432 Higher Engineering Mathematics
Hence
cos
4 x dx
=
1
4
cos
3 x sin x +
3
4
1
2
cos x sin x +
1
2
x
=
1
4
cos
3 x sin x +
3
8
cos x sin x +
3
8
x + c
Problem 12. Determine a reduction formula
for
π
2
0
cos
n x dx and hence evaluate
π
2
0
cos
5 x dx
From equation (5),
cos
n x dx =
1
n
cos
n−1 x sin x +
n −1
n
I n−2
and hence
π
2
0
cos
n x dx =
1
n
cos n−1 x sin x
π
2
0
+
n −1
n
I n−2
= [0 −0] +
n − 1
n
I n−2
i.e.
π
2
0
cos
n x dx = I n =
n − 1
n
I n−2
(6)
(Note that this is the same reduction formula as for
π
2
0
sin
n x dx (in Problem 10) and the result is usually
known as Wallis’s formula).
Thus, from equation (6),
π
2
0
cos
5 x dx =
4
5
I 3 ,
I 3 =
2
3
I 1
and
I 1 =
π
2
0
cos
1 x dx
= [sin x]
π
2
0 = (1 − 0) = 1
Hence
π
2
0
cos
5 x dx =
4
5
I 3 =
4
5
2
3
I 1
=
4
5
2
3
(1)
=
8
15
Now try the following exercise
Exercise 171 Further problems on
reduction formulae for integrals of the form
sin n x dx and
cos n x dx
1. Use a reduction formula to determine
sin 7 x dx.
⎡
⎢
⎣
−
1
7
sin
6 x cos x −
6
35
sin
4 x cos x
−
8
35
sin 2 x cos x −
16
35
cos x + c
⎤
⎥
⎦
2. Evaluate
π
0 3 sin 3 x dx using a reduction
formula.
[4]
3. Evaluate
π
2
0
sin
5 x dx using a reduction
formula.
8
15
4. Determine, using a reduction formula,
cos
6 x dx.
⎡
⎢
⎣
1
6
cos 5 x sin x +
5
24
cos 3 x sin x
+
5
16
cos x sin x +
5
16
x + c
⎤
⎥
⎦
5. Evaluate
π
2
0
cos
7 x dx.
16
35
44.5 Further reduction formulae
The following worked problems demonstrate further
examples where integrals can be determined using
reduction formulae.
Problem 13. Determine a reduction formula for
tan n x dx and hence find
tan 7 x dx.
Let I n =
tan
n x dx ≡
tan
n−2 x tan
2 x dx
by the laws of indices
=
tan
n−2 x(sec
2 x − 1) dx
since 1 + tan 2 x = sec 2 x
=
tan
n−2 x sec
2 x dx −
tan
n−2 x dx