Understanding Engineering Mathematics

9.2.6 Thedu=f′.x/dxsubstitution

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Integration by substitution is often difficult. The substitution to use is not always obvious.
One case when it is fairly easy to spot the substitution occurs when the integrand contains
afunction f.x/and its derivative f′.x/. One then tries substituting for the function,
u=f(x). This is the reverse of the function of a function rule of differentiation. This
approach relies on ‘noticing’ the derivative – i.e. on good facility with differentiation. An
important general example of this approach is the integral

∫

f′(x)dx

f(x)

=ln|f(x)|+C

because if we putu=f(x)we getdu=f′(x)dx, which occurs on the top and the
integral becomes ∫
du
u

=ln|u|+C=ln|f(x)|+C

Similarly, for example

∫

cos(f (x))f′(x)dx=sin(f (x))+C

by puttingu=f(x).

In general, to integrate

∫

g(f (x))f′(x)dxputu=f(x)and convert it to

∫

g(u)du,

sometimes written

∫

g(f (x))df(x), and hope that you can integrate this. We illustrate

the process by a further example:

Given

∫

xex

2

dx

Note that the derivative ofx^2 is 2x, and we have^12 ( 2 x)=xin the integrand. We know

the derivative ofexisex, so maybe the integral looks something likeex
2
? So differentiate
ex
2
and see what we get

d

dx

(ex

2

)=ex

2

×( 2 x)= 2 xex

2

Not quite – the factor 2 is the problem. But that is easily dealt with:

d

dx

(

1

2

ex

2

)

=xex

2

gives us what we want. So

∫

xex

2

dx=

1

2

ex

2

+C

You may need to try this a couple of times – and in the end it may not work, anyway. So,
BE FLEXIBLE AND KNOW YOUR DIFFERENTIATION.