Higher Engineering Mathematics, Sixth Edition

416 Higher Engineering Mathematics

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 +4cosθ

=

∫

(

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,

from 12 of Table 40.1, page 399. Hence

∫

dθ

5 +4cosθ

=

2

3

tan−^1

(

1

3

tan

θ

2

)

+c

Now try the following exercise

Exercise 165 Further problems on the

t=tan

θ

2

substitution

Integratethefollowingwithrespect to thevariable:

1.

∫

dθ

1 +sinθ

⎡

⎢

⎣

− 2

1 +tan

θ

2

+c

⎤

⎥

⎦

2.

∫

dx

1 −cosx+sinx

⎡

⎢

⎣ln

⎧

⎪⎨

⎪⎩

tan

x

2

1 +tan

x

2

⎫

⎪⎬

⎪⎭

+c

⎤

⎥

⎦

3.

∫

dα

3 +2cosα

[

2

√

5

tan−^1

(

1

√

5

tan

α

2

)

+c

]

4.

∫

dx

3sinx−4cosx

⎡

⎢

⎣

1

5

ln

⎧

⎪⎨

⎪⎩

2tan

x

2

− 1

tan

x

2

+ 2

⎫

⎪⎬

⎪⎭

+c

⎤

⎥

⎦

42.3 Further worked problems on the

t=tan

θ

2

substitution

Problem 5. Determine

∫

dx

sinx+cosx

If tan

x

2

then sinx=

2 t

1 +t^2

,cosx=

1 −t^2

1 +t^2

and

dx=

2dt

1 +t^2

from equations (1), (2) and (3).

Thus

∫

dx

sinx+cosx

=

∫

2dt

1 +t^2

(

2 t

1 +t^2

)

+

(

1 −t^2

1 +t^2

)

=

∫

2dt

1 +t^2

2 t+ 1 −t^2

1 +t^2

=

∫

2dt

1 + 2 t−t^2

=

∫

−2dt

t^2 − 2 t− 1

=

∫

−2dt

(t− 1 )^2 − 2

=

∫

2dt

(

√

2 )^2 −(t− 1 )^2

= 2

[

1

2

√

2

ln

{√

2 +(t− 1 )

√

2 −(t− 1 )

}]

+c

(see Problem 11, Chapter 41, page 413),

i.e.

∫

dx

sinx+cosx

=

1

√

2

ln

⎧

⎪⎨

⎪⎩

√

2 − 1 +tan

x

2

√

2 + 1 −tan

x

2

⎫

⎪⎬

⎪⎭

+c

Problem 6.∫ Determine

dx

7 −3sinx+6cosx

From equations (1) and (3),

∫

dx

7 −3sinx+6cosx

=

∫ 2dt

1 +t^2

7 − 3

(

2 t

1 +t^2

)

+ 6

(

1 −t^2

1 +t^2

)