406 Higher Engineering Mathematics

=

a^2

2

(

θ+

sinh2θ

2

)

+c

=

a^2

2

[θ+sinhθcoshθ]+c,

since sinh2θ=2sinhθcoshθ

Sincex=asinhθ,thensinhθ=

x

a

andθ=sinh−^1

x

a

Also since cosh^2 θ−sinh^2 θ= 1

then coshθ=

√

( 1 +sinh^2 θ)

=

√[

1 +

(x

a

) 2 ]

=

√(

a^2 +x^2

a^2

)

=

√

(a^2 +x^2 )

a

Hence

∫ √

(x^2 +a^2 )dx

=

a^2

2

[

sinh−^1

x

a

+

(x

a

)√(x (^2) +a (^2) )
a
]
+c

a^2
2
sinh−^1
x
a

• x
  2
  √
  (x^2 +a^2 )+c
  Now try the following exercise
  Exercise 160 Further problems on
  integration using thesinhθsubstitution

1. Find

∫

2

√

(x^2 + 16 )

dx.

[

2sinh−^1

x

4

+c

]

2. Find

∫

3

√

( 9 + 5 x^2 )

dx.

[

3

√

5

sinh−^1

√

5

3

x+c

]

3. Find

∫ √

(x^2 + 9 )dx.

[

9

2

sinh−^1

x

3

+

x

2

√

(x^2 + 9 )+c

]

4. Find

∫ √

( 4 t^2 + 25 )dt.

[

25

4

sinh−^1

2 t

5

+

t

2

√

( 4 t^2 + 25 )+c

]

5. Evaluate

∫ 3

0

4

√

(t^2 + 9 )

dt. [3.525]

6. Evaluate

∫ 1

0

√

( 16 + 9 θ^2 )dθ. [4.348]

40.8 Worked problems on integration

using thecoshθsubstitution

Problem 24. Determine

∫

1

√

(x^2 −a^2 )

dx.

Letx=acoshθthen

dx

dθ

=asinhθand

dx=asinhθdθ

Hence

∫

1

√

(x^2 −a^2 )

dx

=

∫

1

√

(a^2 cosh^2 θ−a^2 )

(asinhθdθ)

=

∫

asinhθdθ

√

[a^2 (cosh^2 θ− 1 )]

=

∫

asinhθdθ

√

(a^2 sinh^2 θ)

,

since cosh^2 θ−sinh^2 θ= 1

=

∫

asinhθdθ

asinhθ

=

∫

dθ=θ+c

=cosh−^1

x

a

+c,sincex=acoshθ

Itisshownonpage339that

cosh−^1

x

a

=ln

{

x+

√

(x^2 −a^2 )

a

}

which provides as alternative solution to

∫

1

√

(x^2 −a^2 )

dx

Problem 25. Determine

∫

2 x− 3

√

(x^2 − 9 )

dx.