5548993 - Further Pure an 3 Moule FP Further Pure
5548993 - Further Pure an 3 Differentiating inverse trigonometric functions Throughout the course you have graually been increasing the number of functions that you can ifferentiate an integrate. This chapter etens this evelopment to inverse trigonometric functions. When you have complete it, you shoul know the erivatives of tan, sin an cos know the integrals corresponing to these erivatives be familiar with other inverse trigonometric functions an relations between them use these relations to ifferentiate other inverse trigonometric functions.. The inverse tangent The simplest of the inverse trigonometric functions to ifferentiate is tan. You can o this irectly from the efinition, that y = tan is the number such that tan y = an π < y < π. You know, from a general result about inverse functions (see C3 Section.9) that its graph is the reflection in the line y = of the part of the graph of y = tan for which π < < π. This is shown in Fig... y π y = tan π Fig.. The equation tan y = is not in the form y =..., but it can be ifferentiate using the metho for curves efine implicitly escribe in C4 Chapter 8. The erivative with respect to of tan y is sec y y, an the erivative of is, so tan y = sec y y = y = sec y.
5548993 - Further Pure an 3 4 Further Pure But this isn t really satisfactory. When you ifferentiate y = tan, you epect an answer in terms of, not y. However, this is easily ealt with. Since sec y = + tan y, an tan y =, so sec y = +, y = +. tan = +. It is interesting that the erivative of tan is not any sort of trigonometric function, but a rational function. This may remin you of what happens with ln, whose erivative oesn t involve logarithms or eponentials. Notice that, since +, the graient of the graph in Fig.. is less than or equal to throughout its length. Eample.. Differentiate with respect to tan, 3 (b) tan. Both erivatives can be foun by using the chain rule. tan = 3 + ( ) = 3 3 +. 3 9 To write this more simply, multiply top an bottom of the fraction by 9, to get tan 3 = 9 3 9 ( + 9 ) = 3 9 +. (b) tan = = + ( ) +. 4 Eample.. Fin tan In C4 Eample..3, ln was foun by writing the integran as ln an using integration by parts. This works because the erivative of ln is a rational function of, an oesn t involve a logarithm. You can fin tan in a similar way, an for the same reason. Writing u = tan an v =, so that v =, tan = tan +.
5548993 - Further Pure an 3 Differentiating inverse trigonometric functions 5 You shoul recognise this last integral, So f (), which is ln f() +k (see C4 Section.4). So f() + =, as a constant multiple of the form + + = ln( + ) + k. tan = tan ln( + ) k.. Inverse sine an cosine The metho of ifferentiating sin an cos is similar to that for tan, but there are some small complications. The easier an more important is sin, so begin with this. The efinition is that y = sin is the number such that sin y = an π y π. Its graph is shown in Fig... The omain of the function is. y π y = sin π Fig.. Differentiating this equation by the implicit metho gives sin y = cos y y = y = cos y. Again this erivative has to be epresse in terms of, an this time the relation you want is cos y + sin y =, so that cos y =± sin y, which is ±. But shoul the sign be + or? To answer this, look at the graph in Fig... You can see that the graient of the graph is always positive, so the + sign must be chosen. So replacing cos y by, y =.
5548993 - Further Pure an 3 6 Further Pure Eample.. Fin the omains, an the erivatives with respect to, of sin 5, (b) sin ( ). The omain of sin is, so the numbers in the omain of sin 5 must satisfy the inequalities. The omain is therefore 5 5. 5 Using the chain rule, sin = 5 ( ) 5 5 = 5 5 = 5 ( ) 5 =. 5 (b) The numbers in the omain of sin ( ) must satisfy the inequalities an, that is an. The omain is therefore. Using the chain rule, sin ( ) = = = ( ) ( ) ( + ). You can fin the erivative of cos using the same metho as for sin. This is left for you to o for yourself in Eercise A Question 7. But there is an even easier way. y π y = cos π y = 4 π y = sin π Fig..3
5548993 - Further Pure an 3 Differentiating inverse trigonometric functions 7 Figure.3 shows the graphs of y = sin an y = cos rawn using the same aes. You can see that the graphs are reflections of each other in the line y = π. So, for each value of, the 4 graient of y = cos is minus the graient of y = sin. That is, cos = sin =. If < <, sin = an cos =. The erivatives of tan, sin an cos are important an you shoul remember them. You may be surprise that the erivatives of sin an cos are state for < <, an not for the whole omain. It is easy to see why from Fig..3. When = an = the tangents to the graphs are parallel to the y-ais, so that the graient is unefine. Also of course has no meaning for these values of. Eercise A Fin the erivatives of the following with respect to. tan (b) sin 3 (c) tan () (sin ) (e) sin (f) tan ( ) (g) sin Fin the minimum point of the graph of y = 4 tan. 3 State the natural omain (that is, the largest possible omain) of the function f() = 5 3 sin. Fin the turning points on the graph of y = f(), an sketch the graph. Hence fin the range of the function. 4 Repeat Question 3 for the function f() = 5 4 sin. 5 The tangent to y = (tan ) at the point where = cuts the y-ais at the point P. Fin the y-coorinate of P. 6 Fin the maimum value of f() = (sin ) cos in the interval < <. 7 Fin cos by the metho use in Section. to fin sin.
5548993 - Further Pure an 3 8 Further Pure.3 The corresponing integrals The erivatives in Sections. an. can also be interprete as integrals. From tan = it follows that + + = tan + k. An from sin = you can euce = sin + k. You coul also use cos = to obtain = cos + k, but there is no point in using two ifferent forms for the same integral, an sin + k is simpler. What is the connection between k an k? Eample.3. Fin +. + = [ tan ] = tan tan = π 4 = π. 4 Use a graphic calculator to isplay the graph of y = for, an + ientify the area represente by the integral in Eample.3.. It is interesting that the number π appears in calculating an area which has nothing to o with a circle. Eample.3. Fin. This integral nees a little more care. Notice that the integran is not efine when =, because =. So this is an improper integral, an it must be calculate as a limit. Use a graphic calculator to isplay the graph of y =.
5548993 - Further Pure an 3 Differentiating inverse trigonometric functions 9 Begin by fining, for a number s such that < s <, s = [ sin ] s = sin s sin = sin s. The graph of y = sin in Fig.. shows that, as, sin π.so s = lim s = lim(sin s) s = π. You will fin that you often want to fin integrals like those at the beginning of this section in a slightly more general form, as or, where a is a positive number. a + a It is easy to o this by using the substitution = au. Then = a, so u a + = a + a u a u a = a ( + u ) u = a + u u = a tan u + k = a tan a + k, an a = a a u a u = = a a = sin u + k = sin a + k. a a ( u ) u u u You will nee to remember these results, either in the forms given at the beginning of the section or in these more general forms. If a >, a + = a tan a + k, a = sin a + k.
5548993 - Further Pure an 3 Further Pure Eample.3.3 Figure.4 shows the graph of y = for. Fin the volume of the soli forme 4 + when the region boune by this curve an parts of the lines =, = an the -ais is rotate though a complete revolution about the -ais. y y = 4 + Fig..4 The volume is given by the integral π y = π The volume of the soli is 4 π. 4 + ] = π [ tan = π(tan tan ( )) = π( π ( π)) 4 4 = π π = π. 4 Eample.3.4 Fin, where a an b are positive constants. a b If b is written as au, then a b becomes a a u, which simplifies to a u. So, substituting = au b, Eample.3.5 Fin 9 + 6 + 5. a b = a u a b u = b = b sin u + k = b sin b a + k. u u Since 9 + 6 + 5 = (3 + ) + 4, substitute 3 + = u. Then u = ; also when 3 = an, u = an respectively. So the integral becomes 4u + 4 u = 3 6 u + u = [ 6 tan u ] = 6 (tan tan ( )) = 6 (tan + tan ). If you want a numerical answer, on t forget to put your calculator into raian moe. The value is.35, correct to 3 ecimal places.
5548993 - Further Pure an 3 Differentiating inverse trigonometric functions Eercise B Evaluate the following efinite integrals. Give each answer as an eact multiple of π if possible; otherwise give the answer correct to 3 significant figures. () 3 5 (b) 4 (e) 3 + Fin the following infinite integrals. (b) + 3 Fin the following improper integrals. 5 3 (b) 5 3 (c) 5 + (c) 9 + 3 (f) 3 4 (c) 9 4 + 5 + 4 4 Use a substitution of the form = cu for a suitable value of c to fin the following inefinite integrals. (b) 9 + 4 (c) 4 9 4 () (e) (f) + 9 + 3 4 5 5 By completing the square an then using a substitution of the form = a + bu, fin the following inefinite integrals. + + (b) + 6 + 3 (c) 4 + 5 () (e) (f) 5 4 8 + 6 9 6 Evaluate the following efinite integrals. Give your answers to 3 significant figures. In some parts you may nee to use one of the methos escribe in Question 4 an Question 5. () (g) + 5 (b) (e) 9 + 5 3 3 + (h).5 7 Fin 8 Show that the rule.5 (c) + 6 9 (f) 4 (i), where a an b are positive constants. a + b a = sin a + k oesn t. 6 + 5 6 9 4 a + = a tan + k remains true if a <, but that a