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 trape-
zoidal, 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 functionas a power series usingMaclaurin’s
series, and then integrating each algebraic term in turn.
This is demonstrated in thefollowingworked problems.
As a reminder, the general solutionof integrals of the
form
∫
axndx,whereaandnare constants, is given by:
∫
axndx=axn+^1
n+ 1+cProblem 13. Evaluate∫ 0. 4
0. 1 2esinθdθ, correct to
3 significant figures.A power series for esinθis firstly obtained usingMaclau-
rin’s series.
f(θ )=esinθ f( 0 )=esin 0=e^0 = 1f′(θ )=cosθesinθ f′( 0 )=cos 0esin0=( 1 )e^0 = 1f′′(θ )=(cosθ)(cosθesinθ)+(esinθ)(−sinθ),
by the product rule,
=esinθ(cos^2 θ−sinθ);f′′( 0 )=e^0 (cos^20 −sin0)= 1f′′′(θ )=(esinθ)[(2cosθ(−sinθ)−cosθ)]+(cos^2 θ−sinθ)(cosθesinθ)=esinθcosθ[−2sinθ− 1 +cos^2 θ−sinθ]f′′′( 0 )=e^0 cos0[( 0 − 1 + 1 − 0 )]= 0Hence from equation (5):
esinθ=f( 0 )+θf′( 0 )+θ^2
2!f′′( 0 )+θ^3
3!f′′′( 0 )+···= 1 +θ+θ^2
2+ 0Thus∫ 0. 40. 12esinθdθ=∫ 0. 40. 12(
1 +θ+θ^2
2)
dθ=∫ 0. 40. 1( 2 + 2 θ+θ^2 )dθ=[
2 θ+
2 θ^2
2+
θ^3
3] 0. 40. 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∫ 10sinθ
θdθusing
Maclaurin’s series, correct to 3 significant figures.Let f(θ )=sinθ f( 0 )= 0
f′(θ )=cosθ f′( 0 )= 1
f′′(θ )=−sinθ f′′( 0 )= 0
f′′′(θ )=−cosθ f′′′( 0 )=− 1
fiv(θ )=sinθ fiv( 0 )= 0
fv(θ )=cosθ fv( 0 )= 1Hence from equation (5):sinθ=f( 0 )+θf′( 0 )+θ^2
2!f′′( 0 )+θ^3
3!f′′′( 0 )+θ^4
4!fiv( 0 )+θ^5
5!fv( 0 )+···= 0 +θ( 1 )+θ^2
2!( 0 )+θ^3
3!(− 1 )+θ^4
4!( 0 )+θ^5
5!( 1 )+···i.e. sinθ=θ−θ^3
3!+θ^5
5!−···Hence
∫ 10sinθ
θdθ=∫ 10(
θ−θ^3
3!+θ^5
5!−θ^7
7!+···)θdθ=∫ 10(
1 −θ^2
6
+θ^4
120
−θ^6
5040
+···)
dθ