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

460 Chapter 14 Techniques of integration
Solution We let
and so
u=e t
and
dv
dt
=sint
du
dt =et and v=−cost
Applying integration by parts yields
∫
∫
e t sintdt =−e t cost− (−cost)e t dt +c
∫
=−e t cost+
e t costdt +c (14.1)
We now apply integration by parts to ∫ e t costdt. We let
Then
u=e t
and
dv
dt
=cost
du
dt =et and v =sint
So
∫ ∫
e t costdt =e t sint−
e t sintdt (14.2)
Substituting Equation (14.2) into Equation (14.1) yields
∫
∫
e t sintdt =−e t cost+e t sint− e t sintdt+c
Rearranging the equation gives
∫
2 e t sintdt =−e t cost+e t sint+c
from which we see that
∫
e t sintdt = −et cost+e t sint+c
2
Example14.5 Evaluate
∫ 2
0
x n e x dx
forn=3,4,5.
Solution The integral may be evaluated by using integration by parts repeatedly. However, this
is slow and cumbersome. Instead it is useful to develop a reduction formula as is now
illustrated.
Letu =x n and dv
dx = ex . Then du
dx =nxn−1 and v = e x .