Mathematical Methods for Physics and Engineering : A Comprehensive Guide

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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.

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