Understanding Engineering Mathematics

but only on this first example will we explicitly write

u=x

dv

dx

=sinx

du

dx

= 1 v=−cosx

to help you on your way. Note that we choseu=xso that when we

differentiate thexbecomes a much more amenable 1! We have

∫

xsinxdx=x(−cosx)−

∫

d(x)

dx

(−cosx)dx

=−xcosx+

∫

cosxdx

=−xcosx+sinx+C

With practice, you should be able to dispense with explicit use ofu,

v, etc. – although continue with it until you feel confident enough to

do without.

(ii) We may need to use integration by parts more than once (if you do,

remember to keep integrating the same factor). Again, we are going to

chose to differentiate thex^2 to reduce it, so we first integrate theex.

∫

x^2 exdx=x^2 (ex)−

∫

2 xexdx

=x^2 ex− 2

[

xex−

∫

exdx

]

=x^2 ex− 2 xex+ 2 ex+C

(iii) In integrating

∫

exsinxdx we sometimes feel like we are going

round in circles with integration by parts! Write

I=

∫

exsinxdx=exsinx−

∫

excosxdx

(note that it doesn’t matter which ofex,sinxwe integrate)

=exsinx−

[

excosx−

∫

ex(−sinx)dx

]

being careful to keep integrating the same factor

=ex(sinx−cosx)+C−

∫

exsinxdx

=ex(sinx−cosx)−I+C