Higher Engineering Mathematics

406 INTEGRAL CALCULUS

which provides as alternative solution to

∫

1

√

(x^2 −a^2 )

dx

Problem 25. Determine

∫

2 x− 3

√

(x^2 −9)

dx.

∫

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

substitutionu=(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)−3 cosh−^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 cosh 2θ= 1 +2 sinh^2 θ

from Table 5.1, page 45,

=

a^2

2

[

sinh 2θ

2

−θ

]

+c

=

a^2

2

[ sinhθcoshθ−θ]+c,

since sinh 2θ=2 sinhθcoshθ

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 −2 cosh−^1

3

2

)

−(0−2 cosh−^1 1)

Since cosh−^1

x

a

=ln

{

x+

√

(x^2 −a^2 )

a

}

then

cosh−^1

3

2

=ln

{

3 +

√

(3^2 − 22 )

2

}

=ln 2. 6180 = 0. 9624

Similarly, cosh−^11 = 0