Mathematical Methods for Physics and Engineering : A Comprehensive Guide

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.