Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PRELIMINARY CALCULUS

The separation of the functions is not always so apparent, as is illustrated by

the following example.

Evaluate the integralI=

∫

x^3 e−x

2

dx.

Firstly we rewrite the integral as

I=

∫

x^2

(

xe−x

2 )

dx.

Now, using the notation given above, we identifyx^2 withuandxe−x

2

withdv/dx. Hence

v=−^12 e−x

2

anddu/dx=2x,sothat

I=−^12 x^2 e−x

2

−

∫

(−x)e−x

2

dx=−^12 x^2 e−x

2

−^12 e−x

2

+c.

A trick that is sometimes useful is to take ‘1’ as one factor of the product, as

is illustrated by the following example.

Evaluate the integralI=

∫

lnxdx.

Firstly we rewrite the integral as

I=

∫

(lnx)1dx.

Now, using the notation above, we identify lnxwithuand 1 withdv/dx. Hence we have
v=xanddu/dx=1/x,andso

I=(lnx)(x)−

∫ (

1

x

)

xdx=xlnx−x+c.

It is sometimes necessary to integrate by parts more than once. In doing so,

we may occasionally re-encounter the original integralI. In such cases we can

obtain a linear algebraic equation forIthat can be solved to obtain its value.

Evaluate the integralI=

∫

eaxcosbx dx.

Integrating by parts, takingeaxas the first function, we find

I=eax

(

sinbx

b

)

−

∫

aeax

(

sinbx

b

)

dx,

where, for convenience,we have omitted the constant of integration. Integrating by parts
a second time,

I=eax

(

sinbx

b

)

−aeax

(

−cosbx

b^2

)

+

∫

a^2 eax

(

−cosbx

b^2

)

dx.

Notice that the integral on the RHS is just−a^2 /b^2 times the original integralI. Thus

I=eax

(

1

b

sinbx+

a

b^2

cosbx

)

−

a^2

b^2

I.