REDUCTION FORMULAE 425

H

and I 0 =

∫

x^0 exdx=

∫

exdx=ex

Thus

∫

x^3 exdx=x^3 ex−3[x^2 ex− 2 I 1 ]

=x^3 ex−3[x^2 ex−2(xex−I 0 )]

=x^3 ex−3[x^2 ex−2(xex−ex)]

=x^3 ex− 3 x^2 ex+6(xex−ex)

=x^3 ex− 3 x^2 ex+ 6 xex−6ex

i.e.

∫

x^3 exdx=ex(x^3 − 3 x^2 + 6 x−6)+c

Now try the following exercise.

Exercise 170 Further problems on using

reduction formulae for integrals of the form∫

xnexdx

1. Use a reduction formula to determine∫
   x^4 exdx.
   [ex(x^4 − 4 x^3 + 12 x^2 − 24 x+24)+c]


2. Determine

∫

t^3 e^2 tdtusing a reduction for-

mula. [

e^2 t

( 1

2 t

(^3) − 3
4 t
(^2) + 3
4 t−
3
8
)
+c
]

3. Use the result of Problem 2 to evaluate∫ 1
   05 t

(^3) e 2 tdt, correct to 3 decimal places.
[6.493]
44.3 Using reduction formulae for
integrals of the form
∫
xncosxdx
and
∫
xnsinxdx
(a)
∫
xncosxdx
Let In=
∫
xncosxdx then, using integration by
parts:
if u=xnthen
du
dx
=nxn−^1
and if dv=cosxdxthen
v=
∫
cosxdx=sinx
Hence In=xnsinx−
∫
( sinx)nxn−^1 dx
=xnsinx−n
∫
xn−^1 sinxdx
Using integration by parts again, this time with
u=xn−^1 :
du
dx
=(n−1)xn−^2 , and dv=sinxdx,
from which,
v=
∫
sinxdx=−cosx
Hence In=xnsinx−n
[
xn−^1 (−cosx)
−
∫
(−cosx)(n−1)xn−^2 dx
]
=xnsinx+nxn−^1 cosx
−n(n−1)
∫
xn−^2 cosxdx
i.e.
In=xnsinx+nxn−^1 cosx
−n(n−1)In− 2
(2)
Problem 3. Use a reduction formula to deter-
mine
∫
x^2 cosxdx.
Using the reduction formula of equation (2):
∫
x^2 cosxdx=I 2
=x^2 sinx+ 2 x^1 cosx−2(1)I 0
and I 0 =
∫
x^0 cosxdx

∫
cosxdx=sinx
Hence
∫
x^2 cosxdx=x^2 sinx+ 2 xcosx
−2 sinx+c
Problem 4. Evaluate
∫ 2
14 t
(^3) costdt, correct to
4 significant figures.
Let us firstly find a reduction formula for∫
t^3 costdt.