Reduction formulae 433
=
tan
n−2 x sec
2 x dx − I n−2
i.e. I n =
tan n−1 x
n − 1
− I n−2
When n =7,
I 7 =
tan
7 x dx =
tan 6 x
6
− I 5
I 5 =
tan 4 x
4
− I 3 and I 3 =
tan 2 x
2
− I 1
I 1 =
tan x dx = ln(sec x)
from Problem 9, Chapter 39, page 394
Thus
tan
7 x dx =
tan 6 x
6
−
tan 4 x
4
−
tan 2 x
2
− ln(sec x)
Hence
tan
7 x dx
=
1
6
tan
6 x −
1
4
tan
4 x +
1
2
tan
2 x
− ln(sec x) + c
Problem 14. Evaluate, using a reduction formula,
π
2
0
sin
2 t cos
6 t dt .
π
2
0
sin
2 t cos
6 t dt =
π
2
0
(1 − cos
2 t ) cos
6 t dt
=
π
2
0
cos
6 t dt −
π
2
0
cos
8 t dt
If
I n =
π
2
0
cos
n t dt
then
π
2
0
sin
2 t cos
6 t dt = I 6 − I 8
and from equation (6),
I 6 =
5
6
I 4 =
5
6
3
4
I 2
=
5
6
3
4
1
2
I 0
and
I 0 =
π
2
0
cos
0 t dt
=
π
2
0
1 dt = [x]
π
2
0 =
π
2
Hence
I 6 =
5
6
·
3
4
·
1
2
·
π
2
=
15π
96
or
5π
32
Similarly, I 8 =
7
8
I 6 =
7
8
·
5π
32
Thus
π
2
0
sin
2 t cos
6 t dt = I 6 − I 8
=
5π
32
−
7
8
·
5π
32
=
1
8
·
5π
32
=
5π
256
Problem 15. Use integration by parts to
determine a reduction formula for
(ln x) n dx.
Hence determine
(ln x) 3 dx.
Let I n =
(ln x) n dx.
Using integration by parts, let u =(ln x)
n , from which,
du
dx
= n(ln x)
n−1
1
x
and du = n(ln x)
n−1
1
x
dx
and let dv = dx, from which, v =
dx = x
Then I n =
(ln x)
n dx
= (ln x)
n
(x) −
(x)n(ln x)
n−1
1
x
dx