082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

458 Chapter 14 Techniques of integration
Integrating thisequation yields
∫
u dv ∫ ∫ d
dx dx = dx (uv)dx−
v du
dx dx
Recognizing thatintegration and differentiation areinverse processes allows
∫ d(uv)
dx
dx
tobe simplified touv. Hence,
∫
( ) ∫ dv
u dx=uv−
dx
( du
v
dx
)
dx
This isthe formulafor integration by parts.
∫
Example14.1 Find
xsinxdx.
Solution We recognize the integrand as a product of the functions x and sinx. Let u = x,
dv du
=sinx. Then =1, v = −cosx. Using the integration by parts formula we get
dx dx
∫
∫
xsinxdx=x(−cosx)− (−cosx)1dx
=−xcosx+sinx+c
Whendealingwithdefiniteintegralsthecorrespondingformulaforintegrationbyparts
is
∫ b
a
( ∫ dv b
( ) du
u
)dx =[uv] b a
dx
− v dx
dx
a
Example14.2 Evaluate
Solution We let
Then
∫ 2
0
u=x
xe x dx
and
dv
dx = ex
du
=1 and v=ex
dx