Higher Engineering Mathematics, Sixth Edition

Differentiation of inverse trigonometric and hyperbolic functions 343

Hence

d

dx

[coth−^1 (sinx)]=

cosx

[1−(sinx)^2 ]

=

cosx

cos^2 x

since cos^2 x+sin^2 x= 1

=

1

cosx

=secx

Problem 18. Differentiate

y=(x^2 − 1 )tanh−^1 x.

Using the product rule,

dy

dx

=(x^2 − 1 )

(

1

1 −x^2

)

+(tanh−^1 x)( 2 x)

=

−( 1 −x^2 )

( 1 −x^2 )

+ 2 xtanh−^1 x= 2 xtanh−^1 x− 1

Problem 19. Determine

∫

dx

√

(x^2 + 4 )

.

Since

d

dx

(

sinh−^1

x

a

)

=

1

√

(x^2 +a^2 )

then

∫

dx

√

(x^2 +a^2 )

=sinh−^1

x

a

+c

Hence

∫

1

√

(x^2 + 4 )

dx =

∫

1

√

(x^2 + 22 )

dx

=sinh−^1

x

2

+c

Problem 20. Determine

∫

4

√

(x^2 − 3 )

dx.

Since

d

dx

(

cosh−^1

x

a

)

=

1

√

(x^2 −a^2 )

then

∫

1

√

(x^2 −a^2 )

dx=cosh−^1

x

a

+c

Hence

∫

4

√

(x^2 − 3 )

dx= 4

∫

1

√

[x^2 −(

√

3 )^2 ]

dx

=4cosh−^1

x

√

3

+c

Problem 21. Find

∫

2

( 9 − 4 x^2 )

dx.

Since tanh−^1

x

a

=

a

a^2 −x^2

then

∫

a

a^2 −x^2

dx=tanh−^1

x

a

+c

i.e.

∫

1

a^2 −x^2

dx=

1

a

tanh−^1

x

a

+c

Hence

∫

2

( 9 − 4 x^2 )

dx= 2

∫

1

4

( 9

4 −x

2 )dx

=

1

2

∫

1

[(

3

2

) 2

−x^2

]dx

=

1

2

[

1

( 3

2

)tanh−^1

x

( 3

2

)+c

]

i.e.

∫

2

( 9 − 4 x^2 )

dx=

1

3

tanh−^1

2 x

3

+c

Now try the following exercise

Exercise 137 Further problems on

differentiation of inverse hyperbolic

functions

In Problems 1 to 11, differentiate with respect to

the variable.

1. (a) sinh−^1

x

3

(b) sinh−^14 x

[

(a)

1

√

(x^2 + 9 )

(b)

4

√

( 16 x^2 + 1 )

]

2. (a) 2 cosh−^1

t

3

(b)

1

2

cosh−^12 θ

[

(a)

2

√

(t^2 − 9 )

(b)

1

√

( 4 θ^2 − 1 )

]

3. (a) tanh−^1

2 x

5

(b) 3 tanh−^13 x

[

(a)

10

25 − 4 x^2

(b)

9

( 1 − 9 x^2 )

]

4. (a) sech−^1

3 x

4

(b)−

1

2

sech−^12 x

[

(a)

− 4

x

√

( 16 − 9 x^2 )

(b)

1

2 x

√

( 1 − 4 x^2 )

]