Higher Engineering Mathematics, Sixth Edition

Integration using trigonometric and hyperbolic substitutions 407

∫
2 x− 3
√
(x^2 − 9 )

dx=

∫

2 x

√

(x^2 − 9 )

dx

−

∫

3

√

(x^2 − 9 )

dx

The first integral is determined using the algebraic sub-
stitutionu=(x^2 − 9 ), and the second integral is of the

form

∫

1

√

(x^2 −a^2 )

dx(see Problem 24)

Hence

∫

2 x

√

(x^2 − 9 )

dx−

∫

3

√

(x^2 − 9 )

dx

= 2

√

(x^2 −9)−3cosh−^1

x

3

+c

Problem 26.

∫ √

(x^2 −a^2 )dx.

Letx=acoshθthen

dx

dθ

=asinhθand

dx=asinhθdθ

Hence

∫ √

(x^2 −a^2 )dx

=

∫ √

(a^2 cosh^2 θ−a^2 )(asinhθdθ)

=

∫ √

[a^2 (cosh^2 θ− 1 )](asinhθdθ)

=

∫ √

(a^2 sinh^2 θ)(asinhθdθ)

=a^2

∫

sinh^2 θdθ=a^2

∫(

cosh 2θ− 1

2

)

dθ

since cosh2θ= 1 +2sinh^2 θ

from Table 5.1, page 45,

=

a^2

2

[

sinh2θ

2

−θ

]

+c

=

a^2

2

[sinhθcoshθ−θ]+c,

since sinh2θ=2sinhθcoshθ

Sincex=acoshθthen coshθ=

x

a

and

θ=cosh−^1

x

a

Also, since cosh^2 θ−sinh^2 θ=1, then

sinhθ=

√

(cosh^2 θ− 1 )

=

√[

(x

a

) 2

− 1

]

=

√

(x^2 −a^2 )

a

Hence

∫ √

(x^2 −a^2 )dx

=

a^2

2

[√

(x^2 −a^2 )

a

(x

a

)

−cosh−^1

x

a

]

+c

=

x

2

√

(x^2 −a^2 )−

a^2

2

cosh−^1

x

a

+c

Problem 27. Evaluate

∫ 3

2

√

(x^2 − 4 )dx.

∫ 3

2

√

(x^2 − 4 )dx=

[

x

2

√

(x^2 − 4 )−

4

2

cosh−^1

x

2

] 3

2

from Problem 26, whena=2,

=

(

3

5

√

5 −2cosh−^1

3

2

)

−( 0 −2cosh−^11 )

Since cosh−^1

x

a

=ln

{

x+

√

(x^2 −a^2 )

a

}

then

cosh−^1

3

2

=ln

{

3 +

√

( 32 − 22 )

2

}

=ln2. 6180 = 0. 9624

Similarly, cosh−^11 = 0

Hence

∫ 3

2

√

(x^2 − 4 )dx

=

[

3

2

√

5 − 2 ( 0. 9624 )

]

−[0]

= 1. 429 , correct to 4 significant figures.

Now try the following exercise

Exercise 161 Further problemson

integration using thecoshθsubstitution

1. Find

∫

1

√

(t^2 − 16 )

dt.

[

cosh−^1

x

4

+c

]