Higher Engineering Mathematics, Sixth Edition

Thet=tanθ 2 substitution 417

=

∫

2dt

1 +t^2

7 ( 1 +t^2 )− 3 ( 2 t)+ 6 ( 1 −t^2 )

1 +t^2

=

∫

2dt

7 + 7 t^2 − 6 t+ 6 − 6 t^2

=

∫

2dt

t^2 − 6 t+ 13

=

∫

2dt

(t− 3 )^2 + 22

= 2

[

1

2

tan−^1

(

t− 3

2

)]

+c

from 12, Table 40.1, page 399. Hence

∫

dx

7 −3sinx+6cosx

=tan−^1

⎛

⎜

⎝

tan

x

2

− 3

2

⎞

⎟

⎠+c

Problem 7. Determine

∫

dθ

4cosθ+3sinθ

From equations (1) to (3),

∫

dθ

4cosθ+3sinθ

=

∫ 2dt

1 +t^2

4

(

1 −t^2

1 +t^2

)

+ 3

(

2 t

1 +t^2

)

=

∫

2dt

4 − 4 t^2 + 6 t

=

∫

dt

2 + 3 t− 2 t^2

=−

1

2

∫

dt

t^2 −

3

2

t− 1

=−

1

2

∫

dt

(

t−

3

4

) 2

−

25

16

=

1

2

∫

dt

(

5

4

) 2

−

(

t−

3

4

) 2

=

1

2

⎡

⎢

⎢

⎣

1

2

(

5

4

)ln

⎧

⎪⎪

⎨

⎪⎪

⎩

5

4

+

(

t−

3

4

)

5

4

−

(

t−

3

4

)

⎫

⎪⎪

⎬

⎪⎪

⎭

⎤

⎥

⎥

⎦+c

from Problem 11, Chapter 41, page 413

=

1

5

ln

⎧

⎪⎨

⎪⎩

1

2

+t

2 −t

⎫

⎪⎬

⎪⎭

+c

Hence

∫

dθ

4cosθ+3sinθ

=

1

5

ln

⎧

⎪⎨

⎪⎩

1

2

+tan

θ

2

2 −tan

θ

2

⎫

⎪⎬

⎪⎭

+c

or

1

5

ln

⎧

⎪⎨

⎪⎩

1 +2tan

θ

2

4 −2tan

θ

2

⎫

⎪⎬

⎪⎭

+c

Now try the following exercise

Exercise 166 Further problems on the

t=tan

θ

2

substitution

In Problems 1 to 4, integrate with respect to the

variable.

1.

∫

dθ

5 +4sinθ

⎡

⎢

⎣

2

3

tan−^1

⎛

⎜

⎝

5tan

θ

2

+ 4

3

⎞

⎟

⎠+c

⎤

⎥

⎦

2.

∫

dx

1 +2sinx

⎡

⎢

⎣

1

√

3

ln

⎧

⎪⎨

⎪⎩

tan

x

2

+ 2 −

√

3

tan

x

2

+ 2 +

√

3

⎫

⎪⎬

⎪⎭

+c

⎤

⎥

⎦

3.

∫

dp

3 −4sinp+2cosp

⎡

⎢

⎣

1

√

11

ln

⎧

⎪⎨

⎪⎩

tan

p

2

− 4 −

√

11

tan

p

2

− 4 +

√

11

⎫

⎪⎬

⎪⎭

+c

⎤

⎥

⎦