cosh x sinh x So writing t = tan(x/2) we have 6.4 Integration using tan(x/2) 2t 1 + t 2 cos x = 1 t2 sin x =

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1 6.4 Integration using tan/ We will revisit the ouble angle ientities: sin = sin/ cos/ = tan/ sec / = tan/ + tan / cos = cos / sin / tan = = tan / sec / tan/ tan /. = tan / + tan / So writing t = tan/ we have Also so sin = t + t cos = t + t tan = t t. t = sec / = + tan / = + t = t + t. We can use these formulas to calculate integrals of the form a cos + b sin + c by converting them into integrals of rational functions. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn 007 / 6 Eample 6.4.: Integrate + sin. Let t = tan/. Then + sin = + t + t t +t = 6t + t + 6 t = t + t + t = t + t + t = lnt + lnt + + C = ln tan/ + ln tan/ + + C 6. Hyperbolic functions We efine the hyperbolic cosine of by an the hyperbolic sine of by = e + e sinh = e e. These functions turn out to be very similar in certain respects to the usual trigonometric functions. For eample, they satisfy similar ientities. This will be justifie more precisely when we consier comple numbers net term. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn / 6 By analogy with the stanar trig functions we efine an tanh = sinh sech = coth = tanh = sinh. cosech = sinh Although these functions are in some ways very similar to the stanar trig functions, they also have some striking ifferences. For eample, they are not perioic. The graph of : e This is an even function, an cosh 0 =. Note that this is also the minimum value of cosh: if y = then y = e e so y = 0 implies that e e = 0, i.e. e =, so = 0. e Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn / 6 The graph of sinh : e sinh The graph of tanh : e This is an o function, an sinh 0 = 0. There are no stationary points, but there is a point of inflection at 0. Note that the omain of all three functions is R. The range of sinh is R, of cosh is y, an of tanh is y <. Anton Co City University AS0 Week 7 Autumn / 6 Anton Co City University AS0 Week 7 Autumn / 6

2 Last time we claime that hyperbolic functions ha many similarities with trigonometric functions but saw that their graphs were quite ifferent. To justify, in part, our claim, we will now consier various hyperbolic ientities. Eample 6..: Show that sinh = sinh. Eample 6..: Show that cosh sinh =. cosh sinh = 4 e + + e 4 e + e = 4 4 =. sinh = e e e + e = e + e = sinh. The last two eamples are both very similar to the corresponing trig formulas, apart from the minus sign in 6... This is generally true: we can fin new hyperbolic ientities using Anton Co City University AS0 Week 7 Autumn / 6 Anton Co City University AS0 Week 7 Autumn / 6 Eample 6..: Fin a hyperbolic analogue to Osborn s Rule: i Change each trig function in an ientity to the corresponing hyperbolic function. ii Whenever a prouct of two sines occurs, change the sign of that term. This rule oes not prove the ientity; it can only be use to suggest possible ientities, which can then be verifie. Also note that proucts of sines can be isguise: for eample in tan we have sin cos. tan = Osborn s rule suggests that we try The righthan sie equals sinh + sinh cosh tanh = = sinh tan tan. tanh + tanh. cosh cosh + sinh = sinh cosh + sinh. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn 007 / 6 By Eample 6.. this equals sinh cosh + sinh so it is enough to prove that cosh + sinh =. 6.6 Solving hyperbolic equations These are usually simpler to solve than the corresponing trig equations. Eample 6.6.: Solve sinh =. But cosh + sinh = 4 e + + e + 4 e + e = e + e = as require. We have which becomes e e e + e = e e =. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn / 6 Eample 6.6.: Solve Therefore or e e = 0 e + e = 0. e = is impossible, so the only solution is e =, i.e. = ln. Sometimes, as for stanar trig functions, it is best to use an ientity to simplify the equation. cosh + 7 sinh = 4. We use cosh sinh =. Then we have + sinh + 7 sinh = 4 which simplifies to sinh + 44 sinh = 0. So sinh = 4 or sinh = 4. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn / 6

3 If sinh = 4 then e e = 4 i.e. e e = 8, or equivalently e + 8e = 0. Therefore e e + = 0 an hence e = as e = is impossible. So = ln = ln. If sinh = 4 then a similar calculation shows that = ln, an so the solutions to the equation are = ln an = ln. Integrate Note that + + = +. Thus = = = ln tan C. Anton Co City University AS0 Week 7 Autumn / 6 Anton Co City University AS0 Week 7 Autumn / 6 Eample 6..: + + = + + = = + tan + C = tan + + C. Compare with E 6... We can also eal with more complicate rational functions by using these methos together with partial fractions. Finally, we consier the integrals of inverse trigonometric functions. To integrate sin we use integration by parts with u = sin an v =. sin = sin Similarly tan = tan = sin + + C. + = tan ln + + C. Anton Co City University AS0 Week 7 Autumn / 6 Anton Co City University AS0 Week 7 Autumn / Calculus of hyperbolics It is easy to etermine the erivatives of hyperbolic functions. Eample 6.7.: Show that = sinh. = e + e = e e = sinh. Similarly we can show that sinh =. Note: Osborn s Rule oes not apply to calculus. We can now etermine the erivatives of all the other hyperbolic functions. These shoul be memorise. f f sinh sinh tanh sech cosech coth cosech coth cosech sech sech tanh. Reversing the roles of the two columns an remembering to a in the constant! we can euce the integrals of the functions in the right-han column. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn 007 / Inverse hyperbolic functions First consier sinh. From the graph we see that this is injective with image R. Thus it possesses an inverse function for all values of. For R we efine y = sinh if an only if = sinh y. Net consier tanh. This is also injective, but with image set < <. So for < < we efine The function cosh is not injective, so we cannot efine an inverse to the entire function. However, if we only consier cosh y on the omain y 0 then the function is injective, with image set cosh y. So for we efine y = cosh if an only if = cosh y an y 0. We can sketch the graphs of these functions: y = tanh if an only if = tanh y. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn / 6

4 y = sinh y = tanh y = cosh Sometimes these functions are enote by arsinh, arcosh, an artanh. It is easy to ifferentiate these functions. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn / 6 Eample 6.8.: Show that sinh = +. If y = sinh then = sinh y. Now y = cosh y, so y = cosh y. By E 6.., an the fact that cosh y 0 for all y, we have that cosh y = sinh y + = +. So Similarly sinh = +. cosh =. Anton Co City University AS0 Week 7 Autumn / 6 Eample 6.8.: Show that tanh =. If y = tanh then = tanh y, an so we have Osborn s Rule suggests that y = sech y an y = sech y. sech y = tanh y. We can an shoul verify this using the efinitions. Hence y = tanh y =. These three stanar erivatives shoul be memorise; we will see their usefulness in the net lecture. Anton Co City University AS0 Week 7 Autumn / 6 Recall E.. an E.. that we solve integrals of the form or 4 using the ientity cos u = sin u to suggest the substitution = a sin u. From the ientity cosh u sinh u = we can now solve integrals of the form or 4 + by means of the substitution = a cosh u or = a sinh u. Eample 6.8.: Calculate We have an so = = +. Let + = cosh u with u 0. Then u = sinh u an + = sinh u. Anton Co City University AS0 Week 7 Autumn / 6 Anton Co City University AS0 Week 7 Autumn / 6 Now sinh u sinh u u = sinh u u = cosh u u [ sinh u = ] u + C. But sinh u = sinh u cosh u = cosh u cosh u by our assumption on u an so = [ ] + + cosh + + C. Eample 6.8.4: Calculate Let = sinh u so u = cosh u an Then sinh = + sinh u = cosh u. cosh u [ u ] sinh u = cosh u 0 = sinh. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn 007 / 6

5 Generally we can quote an hence shoul know + a = sinh +C a a = cosh +C. a For integrals of the form a + b + c we can now solve by completing the square. Finally, we woul like to have a more eplicit formula for cosh an sinh. As an sinh are efine in terms of e, we might epect a formula involving ln. Let y = sinh, so = sinh y. Then Multiplying by e y we see that an hence = e y e y. e y e y = 0 e y + = 0. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn / 6 Solving for e y we obtain e y = ± +. But + > for all, an e y 0 for all y. Hence e y = + + an so In the same way we can show that sinh = y = ln + +. cosh = ln + recall that we have only efine cosh for. With these results we can now simplify our earlier eamples. Eample 6.8.: In E 6.6. we showe that cosh + 7 sinh = 4 ha solutions sinh = 4 an sinh = 4. By the above results we immeiately obtain an = ln = ln = ln = ln = ln. Anton Co City University AS0 Week 7 Autumn 007 / 6 Anton Co City University AS0 Week 7 Autumn / 6