Calculus: Analytic Geometry and Calculus, with Vectors

8.5 Integration by substituting z = tan 0/2 479

where Pl and Ql are polynomials in z. Several examples appear in the
problems at the end of this section.
Except in special cases, evaluating integrals of the form

I aozm + aizm-i +... + am_lz + am

dx

boz" + biz's-1 + ... + bn

is very tedious business. When n < 2, and in a few special additional

cases, answers can befound in tables of integrals. Some information

about the matter will appear in Section 9.4 of this book.

Problems 8.59

1 By use of the substitution z = tan 2, show that

(a) J sin B

dB = fl dz=log N + c =log I tan

01

+ c

(b) r 1

=

r 2 r1-z+1+z

J cos 0 de l^1 z2

A = 1 (1 - z) (1 + z)A

= J ( 1 + l

1 z) dz = log

1+tan2

+c

1 -tang

1 2

(c),J a+b sinBdeJ az2+2bz+adz

(a) fa+bsin8+ccosOd9f (a-c)z2+2bz+(a+c)dx

sing+cosB z(1 -z2)(1+2z-z2)dz

(e) ) tanB + cot BdB = 4 f (1 + z2)4

2z 2

(f) 1 -- z2+c=sin0 +c=

f cos0dO=2 f (1+z2)2dz

(g) logi +Z2 + c = log sec B + c =ftan 0 d9 = 4f 1 zz4A

(h) log a+ 1 +zz2) + c = log (a + b sin B) + c

b cos 6

J a+bsin0

• 2b 1 - z2 dz
  f(az2 + 2bz + a) (1 + z2)
  sin 6 4z
  (2) f 1 + sin g d9 =

f + z2)(1 + z)2A

(;) f cos 0 f 1 - z2

1+cosBde - 1+z2dz

1+z2z2dz=2tan' tanL -tang+c

f

2