Chapter 42
The t = tan
θ
2 substitution
42.1 Introduction
Integrals of the form
1
a cos θ + b sin θ + c
dθ, where
a, b and c are constants, may be determined by using the
substitution t = tan
θ
2
. The reason is explained below.
If angle A in the right-angled triangle ABC shown in
Fig. 42.1 is made equal to
θ
2
then, since tangent =
opposite
adjacent
, if BC = t and AB = 1, then tan
θ
2
= t .
By Pythagoras’ theorem, AC =
√
1 +t 2
␪
2
A
1
B
Ί1 1 t 2
C
t
Figure 42.1
Therefore sin
θ
2
=
t
√
1 +t 2
and cos
θ
2
=
1
√
1 +t 2
Since
sin 2x = 2 sin x cos x (from double angle formulae,
Chapter 17), then
sin θ = 2 sin
θ
2
cos
θ
2
= 2
t
√
1 + t 2
t
√
1 + t 2
i.e.
sin θ =
2t
(1 + t 2 )
(1)
Since cos 2x = cos
2 θ
2
− sin
2 θ
2
=
1
√
1 + t 2
2
−
t
√
1 + t 2
2
i.e.
cos θ =
1 −t 2
1 +t 2
(2)
Also, since t = tan
θ
2
,
dt
dθ
=
1
2
sec 2 θ
2
=
1
2
1 + tan 2 θ
2
from trigonometric
identities,
i.e.
dt
dθ
=
1
2
(1 + t
2
)
from which,
dθ =
2 dt
1 +t 2
(3)
Equations (1), (2) and (3) are used to determine
integrals of the form
1
a cos θ + b sin θ + c
dθ where
a, b or c may be zero.