1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

8 .2 • TRIGONOMETRIC INTEGRALS 30 5

We solve for cos 28 and sin 28 to obtain the substitutions

1

and sin 28 =

2

i (z^2 - z-^2 ).

Using the identity for cos 28 along with Substitutions (8-4) and (8-5), we rewrite

the integral as

J

! (z2 + z -2) dz= J i(z4+1) dz= J f (z)dz,

5-4 (z+z-^1 ) iz 2z2(z- 2) (2z- 1)

ct (O)^2 ct (OJ ct (OJ

where f (z) = 2 z ~~:~~•-!). The singularities of J lying inside Ci(O) are poles

located at the points 0 and ~-We use Theorem 8.2 to get the residues:

R [f OJ 1. d^2 J ( ) 1. d. ( z

4

+ 1)

es ' = z-o im d-z z z = z-o lill -d z i2 (2 z (^2) - (^5) z + 2)

. .4z^3 (2z^2 - 5z+2) - (4z- 5) (z^4 +1) 5i

= hmi = -

.-o 2 (2z2 - 5z + 2)^2 8

and

[

l]. ( 1). i(z

4

+1) 17i

Res f, -

2

= hm z - -

2

f (z) = h m

4 2

( 2 ) = --

2

·

z-~ z-i Z z- 4

Therefore, we conclude that

[

2

" cos28d8 = 2 7ri (5i _ 17i) = ~-

}0 5-4cos8 8 24 6

-------~EXERCISES FOR SECTION 8.2

Use residues to find

1 r2" 1 dfJ

· Jo 3 cos 8 + 5 ·

2. r2" 1 dfJ.
   Jo 4sin8+ 5

3 Ii2" 1 dfJ

-^0 15sin^2 8 + 1 ·

r2" 1

4. Jo 5cos2B+4d8.