Reduction formulae 429
Problem 6. Use a reduction formula to determine
x 3 sin x dx.
Using equation (3),
x
3 sin x dx = I 3
= −x 3 cos x + 3x 2 sin x − 3(2)I 1
and
I 1 = −x 1 cos x + 1x 0 sin x
= −x cos x + sin x
Hence
x
3 sin x dx = −x 3 cos x + 3x 2 sin x
− 6[−x cos x + sin x]
= −x 3 cos x + 3x 2 sin x
+ 6x cos x − 6 sin x + c
Problem 7. Evaluate
π
2
0
3θ
4 sin θ dθ, correct to 2
decimal places.
From equation (3),
I n = [−x
n cos x + nx
n−1
(sin x)]
π
2
0 − n(n − 1)I n−2
=
−
π
2
n
cos
π
2
+ n
π
2
n−1
sin
π
2
− (0)
− n(n − 1)I n−2
= n
π
2
n−1 − n(n − 1)I n−2
Hence
π
2
0
3θ
4 sin θ dθ = 3
π
2
0
θ
4 sin θ dθ
= 3I 4
= 3
4
π
2
3 − 4(3)I 2
I 2 = 2
π
2
1 − 2(1)I 0 and
I 0 =
π
2
0
θ
0 sin θ dθ = [−cos x]
π
2
0
= [−0 − (−1)] = 1
Hence
3
π
2
0
θ
4 sin θ dθ
= 3I 4
= 3
4
π
2
3 − 4(3)
2
π
2
1 − 2(1)I 0
= 3
4
π
2
3 − 4(3)
2
π
2
1 − 2(1)(1)
= 3
4
π
2
3 − 24
π
2
1 + 24
= 3(15.503 − 37.699 + 24)
= 3(1.8039) = 5.41
Now try the following exercise
Exercise 170 Further problems on
reduction formulae for integrals of the form
x n cos x dx and
x n sin x dx
1. Use a reduction formula to determine
x 5 cos x dx.
⎡
⎢
⎣
x 5 sin x + 5x 4 cos x − 20x 3 sin x
− 60x 2 cos x + 120x sin x
+ 120 cos x + c
⎤
⎥
⎦
2. Evaluate
π
0 x 5 cos x dx, correct to 2 decimal
places.
[−134.87]
3. Use a reduction formula to determine
x 5 sin x dx.
⎡
⎢
⎣
−x 5 cos x + 5x 4 sin x + 20x 3 cos x
− 60x 2 sin x − 120x cos x
+ 120 sin x + c
⎤
⎥
⎦
4. Evaluate
π
0 x 5 sin x dx, correct to 2 decimal
places.
[62.89]
44.4 Using reduction formulae for
integrals of the form
sin n x dx
and
cos
n x dx
(a)
sin n x dx
Let I n =
sin
n x dx ≡
sin
n−1 x sin x dx from laws of
indices.
Using integration by parts, let u = sin n−1 x, from which,