434 Higher Engineering Mathematics
= x(ln x)
n
− n
(ln x)
n−1 dx
i.e. I n = x(ln x)
n
− nI n−1
When n =3,
(ln x)
3 dx = I 3 = x(ln x)
3
− 3I 2
I 2 = x(ln x)
2
− 2I 1 and I 1 =
ln x dx = x(ln x − 1) from
Problem 7, page 422.
Hence
(ln x)
3 dx = x(ln x)
3
− 3[x(ln x)
2
− 2I 1 ] + c
= x(ln x)
3
− 3[x(ln x)
2
− 2[x(ln x − 1)]] + c
= x(ln x)
3
− 3[x(ln x)
2
− 2x ln x + 2x] + c
= x(ln x)
3
− 3x(ln x)
2
+ 6x ln x − 6x + c
= x[(ln x)
3
− 3(ln x)
2
+ 6 ln x − 6] + c
Now try the following exercise
Exercise 172 Further problems on
reduction formulae
1. Evaluate
π
2
0
cos
2 x sin
5 x dx.
8
105
2. Determine
tan
6 x dx by using reduction formulae and hence evaluate
π
4
0
tan
6 x dx.
13
15
−
π
4
3. Evaluate
π
2
0
cos
5 x sin
4 x dx.
8
315
4. Use a reduction formula to determine
(ln x)
4 dx.
x(ln x) 4 − 4x(ln x) 3 + 12x(ln x) 2
− 24x ln x + 24x + c
5. Show that
π
2
0
sin
3
θ cos
4
θ dθ =
2
35