Higher Engineering Mathematics

414 INTEGRAL CALCULUS

=

∫^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 3).

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 18.

Problem 3. Determine:

∫

dx

1 +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

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 +4 cosθ

If t=tan

θ

2

then cosθ=

1 −t^2

1 +t^2

and dx=

2dt

1 +t^2

from equations (2) and (3).

Thus

∫

dθ

5 +4 cosθ

=

∫

(

2dt

1 +t^2

)

5 + 4

(

1 −t^2

1 +t^2

)

=

∫

(

2dt

1 +t^2

)

5(1+t^2 )+4(1−t^2 )

(1+t^2 )

= 2

∫

dt

t^2 + 9

= 2

∫

dt

t^2 + 32

= 2

(

1

3

tan−^1

t

3

)

+c,