Higher Engineering Mathematics, Sixth Edition

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

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 tdt using 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

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.