082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

738 Chapter 23 Fourier series
Using Equation (23.4) wehave
a n
= 1 π
∫ 2π
0
t 2 cosntdt
Integrating by parts, wefind
a n
= 1 π
= − 2
nπ
([
t 2sinnt
n
∫ 2π
0
] 2π
0
−
tsinntdt
∫ 2π
= − 2 ([
−t cosnt ] 2π
nπ n
0
= − 2 ( −2πcos2nπ
+
nπ n
0
2t sinnt
n
)
dt
∫ 2π
cosnt
+
0 n
)
[ sinnt
n 2 ] 2π
0
)
dt
= 4 n 2
Hencea 1
= 4,a 2
= 1,a 3
= 4 ,.... Similarly,
9
b n
= 1 π
∫ 2π
0
t 2 sinntdt
= 1 ([
−t 2cosnt
] 2π
+
π n
0 0
= 1 (− 4π2
π n cos2nπ + 2 n
= 1 π
= 1 π
∫ 2π
(− 4π2
n + 2 ([ tsinnt
n n
(− 4π2
n − 2 [
− cosnt
n 2 n
∫ 2π
0
] 2π
0
2t cosnt
n
] 2π
0
)
dt
)
tcosntdt
∫ 2π
))
sinnt
− dt
0 n
)
= − 4π n
Thus b 1
= −4π, b 2
= −2π,.... Finally, the required Fourier series representation is
given by
f(t)= 4π2
3 + (
4cost+cos2t+ 4 9 cos3t+···)
(
−π 4sint+2sin2t+ 4sin3t
3
)
+···