Reduction formulae 427
From equation (1), I n = x
n e
x
− n I n−1
Hence
x
3 e
x dx = I 3 = x 3 e x − 3I 2
I 2 = x 2 e x − 2I 1
I 1 = x 1 e x − 1I 0
and
I 0 =
x
0 e
x dx =
e
x dx = e
x
Thus
x
3 e
x dx = x 3 e x − 3[x 2 e x − 2I 1 ]
= x 3 e x − 3[x 2 e x − 2(xe x − I 0 )]
= x 3 e x − 3[x 2 e x − 2(xe x − e x )]
= x 3 e x − 3x 2 e x + 6(xe x − e x )
= x 3 e x − 3x 2 e x + 6xe x − 6e x
i.e.
x
3 e
x dx = e x (x 3 − 3x 2 + 6x −6)+ c
Now try the following exercise
Exercise 169 Further problems on using
reduction formulae for integrals of the form
x n e x dx
1. Use a reduction formula to determine
x 4 e x dx.
[e x (x 4 − 4x 3 + 12x 2 − 24x + 24) + c]
2. Determine
t 3 e 2t dt using a reduction formula.
e 2t 1
2 t 3 −
3
4 t 2 +
3
4 t −
3
8
+ c
3. Use the result of Problem 2 to evaluate
1
0 5t 3 e 2t dt, correct to 3 decimal places.
[6.493]
44.3 Using reduction formulae for
integrals of the form
x n cos x dx
and
x n sin x dx
(a)
x
n cos x dx
Let I n =
x n cos x dx then, using integration by parts:
if
u = x n then
du
dx
= nx n−1
and if dv = cos x dx then
v =
cos x dx = sin x
Hence I n = x
n sin x −
(sin x)nx
n−1 dx
= x
n sin x − n
x
n−1 sin x dx
Using integration by parts again, this time with
u = x n−1 :
du
dx
= (n − 1)x n−2 , and dv = sin x dx,
from which,
v =
sin x dx = −cos x
Hence I n = x
n sin x − n
x
n−1
(−cos x)
−
(−cos x)(n − 1)x
n−2 dx
= x n sin x + nx n−1 cos x
− n(n − 1)
x
n−2 cos x dx
i.e.
I n = x n sin x + nx n−1 cos x
− n(n −1)I n−2
(2)
Problem 3. Use a reduction formula to determine
x 2 cos x dx.
Using the reduction formula of equation (2):
x
2 cos x dx = I 2
= x 2 sin x + 2x 1 cos x − 2(1)I 0
and
I 0 =
x
0 cos x dx
=
cos x dx = sin x
Hence
x
2 cos x dx = x 2 sin x +2x cos x − 2 sin x +c
Problem 4. Evaluate
2
1 4t 3 cos t dt , correct to 4
significant figures.
Let us firstly find a reduction formula for
t 3 cos t dt .