Higher Engineering Mathematics, Sixth Edition

Thet=tanθ 2 substitution 415

42.2 Worked problemson the

t=tan

θ

2

substitution

Problem 1. Determine

∫

dθ

sinθ

Ift=tan
θ
2

then sinθ=

2 t

1 +t^2

anddθ=

2dt

1 +t^2

from

equations (1) and (3).

Thus

∫

dθ

sinθ

=

∫

1

sinθ

dθ

=

∫^1

2 t

1 +t^2

(

2 dt

1 +t^2

)

=

∫

1

t

dt=lnt+c

Hence

∫

dθ

sinθ

=ln

(

tan

θ

2

)

+c

Problem 2. Determine

∫

dx

cosx

If tan

x

2

then cosx=

1 −t^2

1 +t^2

and dx=

2dt

1 +t^2

from

equations (2) and (3).

Thus

∫

dx

cosx

=

∫

1

1 −t^2

1 +t^2

(

2dt

1 +t^2

)

=

∫

2

1 −t^2

dt

2

1 −t^2

may be resolved into partial fractions (see

Chapter 2).

Let

2

1 −t^2

=

2

( 1 −t)( 1 +t)

=

A

( 1 −t)

+

B

( 1 +t)

=

A( 1 +t)+B( 1 −t)

( 1 −t)( 1 +t)

Hence 2 =A( 1 +t)+B( 1 −t)

When t= 1 , 2 = 2 A, from which,A= 1

When t=− 1 , 2 = 2 B, from which,B= 1

Hence

∫

2dt

1 −t^2

=

∫

1

( 1 −t)

+

1

( 1 +t)

dt

=−ln( 1 −t)+ln( 1 +t)+c

=ln

{

( 1 +t)

( 1 −t)

}

+c

Thus

∫

dx

cosx

=ln

⎧

⎪⎨

⎪⎩

1 +tan

x

2

1 −tan

x

2

⎫

⎪⎬

⎪⎭

+c

Note that since tan

π

4

=1, the above result may be

written as:

∫

dx

cosx

=ln

⎧

⎪⎨

⎪⎩

tan

π

4

+tan

x

2

1 −tan

π

4

tan

x

2

⎫

⎪⎬

⎪⎭

+c

=ln

{

tan

(π

4

+

x

2

)}

+c

from compound angles, Chapter 17.

Problem 3. Determine

∫

dx

1 +cosx

If tan

x

2

then cos x=

1 −t^2

1 +t^2

and dx=

2dt

1 +t^2

from

equations (2) and (3).

Thus

∫

dx

1 +cosx

=

∫

1

1 +cosx

dx

=

∫

1

1 +

1 −t^2

1 +t^2

(

2dt

1 +t^2

)

=

∫

1

( 1 +t^2 )+( 1 −t^2 )

1 +t^2

(

2dt

1 +t^2

)

=

∫

dt

Hence

∫

dx

1 +cosx

=t+c=tan

x

2

+c

Problem 4. Determine

∫

dθ

5 +4cosθ