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

Solution (a) We use the trigonometric identity
(b)
sin 2 nt = 1 −cos2nt
2
toget
∫ π
sin 2 (nt)dt =
−π
∫ π
−π
1 −cos2nt
2
dt =
using the orthogonal properties of cos(nt).
∫ π
−π
cos 2 (nt)dt =
∫ π
−π
1 −sin 2 (nt)dt
=2π−π=π
∫ π
−π
16.3 Improper integrals 483
1
2 dt=π
It is a simple extension to show that integrating sin 2 (nt) or cos 2 (nt) over any interval
of length 2π yields the same result, namely π. It is also possible to extend the result of
Example 16.4 toshow
∫ T/2
−T/2
sin 2 ( 2nπt
T
)
dt =
Finally, integrating sin 2 ( 2nπt
T
the same result,thatis T 2 .
∫ T/2
−T/2
)
( ) 2nπt
cos 2 dt = T T 2
)
and cos 2 ( 2nπt
T
n∈Z
n≠0
over any interval of lengthT gives
EXERCISES16.2
1 Show f (x) =x 2 andg(x) = 1 −xareorthogonal
[ ]
across
0, 4 3
.
2 Show f(x) = 1 x andg(x) =x2 are orthogonal over
[−k,k].
3 (a)Showf(t)=1−tandg(t)=1+tare
orthogonalover [0, √ 3].
(b)Findanotherinterval over which f (t)andg(t)are
orthogonal.
4 Show f (t) = e t andg(t) = 1 −e −2t are orthogonal
across [−1,1].
5 Show f(x) = √ xandg(x)=1− √ x are orthogonal
[ ]
on 0, 16 .
9
Solutions
3 (b) [− √ 3,0]
16.3 IMPROPERINTEGRALS
There aretwo caseswhen evaluation ofanintegral needs specialcare:
(1) one,orboth,ofthe limits ofanintegral areinfinite;
(2) the integrand becomesinfiniteatone,ormore,points ofthe interval ofintegration.