cosh x sinh x So writing t = tan(x/2) we have 6.4 Integration using tan(x/2) = 2 2t 1 + t 2 cos x = 1 t2 We will revisit the double angle identities:

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1 6.4 Integration using tanx/) We will revisit the ouble angle ientities: sin x = sinx/) cosx/) = tanx/) sec x/) = tanx/) + tan x/) cos x = cos x/) sin x/) tan x = = tan x/) sec x/) tanx/) tan x/). = tan x/) + tan x/) So writing t = tanx/) we have Also so sin x = t + t cos x = t + t tan x = t t. t = sec x/) = + tan x/)) = + t = t + t. We can use these formulas to calculate integrals of the form a cos x + b sin x + c by converting them into integrals of rational functions. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Example 6.4.: Integrate + sin x. Let t = tanx/). Then + sin x = ) + t + t t +t = 6t + t + 6 t = t + )t + ) t = t + t + t = lnt + ) lnt + )) + C = ln tanx/) + ) ln tanx/) + )) + C 6. Hyperbolic functions We efine the hyperbolic cosine of x by an the hyperbolic sine of x by = ex + e x ) sinh x = ex e x ). These functions turn out to be very similar in certain respects) to the usual trigonometric functions. For example, they satisfy similar ientities. This will be justifie more precisely when we consier complex numbers next term. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 4 / By analogy with the stanar trig functions we efine an tanh x = sinh x sechx = coth x = tanh x = sinh x. cosechx = sinh x Although these functions are in some ways very similar to the stanar trig functions, they also have some striking ifferences. For example, 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 = ex e x so y = 0 implies that ex e x = 0, i.e. e x =, so x = 0. e x x Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 6 / The graph of sinh x: e x sinh x The graph of tanh x: e x 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 <. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 7 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 8 /

2 Lecture 6 In the last lecture 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. Example 6..: Show that sinh x = sinh x. Example 6..: Show that cosh x sinh x =. cosh x sinh x = 4 ex + + e x ) 4 ex + e x ) = 4 4 =. sinh x = ex e x ) ex + e x ) = ex + e x ) = sinhx). The last two examples 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 Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 9 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 0 / Example 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 example in tan x we have sin x cos x. tan x = Osborn s rule suggests that we try The righthan sie equals sinh x + sinh x cosh x tanh x = = sinh x tan x tan x. tanh x + tanh x. cosh x cosh x + sinh x = sinh x cosh x + sinh x. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / By Example 6.. this equals sinh x cosh x + sinh x so it is enough to prove that cosh x + sinh x =. 6.6 Solving hyperbolic equations These are usually simpler to solve than the corresponing trig equations. Example 6.6.: Solve sinh x =. But cosh x + sinh x = 4 ex + + e x ) + 4 ex + e x ) = ex + e x ) = as require. We have which becomes ex e x ) ex + e x ) = e x e x =. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 4 / Example 6.6.: Solve Therefore or e x e x = 0 e x + )e x ) = 0. e x = is impossible, so the only solution is e x =, i.e. x = ln. Sometimes, as for stanar trig functions, it is best to use an ientity to simplify the equation. cosh x + 7 sinh x = 4. We use cosh x sinh x =. Then we have + sinh x) + 7 sinh x = 4 which simplifies to sinh x + 4)4 sinh x ) = 0. So sinh x = 4 or sinh x = 4. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 6 /

3 If sinh x = 4 then e x e x = 4 i.e. e x e x = 8, or equivalently e x + 8e x = 0. Therefore e x )e x + ) = 0 an hence e x = as ex = is impossible). So x = ln = ln. If sinh x = 4 then a similar calculation shows that x = ln, an so the solutions to the equation are x = ln an x = ln. Lecture Calculus of hyperbolics It is easy to etermine the erivatives of hyperbolic functions. Example 6.7.: Show that ) = sinh x. ) ) = ex + e x ) = ex e x ) = sinh x. Similarly we can show that sinh x) =. Note: Osborn s Rule oes not apply to calculus. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 7 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 8 / We can now etermine the erivatives of all the other hyperbolic functions. These shoul be memorise. f x) f x) sinh x sinh x tanh x sech x cosech x coth x cosech x coth x cosech x sech x sech x tanh x. 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. 6.8 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 x. For x R we efine y = sinh x if an only if x = sinh y. Next consier tanh. This is also injective, but with image set < x <. So for < x < we efine y = tanh x if an only if x = tanh y. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 9 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 0 / 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 x we efine y = cosh x if an only if x = cosh y an y 0. y = sinh x y = tanh x We can sketch the graphs of these functions: Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / y = cosh x Sometimes these functions are enote by arsinh, arcosh, an artanh. It is easy to ifferentiate these functions. Example 6.8.: Show that sinh x) = x +. If y = sinh x then x = sinh y. Now y = cosh y, so y = cosh y. By Ex 6.., an the fact that cosh y 0 for all y, we have that cosh y = sinh y + = x +. So sinh x) = x +. Similarly cosh x) = x. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 4 /

4 Example 6.8.: Show that tanh x) = x. If y = tanh x then x = 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 = x. These three stanar erivatives shoul be memorise; we will see their usefulness in the next lecture. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Lecture 8 Recall Ex.. an Ex..) that we solve integrals of the form x or 4 x using the ientity cos u = sin u to suggest the substitution x = a sin u. From the ientity cosh u sinh u = we can now solve integrals of the form x or 4 + x by means of the substitution x = a cosh u or x = a sinh u. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 6 / Example 6.8.: Calculate x + 4x 8. We have an so x + 4x 8 = x + ) ) x + 4x 8 = x + ) ). Let x + = cosh u with u 0. Then u = sinh u an x + ) ) = sinh u. 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 x + 4x 8 = [ ) ] x+ x+) cosh x+ + C. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 7 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 8 / Example 6.8.4: Calculate x Let x = sinh u so u = cosh u an Then sinh 0 + 4x = + sinh u = cosh u. cosh u [ u ] sinh u = cosh u 0 = sinh. Generally we can quote an hence shoul know) x + a = x ) sinh +C a x a = x ) cosh +C. a For integrals of the form ax + bx + c we can now solve by completing the square. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 9 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 0 / Finally, we woul like to have a more explicit formula for cosh x an sinh x. As an sinh x are efine in terms of e x, we might expect a formula involving ln. Let y = sinh x, so x = sinh y. Then Multiplying by e y we see that an hence x = e y e y. e y xe y = 0 e y x) x + ) = 0. Solving for e y we obtain e y = x ± x +. But x + > x for all x, an e y 0 for all y. Hence e y = x + x + an so In the same way we can show that sinh x = y = lnx + x + ). cosh x = lnx + x ) recall that we have only efine cosh x for x.) With these results we can now simplify our earlier examples. Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 / Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 /

5 Example 6.8.: In Ex 6.6. we showe that cosh x + 7 sinh x = 4 ha solutions sinh x = 4 an sinh x = 4. By the above results we immeiately obtain x = ln 4 ) = ln ) = ln) an x = ln ) = ln). Anreas Fring City University Lonon) AS0 Lecture -8 Autumn 00 /