]

= ex(sin(x) + cos(x)) – ∫excos(x)dx

Algebra now shows that the integral we seek is given by:

The method of integration by parts was discovered by

English mathematician B

ROOK

T

AYLOR

(1685–1731).

See also

ANTIDIFFERENTIATION

;

INTEGRAL CAL

-

CULUS

;

INTEGRATION BY SUBSTITUTION

;

RECURRENCE

RELATION

.

integration by substitution (change of variable, sub-

stitution rule for integration) This technique of inte-

gration is used to find the integral of a function easily

recognized as a

COMPOSITION

of two simpler functions.

It is essentially the

CHAIN RULE

for differentiation

employed in reverse. Specifically, the chain rule states:

Integrating both sides yields:

Noting that, since fis the antiderivative of f′, the right-

hand side of this formula can be thought of as the

indefinite integral of f′evaluated with uas the variable.

Thus it is valid to write:

This equation is the change-of-variables equation for

integration. For example, we can use it to evaluate

∫2x(1 + x2)4dx. Setting u(x) = 1 + x2, giving = 2x,

the integral reads:

Notice that the notation we used here mimics the prop-

erties of fractions: we are permitted to replace a term

dx under an integral sign by the term dx.

When using this technique, one typically chooses

u(x) to be a function that simplifies a complicated part

of the

INTEGRAND

, and then adjusts matters so that the

factor appears explicitly. For example, to evaluate

∫x2dx, it is natural to set u(x) = x3– 1. Then

= 3x3, and we have: