Higher Engineering Mathematics

H

Integral calculus

42

The t=tan

θ

2

substitution

42.1 Introduction

Integrals of the form

∫

1

acosθ+bsinθ+c

dθ,

wherea,bandcare constants, may be determined

by using the substitutiont=tan

θ

2

. The reason is

explained below.
If angleAin the right-angled triangle ABC shown

in Fig. 42.1 is made equal to

θ

2

then, since tangent=

opposite

adjacent

,ifBC=tandAB=1, then tan

θ

2

=t.

By Pythagoras’ theorem,AC=

√

1 +t^2

C

A B

t

2

1

1 t^2

θ

Figure 42.1

Therefore sin

θ

2

=

t

√

1 +t^2

and cos

θ

2

=

1
√
1 +t^2
Since sin 2x=2 sinxcosx(from double angle for-
mulae, Chapter 18), then

sinθ=2 sin

θ

2

cos

θ

2

= 2

(

t

√

1 +t^2

)(

t

√

1 +t^2

)

i.e. sinθ=

2 t

( 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. cosθ=

1 −t^2

1 +t^2

(2)

Also, sincet=tan

θ

2

,

dt

dθ

=

1

2

sec^2

θ

2

=

1

2

(

1 +tan^2

θ

2

)

from trigonomet-

ric identities,

i.e.

dt

dθ

=

1

2

(1+t^2 )

from which, dθ=

2dt

1 +t^2

(3)

Equations (1), (2) and (3) are used to determine

integrals of the form

∫

1

acosθ+bsinθ+c

dθ

wherea,borcmay be zero.

42.2 Worked problems on thet=tan

θ

2

substitution

Problem 1. Determine:

∫

dθ

sinθ

If t=tan

θ

2

then sinθ=

2 t

1 +t^2

and dθ=

2dt

1 +t^2

from equations (1) and (3).

Thus

∫

dθ

sinθ

=

∫

1

sinθ

dθ