Reduction formulae 433
=
∫
tann−^2 xsec^2 xdx−In− 2
i.e.In=
tann−^1 x
n− 1
−In− 2
Whenn=7,
I 7 =
∫
tan^7 xdx=
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 =
∫
tanxdx=ln(secx)
from Problem 9, Chapter 39, page 394
Thus
∫
tan^7 xdx=
tan^6 x
6
−
[
tan^4 x
4
−
(
tan^2 x
2
−ln(secx)
)]
Hence
∫
tan^7 xdx
=
1
6
tan^6 x−
1
4
tan^4 x+
1
2
tan^2 x
−ln(secx)+c
Problem 14. Evaluate, using a reduction formula,
∫ π
2
0
sin^2 tcos^6 tdt.
∫ π
2
0
sin^2 tcos^6 tdt=
∫ π
2
0
( 1 −cos^2 t)cos^6 tdt
=
∫ π
2
0
cos^6 tdt−
∫ π
2
0
cos^8 tdt
If In=
∫ π
2
0
cosntdt
then
∫ π 2
0
sin^2 tcos^6 tdt=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 tdt
=
∫ π 2
0
1dt=[x]
π
2
0 =
π
2
Hence I 6 =
5
6
·
3
4
·
1
2
·
π
2
=
15 π
96
or
5 π
32
Similarly,I 8 =
7
8
I 6 =
7
8
·
5 π
32
Thus
∫ π
2
0
sin^2 tcos^6 tdt=I 6 −I 8
=
5 π
32
−
7
8
·
5 π
32
=
1
8
·
5 π
32
=
5 π
256
Problem 15. Use integration by parts to
determine a reduction formula for
∫
(lnx)ndx.
Hence determine
∫
(lnx)^3 dx.
LetIn=
∫
(lnx)ndx.
Using integration by parts, letu=(lnx)n, from which,
du
dx
=n(lnx)n−^1
(
1
x
)
and du=n(lnx)n−^1
(
1
x
)
dx
and let dv=dx, from which,v=
∫
dx=x
Then In=
∫
(lnx)ndx
=(lnx)n(x)−
∫
(x)n(lnx)n−^1
(
1
x
)
dx