Reduction Formulae
Eu Wen
Further Pure Mathematics 2020
Here I’ve written solutions to a few questions on Reduction Formulae. To
master this aspect of integration one needs to have sufficient confidence in Al-
gebra and Calculus. Otherwise, one will be daunted by the complexities of the
solutions.
1
Question :
Let Z 1
In = (1 − x)n sinxdx
0
for n ≥ 0. Show that In+2 = 1 − (n + 1)(n + 2)In . Hence find the value of I6 ,
correct to 4 decimal places.
Solution :
By parts on Z 1
(1 − x)n sinxdx :
0
du
u = sinx, = cosx
dx
dv n −1
= (1 − x) , v = (1 − x)n+1
dx n+1
1 Z 1
−1 n+1 1
sinx(1 − x) + (1 − x)n+1 cosxdx
n+1 0 n+1 0
Z 1
1
In = (1 − x)n+1 cosxdx
n+1 0
By parts on Z 1
(1 − x)n+1 cosxdx :
0
1
, du
= −sinx
u = cosx,
dx
dv −1
= (1 − x)n+1 , v = (1 − x)n+2
dx n+2
1 Z 1
1 −1 1 1
cosx(1 − x)n+2 − sinx(1 − x)n+2 dx
n+1 n+2 0 n + 1 0 n + 2
1 1 1 1
In = − In+2
n+1 n+2 n+1 n+2
As required:
In+2 = 1 − (n + 1)(n + 2)In
I6 = 1 − 30I4
I4 = 1 − 12I2
I2 = 1 − 2I0
Z 1
I0 = sinxdx = 0.4596977
0
I2 = 0.0806046, I4 = 0.032745
I6 = 0.0177
2
Question :
Let Z 1
1
In = dx.
0 (1 + x4 )n
By considering
d x
,
dx (1 + x4 )n
show that 4nIn+1 = 21n + (4n − 1)In . Given that I1 = 0.86697, correct to 5
decimal places, find I3 .
Solution :
(1 + x4 )n − x[n(1 + x4 )n−1 (4x3 )]
d x
=
dx (1 + x4 )n (1 + x4 )2n
2
Eu Wen
Further Pure Mathematics 2020
Here I’ve written solutions to a few questions on Reduction Formulae. To
master this aspect of integration one needs to have sufficient confidence in Al-
gebra and Calculus. Otherwise, one will be daunted by the complexities of the
solutions.
1
Question :
Let Z 1
In = (1 − x)n sinxdx
0
for n ≥ 0. Show that In+2 = 1 − (n + 1)(n + 2)In . Hence find the value of I6 ,
correct to 4 decimal places.
Solution :
By parts on Z 1
(1 − x)n sinxdx :
0
du
u = sinx, = cosx
dx
dv n −1
= (1 − x) , v = (1 − x)n+1
dx n+1
1 Z 1
−1 n+1 1
sinx(1 − x) + (1 − x)n+1 cosxdx
n+1 0 n+1 0
Z 1
1
In = (1 − x)n+1 cosxdx
n+1 0
By parts on Z 1
(1 − x)n+1 cosxdx :
0
1
, du
= −sinx
u = cosx,
dx
dv −1
= (1 − x)n+1 , v = (1 − x)n+2
dx n+2
1 Z 1
1 −1 1 1
cosx(1 − x)n+2 − sinx(1 − x)n+2 dx
n+1 n+2 0 n + 1 0 n + 2
1 1 1 1
In = − In+2
n+1 n+2 n+1 n+2
As required:
In+2 = 1 − (n + 1)(n + 2)In
I6 = 1 − 30I4
I4 = 1 − 12I2
I2 = 1 − 2I0
Z 1
I0 = sinxdx = 0.4596977
0
I2 = 0.0806046, I4 = 0.032745
I6 = 0.0177
2
Question :
Let Z 1
1
In = dx.
0 (1 + x4 )n
By considering
d x
,
dx (1 + x4 )n
show that 4nIn+1 = 21n + (4n − 1)In . Given that I1 = 0.86697, correct to 5
decimal places, find I3 .
Solution :
(1 + x4 )n − x[n(1 + x4 )n−1 (4x3 )]
d x
=
dx (1 + x4 )n (1 + x4 )2n
2
