Integration Exercises - Part 3 (Sol'ns) (Hyperbolic Functions) (12 pages; 6/2/18) (The constant of integration has been omitted throughout.

Transcription

1 Integration Eercises - Part (Sol'ns) (Hyperbolic Functions) ( pages; 6//8) (The constant of integration has been omitted throughout.) () cosech Given that d tanh = sech, we could investigate d coth = d cosh = sinh.sinh cosh.cosh sinh sinh so that cosech = coth + c = sinh = cosech () tanh (from st principles) tanh = sinh cosh As sinh = cosh, let u = cosh Then du = sinh, and sinh = du = ln u + c cosh u = ln(cosh) + c [No moduli signs are needed, as cosh > 0 for all ] () sechtanh sechtanh = sinh cosh As sinh = cosh, let u = cosh Then du = sinh, and sinh cosh = u du = u + c

2 = sech + c (4) tanh tanh = sech = tanh + c (from differentiation table) (5) sech tanh As sech = tanh, let u = tanh, so that du = sech, and sech tanh = u du = u + c = tanh + c (6) sech tanh As we have a power of sech, consider d sech = sechtanh, from (i). Then let u = sech, so that du = sechtanh and sech tanh = u du = = u + c = sech + c

3 (7) cosh cosh sinh = & cosh + sinh = cosh, so that cosh = ( + cosh), and cosh = + cosh = + sinh + c 4 (8) cosh cosh = cosh( sinh ) Let u = sinh, so that du = cosh, and cosh( sinh ) = u du = u u + c = sinh sinh + c (9) sech Method sech = cosh = cosh cosh +sinh Let u = sinh, so that du = cosh Then cosh = du +sinh +u = arctan(sinh) + c Method sech = (e +e ) = Let u = e, so that du = e e e +

4 and e = du e + u + = arctan(e ) + c To show that arctan(sinh) + c = arctan(e ) + c : Let θ = arctan (sinh) and φ = arctan (e ) We need to show that θ & φ differ by a constant. tanθ = sinh & tan ( φ ) = e Let t = tan ( φ ) Then sinh = (e e ) = (t t ) = t [We can now use the standard right-angled triangle with sides t, t & + t : tanφ = t t & the hypotenuse follows from Pythagoras.] So sinh = tan ( π φ ) = tan (φ π ) Thus tan ( φ π ) = tanθ, and so φ π & φ differ by a multiple of π; ie θ & φ differ by a constant, as required. t (0) 4 9 = 4 9 ( ) = arcosh ( ( )) + C = arcosh ( ) + C 4

5 () (a) cosh (b) sinh (c) tanh (a) By Parts, cosh = cosh = cosh = cosh ( ) ( ) = cosh (b) By Parts, sinh = sinh = sinh + = sinh ( ) + ( ) = sinh + + (c) By Parts, tanh = tanh = tanh + = tanh + ln ( ) () ( 9) Let = coshu, so that = sinhu du Then I = sinhu 7sinh u du = 9 cosech u du 5

6 As d du tanhu = sech u, we can investigate d coshu du sinhu = sinhusinhu coshucoshu sinh u d du cothu: = sinh u = cosech u coshu Thus I = cothu = = 9 9 cosh u 9 9 = 9 9 Alternative method Let = secu, so that = secutanu Then I = secutanu 7tan u = 9 du = /cosu du = cosu du 9 ( sinu cosu ) 9 sin u ( ) = = = sinu 9 cos u 9 (/) 9 9 () 4 + Let = sinhu, so that = coshu du and I = + sinh u (coshu)du = 4 cosh u du = + cosh(u) du = (u + sinh(u)) = sinh ( ) + sinhucoshu = sinh ( ) + ( ) + = sinh ( ) + ( ) + = sinh ( )

7 [Note: The alternative substitution = tanu, leads to a much more complicated epression.] (4) + Let = sinh u, so that = sinhucoshu du and I = sinhu coshu ( sinhucoshu) du = sinh u du = (cosh(u) ) du = sinh(u) u = sinhu coshu u = ( + ) sinh ( ) (5) 4 Let = coshu, so that = sinhu du and I = sinhu. sinhu du = cosh(u) du 4 = sinh(u) u = sinhucoshu cosh () = 4 4 cosh () (6) + Let = sinhu, so that = coshu du and I = sinh u coshu coshu du 7

8 = sinh u(sinh u + )du [A reduction formula could be found for sinh 4 u du at this point, as an alternative to the following.] = [cosh(u) ] { [cosh(u) ] + } du = 4 [cosh(u) ][cosh(u) + ]du = 4 cosh (u) du = [(cosh(4u) + ] du 4 = sinh(4u) u 8 = 6 sinh(u) cosh(u) u 8 = 8 sinhucoshu(cosh u + sinh u) u 8 = 8 + ( + ) 8 sinh (7) (+) ( + ) = + = ( + ) 4 So I = (+ ) 4 = cosh ( + ) = cosh ( + ) 8

9 (8) 6 Noting that =, and that 6 = ( ), let u =, so that du = and I = u du = cosh ( ) (9) 4+ I = [arsinh ( )] = [ +9 ] + arsinh( 4 ) ( ) 0 = 4(5 ) + ln ( 4 + ( 4 ) + ) = 8 + ln ( (4 + 5)) = 8 + ln (0) Let 4 = sinhu, so that 4 = coshu du I = coshu. coshu du 0 4 = 9 ( + coshu)du 4 0 [since cosh u sinh u = & cosh u + sinh u = coshu] 9

10 = 9 8 [u + sinhu] 0 = 9 8 ( + 4 (e e )) = (e e ) () + Let = sinh y [so that + = cosh y] Then = sinhycoshy dy and I = coshy sinhy = cosh y dy = + coshy dy = y + sinhy. sinhycoshy dy = arsinh( ) + sinhycoshy = arsinh( ) + ( + ) () Method ( +) Let = sinh y, so that 4 = 6sinhycoshy dy Then I = (sinh y+) 6sinhycoshy dy 4 0

11 = sinhycoshy cosh ysinhy dy = sech y dy = tanhy = tanhy As = sinh y, cosh y = +, so that sech y = + = + and tanh y = sech y [applying Osborn's rule] so that tanhy = + = + = + Method Let = tan θ, so that 4 = 6tanθsec θ dθ Then I = (tan θ+) 6tanθsec θ 4 dθ = sec θ tanθsec θ tanθ dθ = cosθ dθ = sinθ As = tan θ, sec θ = +,

12 so that cos θ = + = + and sinθ = + = + = + () cosech()coth() Note that d (cosec) = cosec cot So suppose that the solution is of the form Acosech(). Then d (Acosech()) = A d ((sinh()) ) = A( )(sinh()) (cosh()) = Acoth()cosech() So we require A = and I = cosech()