Page 61 - Calculus with Complex Numbers

M aking the substitution z = eit we obtain

z,c 1 dz 1 1
sin lt cos 3/ dt = .:2 - . / +
0 y
- k- / ? z c p
1 j Jc (g .. .y .1 . 1 j
'

/

'

j
'
1 j ( c - 1 .y sl .. sl j ty .
= - g / 4 z
since the integrand is a Laurent expansion with no term in 1/z.
Hence we have
2x
sin lt cos 3/ dt = 0.
0

Exam ples

Hvaluate the following integrals.

(2n./.$,,4)

dt
3 + 2 sin t

2n. sin 5t
dt
0 Sin t
2n.
cos6 t dt
0