Hperbolic Functions Section : The inverse hperbolic functions Notes nd Emples These notes contin subsections on The inverse hperbolic functions Integrtion using the inverse hperbolic functions Logrithmic form of the inverse hperbolic functions Solving equtions The inverse hperbolic functions The inverse hperbolic functions sinh, cosh nd ver similr w to the inverse trigonometric functions. tnh Let sinh, then sinh Differentiting implicitl: cosh d d cosh + sinh +, operte in Hence Similrl nd d sinh d + d cosh d d tnh d These results re given in our formul book under Differentition. To differentite epressions such s sinh e ou cn either use substitution e u or ou cn repet the bove differentition. Both methods re shown in the emple below; use whichever ou prefer. Emple Find d when sinh e d (Method ) Let e u then sinh u. MEI, /0/08 /0
Differentiting both: Using the chin rule: d d du + u du d du d e + u d u e nd e + (e ) e + e 6 (Method ) sinh e sinh e Differentiting implicitl: cosh e d e d cosh e + sinh e + e 6 Integrtion using the inverse hperbolic functions B similr method to the bove, the generl formule which re given in the formul booklet (under Integrtion ) cn be obtined. If sinh, sinh nd hence cosh d d cosh + + Thus d + d sinh + c cosh + c MEI, /0/08 /0
Emple Find d 5 + In the formul for sinh, 5 nd hence 5 d sinh + c 5 + 5 You will probbl hve more complicted epressions to integrte. The ke is, where ever possible, to ensure tht the coefficient of ² is. Emple Find d 6 + 9 d d 6 6 + 9 9 + sinh + c 4 sinh + c 4 This is of the form d 4 with + In some cses ou m need to complete the squre, just s ou did for integrls which gve inverse trigonometricl functions in the lst chpter. Emple 4 Integrte d 6 + 4 + Initill ensure tht the coefficient of is, nd then complete the squre for the denomintor. 6 + 4+ + + 8 { } { } { } ( + ) + 8 ( + ) + 7 This is of the form d + with + for nd 7 MEI, /0/08 /0
d d 6 + 4+ 7 + ( + ) + sinh 7 + c As long s the coefficient of is, the stndrd formule will operte. As with inverse trigonometric functions, in more complicted integrls ou will find tht there is sometimes n term, or even n term in the numertor. The stndrd method is to obtin multiple of the differentil of the denomintor on the top (see Emple 5) or to divide the numertor b the denomintor. Emple 5 Integrte 8 + 4 + 6 + d The derivtive of 4+ 6 + is 6+. 8+ 4(6+ ) d d 4+ 6+ 4+ 6+ 6+ 4 d d 4+ 6+ 4+ 6+ 6+ 4 d d 4+ 6 + ( + ) 5 + 8 4 6 cosh c 5 + + + For the first term, remember tht d f {f }., hence d f f( ) d f( ) + c f The Hperbolic Differentition dominoes nd Hperbolic Integrtion dominoes ctivities give some etr prctice in differentition nd integrtion of functions involving hperbolic nd inverse hperbolic functions. MEI, /0/08 4/0
Logrithmic form of the inverse hperbolic functions Inverse hperbolic functions cn be converted into logrithmic form. nd cosh ln + ( ) sinh ln + + + tnh ln ( < ) The reson for the restrictions on the vlues of for the inverse cosh nd inverse tnh functions cn esil be seen b glnce t the grphs of the inverse hperbolic functions (given below). While sinh - is defined for ll vlues of, both cosh - nd tnh - hve limited domins. sinh cosh 6 4 4 6 6 4 4 6 tnh To help ou recognise the grphs of the inverse hperbolic functions nd the logrithmic forms, tr the Inverse Hperbolic grphs mtching ctivit. The Hperbolic dominoes ctivit tests ou on hperbolic identities nd lso the logrithmic forms of the inverse hperbolic functions. MEI, /0/08 5/0
Note tht these logrithmic formule give the principl vlues of the inverse hperbolic functions. In the cse of sinh - nd tnh -, there is onl one possible vlue of the inverse hperbolic function for ech vlue of, since both sinh nd tnh re one-to-one functions. However, for cosh -, becuse the grph of cosh is smmetricl bout the -is, there re two possible vlues for cosh - for ech vlue of, one positive nd one negtive. This cn be importnt in solving equtions (see Emple lter in these notes). Since ou re required to lern ech of the proofs of these identities the re given below. The formule re ll given in the formul booklet. Emple 6 Prove tht cosh ln + Let cosh. Multipl b e cosh ( e + e ) (e ) e + 0 4 e ± ± cosh ln + or cosh ln ( )( ) ( ( ) ( + ) so ln ( ) ln + Therefore cosh ± ln + + ) 4 Use (e ) e is qudrtic in e, nd this If ou require onl the principl vlue of cosh, then ou cn tke just the first solution. However, if ou need both nswers then ou will need to complete the proof s shown. The equivlent proof for the inverse sinh does not hve this difficult s ou cn see in Emple 7, one of the possible nswers does not eist nd so cn be rejected. Emple 7 Prove tht sinh ln + + MEI, /0/08 6/0
Let sinh sinh ( e e ) e + (e ) e 0 4 e sinh ln + + ± + ± + 4 Multipl b e Use (e ) e is qudrtic in e, nd this Emple 8 + Prove tht tnh ln. Let tnh sinh (e e ) tnh cosh (e + e ) e e e + e e( ) e ( + ) e + + tnh ln These results cn then be pplied to the generl integrtion formule, replcing b. d + + sinh ln + + + c + + ln + c c ln + + ln + c MEI, /0/08 7/0
Now, since ln is just number, it cn be included in the constnt of integrtion, giving the result d sinh + ln + + + c + c Similrl, d + + cosh c ln + c These formule pper in our formul book. Emple 9 shows how n integrtion which cn be done using prtil frctions result in n inverse tnh function. Emple 9 Find d d Using prtil frctions, d + + {ln( + ) ln( )} + c + ln + c tnh This is nother result which ppers in our formul book. Solving equtions Equtions contining hperbolic functions cn often be solved either b using the originl definitions (s in emple 0) or b using hperbolic identities in similr w to solving trigonometric functions (see emples nd ). Emple 0 Solve the eqution 5 cosh 7sinh Using the definitions of cosh nd sinh MEI, /0/08 8/0
5 e + e 7 e e e + e 4 e + e 0 (e + 4)(e ) 0 e 4 e or ln Emple Solve the eqution 4cosh sinh 9 Using cosh + sinh, the eqution becomes 4( + sinh ) sinh 9 8sinh sinh 5 0 (8sinh + 5)(sinh ) 0 sinh or Using sinh ln( + + ) ln ( + ) or ln ( ) 5 5 8 + 64 i.e. ln ( + ) or 89 5 ln 8 5 8 Emple involved finding n inverse sinh, which hs just one possible vlue. However, when n inverse cosh is needed, then there will usull be more thn one solution. This is shown in Emple. Emple Solve the eqution co sh cosh cosh cosh cosh cosh cosh cosh 0 ( cosh )(cosh + ) 0 cosh or MEI, /0/08 9/0
Since cosh, cosh cosh ± ln + ± ln + 5 4 ( ) ± ln ( + 5) MEI, /0/08 0/0