Examples from Section 6.10
Hyperbolic Functions

p. 525

Hyperbolic Function Values and Identities

1.

sinh(x)  =   -  (3/4). Since (cosh^2)(x)  -  (sinh^2)(x)  =  1, \(TraditionalForm
      \`cosh(x)\  = \ 
        \(\@\(1\  + \ \(sinh\^2\)(x)\ \) = \ 
          \(\@\(1\  + \ 9\/16\) = \ \(\@\(25\/16\) = \ 5\/4\)\)\)\). And therefore,
\(TraditionalForm
      \`tanh(x)\  = \ 
        \(\(sinh(x)\)\/\(cosh(x)\) = \ 
          \(\(\(-3\)/4\)\/\(5/4\) = \(-\ \(3\/5\)\)\)\)\).

x  =  ArcSinh[-3/4];

Cosh[x]

Tanh[x]

Sech[x]

Coth[x]

Csch[x]

Derivatives

13.

(∂_x) (6 Sinh[(x/3)])

15.

(∂_t) (2 (Sqrt[t]) Tanh[(Sqrt[t])])

17.

(∂_z) (Log[Sinh[z]])

27.

(∂_θ) ((1 - θ) ArcTanh[θ] )

Integrals

43.

∫6 Cosh[(x/2)  -  Log[3]] \[DifferentialD]x

53.

2  (E^θ) Cosh[θ]  /.  
        Cosh[θ]  ->  
          ((E^θ) +  (E^(-θ)))/2  // Expand

∫2  (E^θ) Cosh[θ] 
        \[DifferentialD]θ

(∫_(-Log[4])^(Log[2]))2  (E^θ) Cosh[θ] 
        \[DifferentialD]θ

67.

∫(1/(Sqrt[4  +  (x^2)])) \[DifferentialD]x

(∫_0^(2 (Sqrt[ 3])))(1/(Sqrt[4  +  (x^2)])) 
        \[DifferentialD]x

69.

∫(1/(1 - (x^2))) \[DifferentialD]x

(∫_(5/4)^2) (1/(1 - (x^2))) \[DifferentialD]x

Applications and Theory

86.

The hyperbolic in hyperbolic functions.

\(ParametricPlot[{Cosh[t], \ Sinh[t]}, \ {t, \(-2\), 2}, 

      PlotRange -> {{0, 3}, {\(-3\), 3}}]\)

The point (x, y)  =  (cosh (t),  sinh (t)) lies on the hyperbola defined by (x^2) -  (y^2) =  1 for all values of t:

Cosh[t]^2 -  Sinh[t]^2  // Simplify

Since cosh (t)  >  0 for all t, we only get the right hand branch of this hyperbola.

(refer to figure 6.35 on p.528). The area A (u) of sector AOP is given by the formula
A (u) = (1/2) cosh (u) sinh (u)  -  
          (∫_1^(cosh (u)))(Sqrt[(x^2) -  1]) 
              dx
This is just the difference of the area of a triangle and the area under the curve y  =  (Sqrt[(x^2) -  1 ])from x  =  0 to x  =  cosh (u). Taking the derivative of both sides yields:
A' (u)  =  
          (1/2) (sinh^2) (u) +  (1/2) (cosh^2) (u)  -  
            (Sqrt[(cosh^2) (u)  -  1]) sinh (u)
The last term simplifies to -(sinh^2) (u)so A' (u)  =  (1/2) (cosh^2) (u)- (1/2) (sinh^2) (u)= (1/2).
This implies that the area is A (u)  =  (1/2) u, one-half the value of the parameter (`angle') u.