Higher Engineering Mathematics, Sixth Edition

Integration using trigonometric and hyperbolic substitutions 405

Hence

∫

1

√

(x^2 +a^2 )

dx

=

∫

1

√

(a^2 sinh^2 θ+a^2 )

(acoshθdθ)

=

∫

acoshθdθ

√

(a^2 cosh^2 θ)

,

since cosh^2 θ−sinh^2 θ= 1

=

∫

acoshθ

acoshθ

dθ=

∫

dθ=θ+c

=sinh−^1

x

a

+c,sincex=asinhθ

Itisshownonpage339that

sinh−^1

x

a

=ln

{

x+

√

(x^2 +a^2 )

a

}

,

which provides an alternative solution to

∫

1

√

(x^2 +a^2 )

dx

Problem 21. Evaluate

∫ 2

0

1

√

(x^2 + 4 )

dx, correct

to 4 decimal places.

∫ 2

0

1

√

(x^2 + 4 )

dx=

[

sinh−^1

x

2

] 2

0

or

[

ln

{

x+

√

(x^2 + 4 )

2

}] 2

0

from Problem 20, wherea= 2

Using the logarithmic form,
∫ 2

0

1

√

(x^2 + 4 )

dx

=

[

ln

(

2 +

√

8

2

)

−ln

(

0 +

√

4

2

)]

=ln2. 4142 −ln1= 0. 8814 ,

correct to 4 decimal places.

Problem 22. Evaluate

∫ 2

1

2

x^2

√

( 1 +x^2 )

dx,

correct to 3 significant figures.

Since the integral contains a term of the form√

(a^2 +x^2 ),thenletx=sinhθ, from which

dx

dθ

=coshθand dx=coshθdθ

Hence

∫

2

x^2

√

( 1 +x^2 )

dx

=

∫

2 (coshθdθ)

sinh^2 θ

√

( 1 +sinh^2 θ)

= 2

∫

coshθdθ

sinh^2 θcoshθ

,

since cosh^2 θ−sinh^2 θ= 1

= 2

∫

dθ

sinh^2 θ

= 2

∫

cosech^2 θdθ

=−2cothθ+c

cothθ=

coshθ

sinhθ

=

√

( 1 +sinh^2 θ)

sinhθ

=

√

( 1 +x^2 )

x

Hence

∫ 2

1

2

x^2

√

1 +x^2 )

dx

=−[2cothθ]^21 =− 2

[√

( 1 +x^2 )

x

] 2

1

=− 2

[√

5

2

−

√

2

1

]

= 0. 592 ,

correct to 3 significant figures

Problem 23. Find

∫√

(x^2 +a^2 )dx.

Letx=asinhθthen

dx

dθ

=acoshθand

dx=acoshθdθ

Hence

∫√

(x^2 +a^2 )dx

=

∫√

(a^2 sinh^2 θ+a^2 )(acoshθdθ)

=

∫√

[a^2 (sinh^2 θ+ 1 )](acoshθdθ)

=

∫ √

(a^2 cosh^2 θ)(acoshθdθ),

since cosh^2 θ−sinh^2 θ= 1

=

∫

(acoshθ)(acoshθ)dθ=a^2

∫

cosh^2 θdθ

=a^2

∫ (

1 +cosh 2θ

2

)

dθ