Calculus of hyperbolic functions
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Osborn's rule states that when converting trigonometric identities to hyperbolic identities, cosine should be replaced with hyperbolic cosine and sine with hyperbolic sine, except when there is a product of two sines, where the sign must be changed. Care must be taken as the product of two sines is sometimes disguised in identities like tanx = sinx/cosx. The derivatives of inverse hyperbolic functions can be found by taking the derivative of the inverse function formula with respect to x. Integration identities can be used to evaluate integrals involving inverse hyperbolic functions and powers of hyperbolic functions.
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Using Eulers formula, exp(ix)=cos(x)+isin.docx
Using Euler's formula,
exp
(
i
x
)
=
cos
(
x
)
+
i
sin
(
x
)
,
and the usual rules of exponents, establish De Moivre's formula,
(
cos
(
n
θ
)
+
i
sin
(
n
θ
)
)
=
(
cos
(
θ
)
+
i
sin
(
θ
)
)
n
.
Use DeMoivre's formula to write the following in terms of
sin
(
θ
)
and
cos
(
θ
)
.
cos
(
6
θ
)
sin
(
6
θ
)
One of the properties of the
sin
(
x
)
and
cos
(
x
)
that I hope you recall from trigonometry is that
cos
(
x
)
is an even function, i.e.
cos
(
−
x
)
=
cos
(
x
)
,
while
sin
(
x
)
is an odd function, i.e.
sin
(
−
x
)
=
−
sin
(
x
)
.
We will see that any function can be split into pieces with these symmetries.
Given a general function
f
(
x
)
,
define
f
e
(
x
)
=
f
(
x
)
+
f
(
−
x
)
2
and
f
o
(
x
)
=
f
(
x
)
−
f
(
−
x
)
2
.
. Show
f
(
x
)
=
f
e
(
x
)
+
f
o
(
x
)
,
and
f
e
(
x
)
is an even function and
f
o
(
x
)
is an odd function.
So every function can be split into even and odd pieces.
Given that
f
(
x
)
=
f
e
(
x
)
+
f
o
(
x
)
where
f
e
(
x
)
is an even function and
f
o
(
x
)
is an odd function, show that
f
e
(
x
)
=
f
(
x
)
+
f
(
−
x
)
2
,
and
f
o
(
x
)
=
f
(
x
)
−
f
(
−
x
)
2
,
and
This shows the decomposition in the previous problem is the unique way to cut a function into even and odd pieces.
If we apply the decomposition developed in the previous two problems to the exponential function, we get the
hyperbolic functions
,
cosh
(
x
)
sinh
(
x
)
=
exp
(
x
)
+
exp
(
−
x
)
2
=
exp
(
x
)
−
exp
(
−
x
)
2
These functions are covered in your calculus text, but sometimes that section is skipped. They are closely related to the usual trigonometric functions and you can define hyperbolic tangent, secant, cosecant, and cotangent in the obvious way (e.g.
tanh
(
x
)
=
sinh
(
x
)
cosh
(
x
)
). The next few problems give a quick overview of some of their properties.
Show
cosh
(
i
x
)
=
cos
(
x
)
,
and
sinh
(
i
x
)
=
i
sin
(
x
)
.
So the hyperbolic functions are just rotations of the usual trigonometric functions in the complex plane.
Verify the following hyperbolic trig identities.
cosh
2
(
x
)
−
sinh
2
(
x
)
=
1
cosh
(
s
+
t
)
=
cosh
(
s
)
cosh
(
t
)
+
sinh
(
s
)
sinh
(
t
)
Note that these are almost the same as the corresponding identities for the regular trig functions, except for changes in the signs. You can derive hyperbolic identities corresponding to all the different identities you learned in trigonometry.
Invert the formula
sinh
(
x
)
=
exp
(
x
)
−
exp
(
−
x
)
2
to write
sinh
−
1
(
x
)
in terms of
log
(
x
)
.
Note that you will need to use the quadratic formula to get the inverse.
Verify the following differentiation rules for the hyperbolic functions
d
sinh
(
x
)
d
x
=
cosh
(
x
)
,
and
d
cosh
(
x
)
d
x
=
sinh
(
x
)
.
So for the hyperbolic functions you don't have to try to remember which derivative gets a minus sign.
Use the substitution
x
=
sinh
(
u
)
to evaluate the integral
∫
d
x
1
+
x
2
‾
‾
‾
‾
‾
‾
√
.
This integral can also be evaluated with a trig substitution, but using a hype.
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Similar to Calculus of hyperbolic functions
Using Eulers formula, exp(ix)=cos(x)+isin.docx
Using Euler's formula,
exp
(
i
x
)
=
cos
(
x
)
+
i
sin
(
x
)
,
and the usual rules of exponents, establish De Moivre's formula,
(
cos
(
n
θ
)
+
i
sin
(
n
θ
)
)
=
(
cos
(
θ
)
+
i
sin
(
θ
)
)
n
.
Use DeMoivre's formula to write the following in terms of
sin
(
θ
)
and
cos
(
θ
)
.
cos
(
6
θ
)
sin
(
6
θ
)
One of the properties of the
sin
(
x
)
and
cos
(
x
)
that I hope you recall from trigonometry is that
cos
(
x
)
is an even function, i.e.
cos
(
−
x
)
=
cos
(
x
)
,
while
sin
(
x
)
is an odd function, i.e.
sin
(
−
x
)
=
−
sin
(
x
)
.
We will see that any function can be split into pieces with these symmetries.
Given a general function
f
(
x
)
,
define
f
e
(
x
)
=
f
(
x
)
+
f
(
−
x
)
2
and
f
o
(
x
)
=
f
(
x
)
−
f
(
−
x
)
2
.
. Show
f
(
x
)
=
f
e
(
x
)
+
f
o
(
x
)
,
and
f
e
(
x
)
is an even function and
f
o
(
x
)
is an odd function.
So every function can be split into even and odd pieces.
Given that
f
(
x
)
=
f
e
(
x
)
+
f
o
(
x
)
where
f
e
(
x
)
is an even function and
f
o
(
x
)
is an odd function, show that
f
e
(
x
)
=
f
(
x
)
+
f
(
−
x
)
2
,
and
f
o
(
x
)
=
f
(
x
)
−
f
(
−
x
)
2
,
and
This shows the decomposition in the previous problem is the unique way to cut a function into even and odd pieces.
If we apply the decomposition developed in the previous two problems to the exponential function, we get the
hyperbolic functions
,
cosh
(
x
)
sinh
(
x
)
=
exp
(
x
)
+
exp
(
−
x
)
2
=
exp
(
x
)
−
exp
(
−
x
)
2
These functions are covered in your calculus text, but sometimes that section is skipped. They are closely related to the usual trigonometric functions and you can define hyperbolic tangent, secant, cosecant, and cotangent in the obvious way (e.g.
tanh
(
x
)
=
sinh
(
x
)
cosh
(
x
)
). The next few problems give a quick overview of some of their properties.
Show
cosh
(
i
x
)
=
cos
(
x
)
,
and
sinh
(
i
x
)
=
i
sin
(
x
)
.
So the hyperbolic functions are just rotations of the usual trigonometric functions in the complex plane.
Verify the following hyperbolic trig identities.
cosh
2
(
x
)
−
sinh
2
(
x
)
=
1
cosh
(
s
+
t
)
=
cosh
(
s
)
cosh
(
t
)
+
sinh
(
s
)
sinh
(
t
)
Note that these are almost the same as the corresponding identities for the regular trig functions, except for changes in the signs. You can derive hyperbolic identities corresponding to all the different identities you learned in trigonometry.
Invert the formula
sinh
(
x
)
=
exp
(
x
)
−
exp
(
−
x
)
2
to write
sinh
−
1
(
x
)
in terms of
log
(
x
)
.
Note that you will need to use the quadratic formula to get the inverse.
Verify the following differentiation rules for the hyperbolic functions
d
sinh
(
x
)
d
x
=
cosh
(
x
)
,
and
d
cosh
(
x
)
d
x
=
sinh
(
x
)
.
So for the hyperbolic functions you don't have to try to remember which derivative gets a minus sign.
Use the substitution
x
=
sinh
(
u
)
to evaluate the integral
∫
d
x
1
+
x
2
‾
‾
‾
‾
‾
‾
√
.
This integral can also be evaluated with a trig substitution, but using a hype.
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Calculus of hyperbolic functions
- 1. Osborn’s rule From previous examples we can see that the close comparison between identities in trigonometric functions and hyperbolic functions can be converted into a formulae known as Osborn’s rule, which states that the cos should be converted to cosh and sin converted to sinh, except when there is a product of two sines, we must change the sign. 1sincos 22 xx 1sinhcosh 22 xx However, whenever using Osborn’s rule care must be taken as the product of two sines is sometimes disguised eg x x x 2 2 2 cos sin tan
- 2. Calculus of Hyperbolic Functions If xx eexxf 2 1 cosh then xeexf xx sinh 2 1 ' xx dx d sinhcosh
- 3. If xx eexxf 2 1 sinh then xeexf xx cosh 2 1 ' xx dx d coshsinh Similarly
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- 8. Example Integrate with respect to x xex cosh
- 9. If then Inverse Hyperbolic functions The inverse hyperbolic functions are defined in a similar manner to the inverse of trigonometric function. xy sinhIf then yx 1 sinh xy cosh yx 1 cosh
- 10. The graphs of inverse hyperbolic functions are obtained from those of the hyperbolic functions by interchanging the x and y axes. 0 x y 0 x y xy 1 sinh xy 1 cosh This function is a one to one function This function is a one to many function
- 11. Example Let so a) Express in the terms of and hence show that b) Deduce that xy 1 tanh yx tanh ytanh y e x x e y 1 12 x x x 1 1 ln 2 1 tanh 1
- 12. Derivatives of inverse Hyperbolic functions Example Find the derivative of with respect of x. a x1 sinh
- 13. Example Find the derivative of with respect of x. a x1 cosh
- 17. Use of Hyperbolic functions in integration Example Using the previous results write down the values of (i) (ii) dx x 1 1 2 dx x 1 1 2
- 18. Example a) Differentiate with respect to x b) Hence find dx x 9 1 2 3 sinh 1 x
- 19. Example Use the substitution of to show thatux cosh2 c x dx x 2 cosh 4 1 1 2
- 20. In general c a x dx ax c a x dx ax 1 22 1 22 cosh 1 sinh 1
- 21. Example Evaluate 13 2 x dx
- 22. Example a) Express in the form where A, B and C are constants. b) Evaluate in terms of natural logarithms 584 2 xx CBxA 2 dx xx 7 4 2 584 1
- 23. Example Evaluate Leaving your answer in terms of natural logarithms. 1 3 2 136 dxxx