Maclaurin’ s series 73
8.5 Numerical integration using
Maclaurin’s series
The value of many integrals cannot be determined using
the various analytical methods. In Chapter 45, the trapezoidal, mid-ordinate and Simpson’s rules are used to
numerically evaluate such integrals. Another method of
finding the approximate value of a definite integral is to
express the function as a power series using Maclaurin’s
series, and then integrating each algebraic term in turn.
This is demonstrated in the following worked problems.
As a reminder, the general solution of integrals of the
form
ax n dx, where a and n are constants, is given by:
ax
n dx =
ax n+1
n + 1
+ c
Problem 13. Evaluate
0.4
0.1 2 e sin θ dθ, correct to
3 significant figures.
A power series for e sin θ is firstly obtained using Maclaurin’s series.
f (θ) = e
sin θ
f (0) = e
sin 0
= e
0
= 1
f
(θ) = cos θ e
sin θ
f
(0) = cos 0 e
sin 0
= (1)e
0
= 1
f
(θ) = (cos θ)(cos θ e
sin θ
) + (e
sin θ
)(−sin θ),
by the product rule,
= e
sin θ
(cos
2
θ − sin θ);
f
(0) = e
0
(cos
2 0 − sin 0) = 1
f
(θ) = (e
sin θ
)[(2 cos θ(−sin θ) − cos θ)]
+ (cos
2
θ − sin θ)(cos θ e
sin θ
)
= e
sin θ cos θ[−2 sinθ − 1 + cos
2
θ − sin θ]
f
(0) = e
0 cos 0[(0 − 1 + 1 − 0)] = 0
Hence from equation (5):
e
sin θ
= f (0) + θ f
(0) +
θ 2
2!
f
(0) +
θ 3
3!
f
(0) + · · ·
= 1 + θ +
θ 2
2
+ 0
Thus
0.4
0.1
2 e
sin θ dθ =
0.4
0.1
2
1 + θ +
θ
2
2
dθ
=
0.4
0.1
(2 + 2θ + θ
2
)dθ
=
2θ +
2θ
2
2
+
θ
3
3
0.4
0.1
=
0.8 + (0.4)
2
+
(0.4) 3
3
−
0.2 + (0.1)
2
+
(0.1) 3
3
= 0.98133 − 0.21033
= 0.771, correct to 3 significant figures.
Problem 14. Evaluate
1
0
sin θ
θ
dθ using
Maclaurin’s series, correct to 3 significant figures.
Let f (θ) = sin θ
f (0) = 0
f (θ) = cos θ
f (0) = 1
f (θ) = −sin θ f (0) = 0
f (θ) = −cos θ f (0) = −1
f iv (θ) = sin θ
f iv (0) = 0
f v (θ) = cos θ
f v (0) = 1
Hence from equation (5):
sin θ = f (0) + θ f (0) +
θ 2
2!
f (0) +
θ 3
3!
f (0)
+
θ 4
4!
f iv (0) +
θ 5
5!
f v (0) + · · ·
= 0 + θ(1) +
θ 2
2!
(0) +
θ 3
3!
(−1)
+
θ 4
4!
(0) +
θ 5
5!
(1) + · · ·
i.e. sin θ = θ −
θ 3
3!
+
θ 5
5!
− · · ·
Hence
1
0
sin θ
θ
dθ
=
1
0
θ −
θ 3
3!
+
θ 5
5!
−
θ 7
7!
+ · · ·
θ
dθ
=
1
0
1 −
θ 2
6
+
θ 4
120
−
θ 6
5040
+ · · ·
dθ
8.5 Numerical integration using
Maclaurin’s series
The value of many integrals cannot be determined using
the various analytical methods. In Chapter 45, the trapezoidal, mid-ordinate and Simpson’s rules are used to
numerically evaluate such integrals. Another method of
finding the approximate value of a definite integral is to
express the function as a power series using Maclaurin’s
series, and then integrating each algebraic term in turn.
This is demonstrated in the following worked problems.
As a reminder, the general solution of integrals of the
form
ax n dx, where a and n are constants, is given by:
ax
n dx =
ax n+1
n + 1
+ c
Problem 13. Evaluate
0.4
0.1 2 e sin θ dθ, correct to
3 significant figures.
A power series for e sin θ is firstly obtained using Maclaurin’s series.
f (θ) = e
sin θ
f (0) = e
sin 0
= e
0
= 1
f
(θ) = cos θ e
sin θ
f
(0) = cos 0 e
sin 0
= (1)e
0
= 1
f
(θ) = (cos θ)(cos θ e
sin θ
) + (e
sin θ
)(−sin θ),
by the product rule,
= e
sin θ
(cos
2
θ − sin θ);
f
(0) = e
0
(cos
2 0 − sin 0) = 1
f
(θ) = (e
sin θ
)[(2 cos θ(−sin θ) − cos θ)]
+ (cos
2
θ − sin θ)(cos θ e
sin θ
)
= e
sin θ cos θ[−2 sinθ − 1 + cos
2
θ − sin θ]
f
(0) = e
0 cos 0[(0 − 1 + 1 − 0)] = 0
Hence from equation (5):
e
sin θ
= f (0) + θ f
(0) +
θ 2
2!
f
(0) +
θ 3
3!
f
(0) + · · ·
= 1 + θ +
θ 2
2
+ 0
Thus
0.4
0.1
2 e
sin θ dθ =
0.4
0.1
2
1 + θ +
θ
2
2
dθ
=
0.4
0.1
(2 + 2θ + θ
2
)dθ
=
2θ +
2θ
2
2
+
θ
3
3
0.4
0.1
=
0.8 + (0.4)
2
+
(0.4) 3
3
−
0.2 + (0.1)
2
+
(0.1) 3
3
= 0.98133 − 0.21033
= 0.771, correct to 3 significant figures.
Problem 14. Evaluate
1
0
sin θ
θ
dθ using
Maclaurin’s series, correct to 3 significant figures.
Let f (θ) = sin θ
f (0) = 0
f (θ) = cos θ
f (0) = 1
f (θ) = −sin θ f (0) = 0
f (θ) = −cos θ f (0) = −1
f iv (θ) = sin θ
f iv (0) = 0
f v (θ) = cos θ
f v (0) = 1
Hence from equation (5):
sin θ = f (0) + θ f (0) +
θ 2
2!
f (0) +
θ 3
3!
f (0)
+
θ 4
4!
f iv (0) +
θ 5
5!
f v (0) + · · ·
= 0 + θ(1) +
θ 2
2!
(0) +
θ 3
3!
(−1)
+
θ 4
4!
(0) +
θ 5
5!
(1) + · · ·
i.e. sin θ = θ −
θ 3
3!
+
θ 5
5!
− · · ·
Hence
1
0
sin θ
θ
dθ
=
1
0
θ −
θ 3
3!
+
θ 5
5!
−
θ 7
7!
+ · · ·
θ
dθ
=
1
0
1 −
θ 2
6
+
θ 4
120
−
θ 6
5040
+ · · ·
dθ