Reduction formulae 427
From equation (1),In=xnex−nIn− 1
Hence
∫
x^3 exdx=I 3 =x^3 ex− 3 I 2
I 2 =x^2 ex− 2 I 1
I 1 =x^1 ex− 1 I 0
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 169 Further problems on using
reduction formulae for integrals of the form∫
xnexdx
- Use a reduction formula to determine∫
x^4 exdx.
[ex(x^4 − 4 x^3 + 12 x^2 − 24 x+ 24 )+c] - Determine
∫
t^3 e^2 tdt using a reduction for-
mula.
[
e^2 t
( 1
2 t
(^3) − 3
4 t
(^2) + 3
4 t−
3
8
)
+c
]
- 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
LetIn=
∫
xncosxdxthen, 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 ,anddv=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 determine
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−2sinx+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.