Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

2.2 INTEGRATION


Rearranging this expression to obtainIexplicitly and including the constant of integration
we find


I=

eax
a^2 +b^2

(bsinbx+acosbx)+c. (2.37)

Another method of evaluating this integral, using the exponential of a complex number,
is given in section 3.6.


2.2.9 Reduction formulae

Integration using reduction formulae is a process that involves first evaluating a


simple integral and then, in stages, using it to find a more complicated integral.


Using integration by parts, find a relationship betweenInandIn− 1 where

In=

∫ 1


0

(1−x^3 )ndx

andnis any positive integer. Hence evaluateI 2 =

∫ 1


0 (1−x

(^3) ) (^2) dx.
Writing the integrand as a product and separating the integral into two we find
In=


∫ 1


0

(1−x^3 )(1−x^3 )n−^1 dx

=


∫ 1


0

(1−x^3 )n−^1 dx−

∫ 1


0

x^3 (1−x^3 )n−^1 dx.

The first term on the RHS is clearlyIn− 1 and so, writing the integrand in the second term
on the RHS as a product,


In=In− 1 −

∫ 1


0

(x)x^2 (1−x^3 )n−^1 dx.

Integrating by parts we find


In=In− 1 +

[x

3 n

(1−x^3 )n

] 1


0


∫ 1


0

1


3 n

(1−x^3 )ndx

=In− 1 +0−

1


3 n

In,

which on rearranging gives


In=

3 n
3 n+1

In− 1.

We now have a relation connecting successive integrals. Hence, if we can evaluateI 0 ,we
can findI 1 ,I 2 etc. EvaluatingI 0 is trivial:


I 0 =

∫ 1


0

(1−x^3 )^0 dx=

∫ 1


0

dx=[x]^10 =1.

Hence


I 1 =

(3×1)


(3×1) + 1


×1=


3


4


,I 2 =


(3×2)


(3×2) + 1


×


3


4


=


9


14


.


Although the first fewIncould be evaluated by direct multiplication, this becomes tedious
for integrals containing higher values ofn; these are therefore best evaluated using the
reduction formula.

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