z=cos 9 +isin9
-4-<------>--<~ 9 -
0 2x(a) The interval [0, 2x] of
integration for F(oos 9, sin (/).8.2 • TRIGONOMETRIC INTEGRALS 303y(b) The contour C of
integration for f(z).
Figure 8.2 The change of variables from a definite integral on [O, 2 ... J to a contour
integral around C.
singular points satisfy the relation z^2 = -lO±~ = -^53 ±^4. Hence the only
singularities that lie inside the unit circle are simple poles corresponding to the
solutions of z^2 =-!, which are the two points z 1 = 7:i and z2 = -?:i· We use
Theorem 8.2 and L'Hopital's rule to get the residues at zk, fork= 1, 2:
. - i4z (z - Zk)
Res [f, Zk] = hm
3 4 0~
z - z • z + 1 z• + 31
=. un ~~~~~ -i4 (Zz - z1;)z-z• 12z3 + 20z
- i4Zk
= 1 2z~ + 20zk
-i
= 3zz + s·As zk = ~ a.nd zi = -! , the residues are given by Res[/ , zk] = 3 (- h+s = -~.
We now use Equation (8-6) to compute the value of the integral:
[2
" dO (-i -i)
}
0
1+3cos 20 = Zni 4 + 4 = ....- EXAMPLE 8.11 Evaluate Jt t+a~oo;> t dt by using a computer algebra sys-
tem.
Solution We can obtain the antiderivative of 1+ 3 !.,. 2 t by using software such
as Mathematica or MAPLE · It is J 1+ 3cos2^1 t dt = -Arctan( 2 2 cott) = g (t) · Since
cotO and cot2n are not defined, the computations for both g (O) and g (Zn)
are indeterminate. The graph s = g (t) shown in Figure 8.3 reveals another