7.3 Hyperbolic Functions 1 Chapter 7. Integrals and Transcendental Functions 7.3. Hyperbolic Functions Note. Recall that cos x and sin x are sometimes called the circular func- tions. This is because we can plot a point P on the circle x2 + y2 =1by letting P = (cos u, sin u) where u is twice the area of the sector determined by A = (1, 0), O = (0, 0) and P : Page 444 Figure from Exercise 86 7.3 Hyperbolic Functions 2 Note. We can similarly define the hyperbolic functions. Consider the hyperbola x2 y2 = 1. Choose a point P on the hyperbola and define − u as twice the (signed) area determined by the sector A = (1, 0), O = (0, 0), and P . Now use the coordinates of P to define the hyperbolic trigonometric functions: P = (cosh u, sinh u). Page 444 Figure from Exercise 86 7.3 Hyperbolic Functions 3 Note. We will use the exponential function to define the hyperbolic trig functions. Definition. We define ex + e x Hyperbolic cosine of x: cosh x = − 2 x x e e− Hyperbolic sine of x: sinh x = − 2 x x sinh x e e− Hyperbolic tangent x: tanh x = = x − x cosh x e + e− cosh x ex + e x Hyperbolic cotangent of x: coth x = = − sinh x ex e x − − 1 2 Hyperbolic secant of x: sech x = = x x cosh x e + e− 1 2 Hyperbolic cosecant of x: csch x = = sinh x ex e x − − 7.3 Hyperbolic Functions 4 The graphs are: From Table 7.3 page 436 7.3 Hyperbolic Functions 5 Theorem. (Table 7.4) We have the following identities: 2 2 cosh x sinh x =1 − sinh2x = 2sinh x cosh x 2 2 cosh 2x = cosh x + sinh x 2 cosh2x +1 cosh x = 2 2 cosh2x 1 sinh x = − 2 2 2 tanh x =1 sech x − 2 2 coth x =1+ csch x Example. Page 441 number 2. Theorem. (Table 7.5) We have the following differentiation properties: d du [sinh u] = cosh u dx dx d du [cosh u] = sinh u dx dx d 2 du [tanh u] = sech u dx dx d 2 du [coth u]= csch u dx − dx d du [ sech u]= sech u tanh u dx − dx 7.3 Hyperbolic Functions 6 d du [ csch u]= csch u coth u dx − dx Example. Prove some of the results in the above theorem. Theorem. (Table 7.6) We have the following integral properties: sinh u du = cosh u + C Z cosh u du = sinh u + C Z 2 sech u du = tanh u + C Z 2 csch u du = coth u + C − Z sech u tanh u du = sech u + C − Z csch u coth u du = csch u + C − Z Proof. These are just the integral versions of the results in Table 7.5. Q.E.D. Examples. Page 442 numbers 20 and 50. 7.3 Hyperbolic Functions 7 d Note. Since [sinh x] = cosh x > 0, then sinh x is an INCreasing dx function and so is one-to-one. The function cosh x is not one-to-one as we can see from the graph. The function sech x =1/ cosh x is also not one- to-one. Therefore, to define the inverse functions of cosh x and sech x, we must restrict the domains. Definition. We define some inverse hyperbolic trig functions. 1 Define y = sinh− x if x = sinh y. (The domain is then x ( , ).) ∈ −∞ ∞ 1 Define y = cosh− x if x = cosh y and y [0, ). (The domain is then ∈ ∞ x [1, ).) ∈ ∞ 1 Define y = sech − x if x = sech y and y [0, ). (The domain is ∈ ∞ then x (0, 1].) ∈ 7.3 Hyperbolic Functions 8 Note. The graphs of the above defined three inverse hyperbolic trig functions are: Page 439 Figure 7.5 Note/Definition. The hyperbolic tangent, cotangent, and cosecant are one-to-one on their domains and therefore have inverses, denoted by 1 1 1 y = tanh− x, y = coth− x, y = csch − x. The graphs of these functions are: 7.3 Hyperbolic Functions 9 Page 439 Figure 7.6 Theorem. We can express the inverse hyperbolic trig functions in terms of the natural logarithm function as follows: 1 2 sinh− x = ln(x + √x + 1), x ( , ). ∈ −∞ ∞ 1 2 cosh− x = ln(x + x 1), x [1, ). − ∈ ∞ 1 1 1+px tanh− x = ln , x ( 1, 1). 2 1 x ∈ − − 2 1 1+ √1 x sech − x = ln − , x (0, 1]. x ! ∈ 2 1 1 √1+ x csch − x = ln + , x ( , 0) (0, ). x x ∈ −∞ ∪ ∞ | | ! 7.3 Hyperbolic Functions 10 1 1 x +1 coth− x = ln , x ( , 1) (1, ). 2 x 1 ∈ −∞ − ∪ ∞ − Theorem. (Table 7.7) We can verify the following identities: 1 1 1 sech − x = cosh− x 1 1 1 csch − x = sinh− x 1 1 1 coth− x = tanh− x Theorem. (Table 7.8) The inverse hyperbolic trig functions are differ- entiated as follows: d 1 1 du sinh− u = dx √1+ u2 dx d 1 1 du cosh− u = , u (1, ) dx √u2 1 dx ∈ ∞ − d 1 1 du tanh− u = , u ( 1, 1) dx √1 u2 dx ∈ − − d 1 1 du coth− u = , u ( , 1) (1, ) dx √1 u2 dx ∈ −∞ − ∪ ∞ − d 1 1 du sech − u = − , u (0, 1) dx u√1 u2 dx ∈ − d 1 1 du csch − u = − , u ( , 0) (0, ) dx u √1+ u2 dx ∈ −∞ ∪ ∞ | | 7.3 Hyperbolic Functions 11 Theorem. (Table 7.9) We have the following integrals involving inverse hyperbolic trig functions: du 1 u = sinh− + C, a> 0 √a2 + u2 a Z du 1 u = cosh− + C, u>a> 0 √a2 + u2 a Z 1 1 u 2 2 du a tanh− a + C if u <a = a2 u2 1 1 u 2 2 Z − a coth− a + C if u >a du 1 1 u = sech − + C, 0 <u<a u√a2 u2 −a a Z − du a 1 u = csch − + C, u =0 u√a2 + u2 −a a 6 Z Examples. Page 442 number 70 and page 443 number 79. Corrected 1/18/2020.