Chapter 6: Integration: partial fractions and improper integrals

Chapter 6: Integration: partial fractions an improper integrals Course S3, 006 07 April 5, 007 These are just summaries of the lecture notes, an few etails are inclue. Most of what we inclue here is to be foun in more etail in Anton. 6. Remark. Here is a quick reminer of the basics of integration, before we move on to partial fractions. 6. Remark. Unlike ifferentiation where we can ifferentiate almost anything we can write own using the basic rules (incluing the chain rule, prouct rule an quotient rule), with integration it is easy to come across simple-looking things we will not be able to o. One example is sin x x. For each ifferentiation formula, we have a corresponing integration formula. Derivative formula x xn nx n x ex e x sin x cos x x cos x sin x x x tan x sec x sec x sec x tan x x We alreay see a gap an we nee is Integration formula x n x n + xn+ + C if n e x x e x + C cos x x sin x + C sin x x cos x + C sec x x tan x + C sec x tan x x sec x + C x ln x + C. One way to efine the natural logarithm x ln x x t (x > 0) t an then the fact that its erivative is /x (for x > 0) is an immeiate consequence of the Funamental theorem. Essentially the antierivative of /x is new thing (much more complicate

S3 006 07 R. Timoney than /x) an there is almost no limit to the new functions we can come across this way new in the sense that they cannot be written in terms of familiar functions. The natural logarithm function is a useful thing in many ways. Apart from these basic integrals, there are integration formulae that follow from the chain rule for ifferentiation an the prouct rule for ifferentiation. Essentially, when we rearrange the integral of the two formulae, we get the metho of substitution an integration by parts. Asie from these there are no real methos but there is a compenium of what are essentially tricks to make the methos useful in given situations. Inclue amongst these are ways of integrating (many) combinations of trigonometric functions (evaluate at x, or maybe ax for a constant a rather than at more complicate expressions) an inverse trigonometric substitutions. These latter have the effect of introucing trigonometry into integration problems that seem to be unrelate to trigonometry. A lot of this was covere in S. If in oubt look in Anton, or at the appenix to this part of the notes. Not all of the trigonometric an inverse trigonometric (an inverse hyperbolic) substitutions were one in S an so we fill in the gaps here. 6.3 Trigonometric Integrals. For the first 3 items we are just listing them for context. (They were covere in S an also you can fin more etail in the appenix below.) (i) Powers of sin x times powers of sin x with one power o Metho: For sin n x cos m x x if n the power of sin x is o, substitute u cos x if m the power of cos x is o, substitute u sin x (ii) Powers of sin x times powers of sin x with both powers even Metho: use the trigonometric ientities sin x ( cos x), cos x ( + cos x) (iii) Powers of sin x an cos x Metho: Use the previous two methos treating sin n x x sin n x(cos x) 0 x an similarly for cos m x x (that is treat the secon power as the zeroth power). There is another way to o these problems, an it is given in the book by Anton. Possibly also some of you know it alreay.

Chapter 6 Techniques of Integration 3 Alternative metho: Use the reuction formulae sin n x x n sinn x cos x + n n cos m x x m cosm x sin x + m m sin n x x cos m x x (vali for n an m ) to express the integrals of a power in terms of integrals where the power is reuce by. Using the formula enough times, will get own to power or zero eventually. (iv) Powers of tan x The first power tan x x We can work this out using the o powers of sin x metho. sin x tan x x cos x x Let u cos x u sin x x sin x u u sin x u u ln u + C ln cos x + C Usually this is rewritten using ln cos x ln cos x ln cos x ln sec x an the result is tan x x ln sec x + C The secon power tan x x This is really quite easy if we recall the ientity + tan x sec x. We get tan x x sec x x tan x x + C Higher powers tan n x x (n 3)

4 S3 006 07 R. Timoney What we o here is to split off two powers (that is tan x sec x ) tan n x x tan n x tan x x tan n x(sec x ) x tan n x sec x x tan n x x In the first integral, substitute u tan x u sec x x u n u tan n x x n un tan n x x n tann x tan n x x This is a reuction metho. It expresses tan n x x in terms of a similar integral with the power reuce by. For example tan 5 x x 5 tan5 x tan 5 x x 4 tan4 x tan 3 x x ( ) 4 tan4 x 3 tan x tan x x an we know tan x x. (v) Powers of sec x The first power: sec x x There is no really nice way to fin this out. What we o is write own the answer sec x x ln sec x + tan x + C an show that it works. That is we check that the right han sie is an antierivative for sec x: x ln sec x + tan x sec x + tan x (sec x tan x + sec x) sec x(tan x + sec x) sec x + tan x sec x

Chapter 6 Techniques of Integration 5 The secon power: This is a basic formula Higher powers: sec x x sec n x x (n 3) sec x x tan x + C What we o here is split off two powers (sec x) an use integration by parts sec n x x sec n x sec x x Let u sec n x v sec x x u (n ) sec n 3 x sec x tan x x v tan x u v uv v u sec n x tan x tan x(n ) sec n 3 x sec x tan x x sec n x tan x (n ) sec n x tan x x sec n x tan x (n ) sec n x(sec x ) x sec n x tan x (n ) sec n x x + (n ) sec n x x We can solve this equation for our unknown integral sec n x x (n ) sec n x x sec n x tan x + (n ) sec n x x sec n x x n secn x tan x + n sec n x x n This is a reuction formula an can be use as the other reuction formulae were use. 6.4 Recap about inverse trigonometric an hyperbolic functions. In S you learne about inverse functions. Here we will review the basic ieas briefly an spell out some etails about inverses of trigonometric an hyperbolic functions. Recall that the inverse of a function y f(x) is a rule that fins what x must be when you know y. That is, the inverse function applie to y, which we write f (y), is the value of x with f(x) y. So f (y) x shoul be equivalent to y f(x). Another way to say it is that the inverse function shoul exactly uno whatever the function oes (thinking of f(x) as the rule f oing something to x).

6 S3 006 07 R. Timoney A snag that arises is that there is often no way to fin x just knowing y. For example, in the simple case y f(x) x, we cannot fin x exactly if we know y 4. The equation f(x) 4 (or x 4) has two solutions x an x. Since a function can only have one value, there is no goo way to efine f (4). What we always o in this example is to efine y (for y 0) to be the positive square root of y, so that 4 (not ). The squre root function is a function that partly unoes what the function f(x) x oes. It is true that ( y) y, but x x is the same as x only when x 0. What we can o is consier a (somewhat artificially) restricte version of the function y f(x) x where we allow only x 0. Also the values in this case are y 0 an we can say we are consiering the function f : [0, ) [0, ) f(x) x where the target set (or co-omain set) on the right is ajuste to be equal to the range. Then we will be able to fin x unambiguously from knowing y f(x). It will be right then that the inverse function f : [0, ) [0, ) f (y) y. The graph of y f (x) is sort of the same as the graph x f (y) but it is not really the same because the axes are swoppe. The graph x f (y) is really ientical with the graph y f(x) but the graph y f (x) is relate to it by taking a reflection in the line y x. Here are the graphs of y x (x 0) an y x an here are the two graphs together an with the line y x, showing that one graph is the refelection of the other in the line y x

Chapter 6 Techniques of Integration 7 From the picture an the fact that the tangent line to the graph y f(x) becomes the tangent line to the graph y f (x) of the inverse function when you reflect in the line y x, it is not har to see that the slope of one tangent line is the reciprocal of the slope of the other. Since the slope is got by m y y x x for two points (x, y ) an (x, y ) on the line, you can see that interchanging x an y (which is what happens wnen you reflect in the line y x) turns the fraction for the slope upsie own. The Liebniz y notation for erivatives also suggests the same fact. If we write y f(x) an x x f (y), then y f (x) an x (f ) (y). There is a theorem that if y f (x) exists x y x an is not zero then the reivative of the inverse function will exists an will be given by (f ) (y) x y ( y ) In the case of the f(x) x (with x 0) we have f (x) x 0 as long as we avoi x 0. So the erivative of the inverse function f (y) y is The only snag with this is that we have got x x (f ) (y) x y ( y ) x y y x an usually we want y y in terms of y, not x. But we have x y f (y) here an so we get the expecte formula y y x y.

8 S3 006 07 R. Timoney Finally, there is one more trick. The y f(x) x an x f (y) y notation is hany for a while, but eventually we woul want to consier the square root function as a thing on its own, not necessarily thinking about how it relates to the function f(x) x. That is, we might look at g(x) x for x 0. We can change y to x throughout an get y y y x x x We now run through the inverse trig functions briefly, using the same approach as what we ha above for the square root example. sin : The function x sin θ certainly oes not have an inverse because sin θ sin(θ + π) sin(π θ) an so knowing only x sin θ we cannot know θ. To get an inverse function, by convention we artificially restrict θ to lie in the range π θ π an consier the function [ f : π, π ] [, ] x f(θ) sin θ The inverse function of this oes make sense an is sometimes calle arcsin x to emphasise that is is not the inverse of the orinary sin θ function, rather a restricte version of sin θ. However we will write is as sin x (another convention that is use at least as often). In summary, sin x only makes sense if x an sin x θ means x sin θ an π θ π To work out the erivative, we have for x sin θ, x θ cos θ an so θ x x sin x cos θ. To get the answer in terms of x we use cos θ + sin θ, hence cos θ sin θ x. Since π θ π we have cos θ 0 an so cos θ x. Thus we get finally x sin x ( < x < ). x Mathematica uses ArcSin[x] for this function.

Chapter 6 Techniques of Integration 9 (We have to rule out x an x to avoi ivision by zero.) The integration version of this is x x sin x + C Here are the graphs of y sin θ ( π/ x π/) an y sin x ( x ) tan x: Since tan(θ + π) tan θ we have a similar problem with fining θ knowing x tan θ an we solve the problem by restrictiing θ to be in the range π/ < θ < π/. That is, we consier the function f : ( π, π ) R x f(θ) tan θ The inverse function of this oes make sense an is sometimes calle arctan x but we will use the alternative tan x. In summary tan x θ means x tan θ an π < θ < π. Here are the graphs of y tan θ ( π/ < x < π/) an y tan x We can work out the erivative x tan x + x

0 S3 006 07 R. Timoney in a similar way to the metho use for the erivative of sin x. integration formula is + x x tan x + C. The corresponing cosh x: Recall that cosh x (e x e x )/. (We introuce it in the appenix to Chapter, in connection with parametric plotting of a hyperbola.) We coul point out that, now we know x ex e x from S, we can say t cosh t e t + e t t et e t sinh t (This iffers by a sign from cos θ sin θ. It is easy to check that sinh t cosh t, θ t which looks the same as the familiar sin θ cos θ.) θ Also cosh( t) cosh t is easy to see an so we cannot hope to fin t knowing only x cosh t. However, if we restrict to t 0 we can solve for t in terms of x as long as x is in the range of cosh t (which means x ). One way to see cosh t always is to notice cosh t sinh t so that cosh t must be true. So cosh x only makes sense if x an t cosh x means cosh t x an t 0. Here are the graphs of y cosh t (t 0) an y cosh x (x ) We can work out the erivative x cosh x x in a similar way to the metho use for the erivative of sin x an tan x. The corresponing integration formula is x x cosh x + C.

Chapter 6 Techniques of Integration A new feature of the function cosh x is that we can write a formula for it using the natural logarithm. (Maybe this is not totally surprising since cosh t is escribe by exponentials.) The iea is that we can solve x cosh t et + e t et + e t for e t as follows: x e t + e t xe t (e t ) + 0 (e t ) xe t + (e t ) xe t + 0 This is a quaratic equation for e t an so e t x ± 4x 4 x ± x Since e t e 0 an x we must have the plus sign e t x + x an this says So we en up with t ln(x + x ) cosh x ln(x + x ) sinh x: The function y sinh t is genuinely invertible (no nee to introuce any restrictions on the omain) an the range is all of R. So t sinh x just means exactly sinh t x Here are the graphs of y sinh t an y sinh x

S3 006 07 R. Timoney We can work out the erivative x sinh x x + in a similar way to the metho use previously. The corresponing integration formula is x + x sinh x + C. We can also solve x sinh t for t in terms of x in a way rather similar to that use for cosh t above an get sinh x ln(x + x + ) 6.5 Inverse trigonometric substitutions. We now consier a class of substitutions which seem quite counter intuitive. They are relate to the erivative formulae for the inverse trigonometric an inverse hyperbolic functions. Here are the erivative formulae from above summarise: x sin x x tan x x cosh x x sinh x x + x x x + The corresponing substitution methos are: integrals involving x, substitute x sin θ. More generally, integrals involving a u (with a constant) try substituting u a sin θ integrals involving try substituting u a tan θ a + u integrals involving u a try substituting u a cosh t integrals involving u + a try substituting u a sinh t Examples: (i) x 4 x

Chapter 6 Techniques of Integration 3 Solution: Accoring the the metho escribe (since 4 x x a u with a an u x) we shoul try substituting u a sin θ that is x sin θ Notice that this looks backwars compare to other substitutions we have one. x is the orignal variable an θ is the new one but we are expressing the ol in terms of the new. Normally we express the new in terms of the ol. Inee, this is what is happening here, but we have not expresse it totally clearly. We can rewrite x sin θ as x sin θ an the way to say what θ is in a way that is not ambiguous, or to say what we really inten, is to have ( x ) θ sin Nevertheless, the way we starte is very convenient for substitution. We have x sin θ x cos θ θ x 4 x cos θ θ 4 ( sin θ) cos θ θ 4 4 sin θ cos θ θ 4( sin θ) cos θ θ 4 cos θ cos θ θ cos θ θ The point of the substitution (an the reason it is often avantageous) is that we get ri of the square root. In this example, everything is very simple, but getting ri of the square root is the essential thing. We get 4 x x θ + C sin (x/) + C using the earlier expression for θ with the inverse function.

4 S3 006 07 R. Timoney (ii) 9 4(x + ) x Solution: In this case, we have a u with a 9, a 3, u (x + ) an so our metho says to try u a sin θ or (x + ) 3 sin θ x 3 cos θ θ x 3 cos θ θ 9 4(x + ) x 9 9 sin θ 3 cos θ θ 9 cos θ 3 cos θ θ 3 cos θ 3 cos θ θ 9 cos θ θ 9 + cos θ θ 4 9 4 (θ + sin θ) + C To get the answer in terms of x, we nee θ in terms of x (x + ) 3 sin θ (x + ) sin θ 3 ( ) θ sin 3 (x + ) an we coul get a correct answer by replacing θ by this everywhere in the answer above. However, there is a tiier way to eal with sin θ using sin θ sin θ cos θ 4 (x + ) cos θ 3 an we can express cos θ in terms of sin θ by looking at a right angle triangle with θ as an angle, (x + ) as the length of the opposite sie an 3 as the length of the hypotenuse. Then Pythagoras theorem gives the ajacent sie as hypoteneuse opposite 9 (x + ) an so cos θ ajacent hypoteneuse 9 4(x + ) 3

Chapter 6 Techniques of Integration 5 We coul also get to this from cos θ sin θ. In any case we en up with a (reasonably tiy) answer 9 4(x + ) x 9 4 θ + 9 sin θ + C 8 9 ( ) 4 sin 3 (x + ) + 9 4 9 4(x + ) (x + ) + C 8 3 3 9 4 sin ( 3 (x + ) ) + (x + ) 9 4(x + ) + C (iii) 4x 6x 7 x Solution: For quaratics insie a square root like this, what we shoul o first is complete the square. That is, rearrange the x an x terms so that (with a suitable constant) we get a multiple of a perfect square 4x 6x 7 4(x + 4x) 7 4(x + 4x + 4 4) 7 4(x + 4x + 4) + 9 9 4(x + ) This means that not only is this problem similar to the previous one, it is in fact the same problem again (now that we complete the square). 6.6 Partial Fractions. Partial fractions are a technique from algebra, but our reason for ealing with them is that they can in principle help fin every integral of the form p(x) q(x) x where p(x) an q(x) are polynomials. Except in a few special cases, we on t yet know how to fin such integrals. One special case, where we on t nee partial fractions, is where q (x) p(x) or q (x) αp(x) for some constant α, because in these cases we can make a substitution u q(x), u q (x) x an it will work out nicely. In fact substitution woul also work if q(x) r(x) n for some n an r (x) αp(x) for a constant α we can substitute u r(x), u r (x) x αp(x) x, p(x) q(x) x p(x) p(x) r(x) x n u n u αp(x) (/α) u n u

6 S3 006 07 R. Timoney The iea of partial fractions is to rewrite p(x) as a sum of fractions with simple enominators q(x) an numerators that are somehow small compare to the enominator. We nee to explain exactly how it goes. To try an get the iea, let us first look at what we can o with orinary numerical fractions. Consier 98. What we can see first is that it is not a proper fraction. 45 ivies into 98 times 45 with a remainer 8. So we can write 98 45 + 8 45 8 45 an the part is not a fraction at all. We concentrate on the proper faction part 8/45. By simple enominator in the case of numerical fractions we mean enominator that is a power of a prime number. An by enominator that is small compare to the enominator, we mean smaller in absolute value than the prime number that occurs in the enominator. For example, in the case of 8/45, the enominator factors 45 5 9 5 3 3 5 3, an the prime powers that are possibly neee in this case are 5, 3 an 3. We shoul be able to write any proper fraction with enominator 45 as a sum a 5 + b 3 + c 3 where a < 5, b < 3 an c < 3. In our example, 8 45 5 + 0 3 + 3 Other examples are 34 45 5 + 3 + 3 an 45 4 5 3 5 + 3 3 5 + 3 + 0 3. (In the last example there is no nee for a enominator 3 because the fraction /45 4/5 in lowest terms.) One can write all proper numerical fractions in this way. You coul maybe experiment with fractions n/0 if you like, but we are not going to nee to o this. It is just here as a sort of motivation for oing something similar for fractions of polynomials p(x)/q(x) We nee to talk about factoring polynomials an about what woul be the corresponing thing to a prime number for polynomials. To start with, a polynomial is an expression you get by taking a finite number of powers of x with constant coefficients in front an aing them up. For example p(x) 4x x + 7 or p(x) 7x + 5x 0 x 9 + x 8 + x + 5

Chapter 6 Techniques of Integration 7 are polynomials. The highest power of x that has a nonzero coefficient in front is calle the egree of the polynomial. The examples above have egree an egree. What is hany to know is that when we multiply polynomials, the egrees a. So (x + )(x + 5)(x + x + ) will have egree + + 4 when it is multiplie out. Constant polynomials have egree 0, except the zero polynomial we are best not giving any egree to the zero polynomial. Now, what kin of polynomial correspons to a prime number? Prime numbers are whole (positive) numbers that cannot be factore except in a very silly way, as themselves times (or minus themselves times ). So the only factorisations of 5 are 5 5 ( 5) ( ). We o not inclue as a prime number. With polynomials, we consier a factorisation to be genuine if the factors each have egree at least. So x + 8x + (x + 4)(x + ) is a genuine factorisation, but x + 4 (x + ) is not consiere genuine, nor is x + (x + /). The factors must contain x to at least the first power. Anything of egree certainly cannot be factore then. Some things of egree can be factore, such as x + 5x + 4 (x + )(x + 4) but other quaratics cannot be factore if we on t allow complex numbers to be use. We cannot factor x + x + (x α)(x β) because if we coul then the roots of x + x + woul be α an β. The roots of x + x + are b ± b 4ac a ± 4 8 ± an these are complex numbers. So α an β woul have to be these complex numbers. A remarkable fact is that every polynomial of egree 3 or more can be factore, at least in theory. It oes not mean it is easy to fin the factors, unfortunately. What you can sometimes rely on to factor polynomials is the Remainer Theorem. It says that if p(x) is a polynomial an you know a root x a (that is a value a so that p(a) 0), then x a must ivie p(x). Using the theory, just as in principle whole numbers can be factore as a prouct of prime numbers, so polynomials p(x) with real coefficients can be factore as a prouct of linear factors like x a an quaratic factors x + bx + c with complex roots. If the coefficient of the highest power of x in p(x) is not, then we also nee to inclue that coefficient. So a complete factorisation of x + 8x + (x + 4)(x + ) (x + )(x + ) (x + ). For 3x 3 + 3x + 6x + 6 3(x 3 + x + x + ), you can check that x is a root an so x ( ) x + must ivie it. We get 3x 3 + 3x + 6x + 6 3(x + )(x + ) from long ivision. Now we can outline how partial fractions works for a fraction p(x) of two polynomials: q(x)

8 S3 006 07 R. Timoney Step 0: (preparatory step). If the egree of the numerator p(x) is not strictly smaller than the egree of the enominator q(x), use long ivision to ivie q(x) into p(x) an obtain a quotient Q(x) an remainer R(x). Then an egree(r(x)) < egree(q(x)). p(x) q(x) Q(x) + R(x) q(x) We concentrate then on the proper fraction part R(x)/q(x). Step : Now factor q(x) completely into a prouct of linear factors x a an quaratic factors x + bx + c with complex roots. Gather up any repeate terms, so that if (say) q(x) (x )(x + )(x + 3)(x + ) we shoul write it as q(x) (x )(x + ) (x + 3). Step : Then the proper fraction R(x) can be written as a sum of fractions of the following q(x) types: (i) (ii) A (x a) m Bx + C (x + bx + c) k where we inclue all possible powers (x a) m an (x + bx + c) k that ivie q(x). The A, B, C stan for constants. As examples, consier the following. We just write own what the partial fractions look like. In each case, we start with a proper fraction where the enominator is completely factore alreay. So some of the har work is alreay one. (i) (ii) (iii) (iv) x + x + 5 (x )(x )(x 3) A x + A x + A 3 x 3 x 3 + x + 7 (x + ) (x 4) A x + + A (x + ) + A 3 x 4 x + x + (x + )(x + x + ) A x + + Bx + C x + x + x 4 + x + (x + )(x + x + ) A x + + B x + C x + x + + B x + C (x + x + ) To make use of these, we have to be able to fin the numbers A, B, C,... that make the equation true. Take the first example x + x + 5 (x )(x )(x 3) A x + A x + A 3 x 3.

Chapter 6 Techniques of Integration 9 To fin the appropriate A, A, A 3, we multiply across by the original enominator (x )(x )(x 3). This has the effect of clearing all the fractions. x + x + 5 A x (x )(x )(x 3) + A (x )(x )(x 3) x + A 3 (x )(x )(x 3) x 3 A (x )(x 3) + A (x )(x 3) + A 3 (x )(x ) There are two avenues to pursue from here. In this case, metho seems easier to me, but in general metho can be as goo. Metho. Plug in the values of x that make the original enominator (x )(x )(x 3) 0. x : + + 5 A ( )( 3) + A (0) + A 3 (0) 7 A A 7/ x : A (0) + A ()( ) + A 3 (0) A A So we get x 3 : 7 0 + 0 + A 3 ()() A 3 7/ x + x + 5 (x )(x )(x 3) 7/ x + x + 7/ x 3. Our interest in this is for integration. We can now easily integrate x + x + 5 (x )(x )(x 3) x 7 Metho. Multiply out the right han sie. 7/ x + x + 7/ x 3 x ln x ln x + 7 ln x 3 + C x + x + 5 A (x )(x 3) + A (x )(x 3) + A 3 (x )(x ) A (x 5x + 6) + A (x 4x + 3) + A 3 (x x + ) (A + A + A 3 )x + ( 5A 4A A 3 )x + (6A + 3A + A )

0 S3 006 07 R. Timoney an compare the coefficients on both sies to get a system of linear equations A + A + A 3 5A 4A A 3 6A + 3A + A 5 These can be solve (by Gaussian elimination, for example) to fin A, A, A 3. Metho is certainly magic in this case, but there are examples where Metho oes not get all the unknown so easily. Another example. Here is one of our previous examples with the numbers worke out. x + x + (x + )(x + x + ) 0x + x + x + x + To fin the integral of this, x + x + (x + )(x + x + ) x x + x ln x + 0x + x + x + x 0x + x + x + x To work out the remaining integral, we use the metho of completing the square x + x + x + x + + (x + ) + an we can then make the substitution x + tan θ. However, there is a trick that saves a lot of simplifying later. The substitution u x + x +, u (x + ) x (x + ) x woul work if the numerator was a multiple of x +. What we can o is write 0x + x + x + x 0x + 0 x + x + x + x + x + x an make the u x + x + substitution in the first half, but the x + tan θ, x sec θ θ in the secon. We get 0x + 0x + 0 x + x + x u u (x + ) + tan θ + sec θ θ 5 u u + θ Finally, our integral works out as 5 ln u + θ + C 5 ln(x + x + ) + tan (x + ) + C x + x + (x + )(x + x + ) x ln x + 5 ln(x + x + ) tan (x + ) + C

Chapter 6 Techniques of Integration 6.7 Improper Integrals. Sometimes, integrals that appear to be infinite in extent can be given a finite value in a way that seems sensible. For example, consier x x If we raw a picture of what this might mean graphically, in the same way as we it for integrals b f(x) x where a an b are finite, we shoul be looking at the area of the region uner the a graph y x x a region that stretches infinitely far into the istance. So it seems infinite an nothing more to be sai. But, before we conclue that it is infinite, suppose we imagine colouring in the area uner that graph with paint, an we o it so that we apply the paint evenly so that we use a fixe amount per square inch. The amount of paint we woul nee shoul be infinite if the area is infinite. We woul never be one painting an infinite area, an so we coul paint a wie but finite section an see how much paint we nee.

S3 006 07 R. Timoney In this picture, we show colouring for x 3, but we coul replace 3 by any finite b >. The area covere up as far as b works out as b [ x x ] b x b ( ) b an we see that, rather than a huge answer, we always get an answer <. In fact as b, the answer approaches. It oes not ten to. So if we have enough paint to paint square unit of area, we will never completely run out although there will be very little paint left when b is large. Maybe there is a case for eciing that /x x shoul have the value. We make this our efinition. We efine x lim x b b x x (which is ). More generally, if y f(x) is efine an continuous for x a we efine the improper integral a f(x) x lim b b f(x) x if this limit exists an is finite. On the other han if the limit oes not exist at all, or is infinite, we say that the improper integral f(x) x fails to converge. a Example. Fin /x x. In this kin of example (an improper integral) we start by using the right efinition. This shows that we realise that there is an issue about the integral making sense, an that we know how the issue is ealt with. We get /x x lim b b /x x lim b [ln x] b lim b ln b ln. If we look at the graph of y ln x we see that this limit is. Because it is not finite, we say that the improper integral x x oes not converge. Other types of improper integral (i) If y f(x) is efine an continuous for x b, then we efine the improper integral b f(x) x b lim a a f(x) x if this limit exists an is finite. If the limit oes not exist, or is infinite, we say that the improper integral b f(x) x oes not converge.

Chapter 6 Techniques of Integration 3 (ii) If y f(x) is continuous on a finite interval a < x b, excluing the left en point a (where it might have an asymptote or other ba behaviour), hen we efine the improper integral b a f(x) x lim c a + b c f(x) x if this limit exists an is finite. If the limit oes not exist, or is infinite, we say that the improper integral oes not converge. Example. Consier 0 /x x. Using the efinition 0 /x x lim c 0 + lim c 0 + c [ x /x x ] lim c 0 + + c As this limit is infinite, the improper integral 0 /x x oes not converge. (iii) If y f(x) is continuous on a finite interval a x < b, excluing the right en point b (where it might have an asymptote or other ba behaviour), hen we efine the improper integral b a f(x) x lim c b c a c f(x) x if this limit exists an is finite. If the limit oes not exist, or is infinite, we say that the improper integral oes not converge. (iv) If an integral is improper for more than one reason, or at a place not one of the enpoints, we ivie it as a sum of improper integrals of the types we have consiere above. If each one of the bits has a finite value, then the value of the whole is the sum. But if any one of the bits fails to converge then the whole is sai not to converge. Examples (a) x 0 x + x +x +x 0 +x (It oes not matter about using 0 as a stopping point. Any finite point woul o, for example x 4 +x x + +x 4 x) +x (b) x 0 x + x x x 0 x Here the problem with the integral is at 0, where the integran has an asymptote. In fact we alreay worke out that x also oes not converge. x 0 /x x oes not converge an so we know that

4 S3 006 07 R. Timoney This is an example where an unthinking use of integration woul prouce a wrong answer. (You might be suspicious if the integral of a positive thing in the left to right irection turne out to be negative.) (c) x x(+x ) x + 0 x(+x ) x + x(+x ) 0 x + x(+x ) x x(+x ) The original is improper because of the limit, the asymptotes at x 0 an the limit. Each of the 4 bits above has just one problem at one en. A Appenix In 006 7, this material was in S. A. Remark. The aim here is to explain how to fin antierivatives (inefinite integrals). By the funamental theorem, this also allows us to fin efinite integrals exactly. A. Remark. (i) Substitution. If we integrate both sies of the chain rule y x y u where y f(u), u u(x), y f(u(x)) u x we get y y x x u x () u x an this means that if we we have an integran we can express like the right han sie as g(u) u x x then we can work it out if we have an antierivative for g(u), that is if we can fin f(u) with f (u) g(u) it will match exactly what we ha in (). It comes own to being able to cancel the x s an g(u) u x x g(u) u Example. Consier the problem of fining 3x cos(x 3 ) x If we notice that u x 3 has u x 3x a factor in the integran, then we can write u 3x cos(x 3 ) x cos u x cos u u x x x cos u u This is something we can o as it is sin u + C. So the answer is sin u + C but we shoul write the answer in terms of x (because the problem was in terms of x) an so we get 3x cos(x 3 ) x sin u + C sin(x 3 ) + C

Chapter 6 Techniques of Integration 5 Normally we write this out in a way like the following an rely on the possibility of treating x like a numerical quantity that can be cancele between numerator an enominator. 3x cos(x 3 ) x Let u x 3 3x cos(x 3 ) x u x 3x u 3x x u 3x x 3x cos u u 3x cos u u sin u + C sin(x 3 ) + C The reason the substitution u x 3 works is because the erivative u/x is a factor in the integran. For a substitution to work, it is usually necessary for u/x to be a factor in the integran (essentially, maybe in a slightly hien way). It is vital that we express all x s an x s in terms of u an u. (ii) Integration by parts If we integrate both sies of the prouct rule u (uv) x x v + uv x we get (uv) x v u x x x + u v x x or uv v u x x + u v x x This allows us a way of transforming integrals of the form of a prouct of one function times the erivative of another u v x uv v u x x x into a ifferent integral (where the ifferentiation has flippe from one factor to the other). The avantage of this comes if we know how to manage the new integral (or at least if it is simpler than the original). The integration by parts formula is usually written with the s s cancele u v uv v u Examples

6 S3 006 07 R. Timoney (a) x ln x x Solution: The two most obvious ways to use integration by parts are u x, v ln x x (Problem with this is we can t fin v) u ln x, v x x It turns out that the secon is goo. x ln x x Let u ln x v x x u x x v x x ln x x u v uv v u (ln x) x x x x x (x /) ln x x (x /) ln x x 4 + C (b) e ln x x Solution: This is one of a very few cases which can be one by taking v x an u the integran. Every integral takes the form u v in that way, but it is rarely a goo way to start integration by parts. Here we can make use of the efinite integral form of the integration by parts formula b u v [uv] b a b a a v u which arises in the same way as the inefinite integral formula (take efinite integrals of the prouct rule for ifferentiation).

Chapter 6 Techniques of Integration 7 e e ln x x Let u ln x v x u x x v x ln x x e u v [uv] e e [(ln x)x] e v u e e ln e ln e e [x] e e (e ) x x x x (c) x cos x x Solution: x cos x x Let u x v cos x x u x x v sin x x cos x x u v uv v u x sin x sin x(x) x x sin x x sin x x The point here is that we have succeee in simplifying the problem. We starte with x times a trigonometric function (cos x) an we have now got to x times a trigonometric function (sin x this time, but that is not so ifferent in ifficulty to cos x). If we continue in the same (or similar) way an apply integration by parts again, we can make the problem even simpler. We use U an V this time in case we might get confuse with

8 S3 006 07 R. Timoney the earlier u an v. x sin x x Let U x V sin x x U x V cos x x sin x x U V UV V U x( cos x) ( cos x) x x cos x + cos x x x cos x + sin x + C Combining with the first stage of the calculation x cos x x x sin x + x cos x sin x C an, in fact C is plus another constant. Since C can be any constant, the answer x cos x x x sin x + x cos x sin x + C is also goo. We will not in fact learn any other techniques than these which are purely integration methos. We will spen some time explaining how to make use of these techniques in specific circumstances (as it is often not at all obvious how to o so). There is one other metho we will come to calle partial fractions, a metho for integrating fractions such as x + (x )(x + x + ) x However, the thing we have to learn about is algebra a way to rewrite fractions like this as sums of simpler ones an there is no new iea that is irectly integration. The algebra allows us to tackle problems of this sort. A.3 Trigonometric Integrals. (i) Powers of sin x times powers of sin x with one power o Metho: For sin n x cos m x x if n the power of sin x is o, substitute u cos x One thing to avoi is U v an V u because this will just unravel what we i to begin with.

Chapter 6 Techniques of Integration 9 if m the power of cos x is o, substitute u sin x Example: sin 3 x cos 4 x x Solution: Let u cos x, u sin x x, x sin 3 x cos 4 x x u sin x sin 3 xu 4 u sin x sin xu 4 u ( cos x)u 4 u ( u )u 4 u u 6 u 4 u 7 u7 5 u5 + C 7 cos7 x 5 cos5 x + C (ii) Powers of sin x times powers of sin x with both powers even Metho: use the trigonometric ientities sin x ( cos x), cos x ( + cos x) Example: Solution: sin 4 x cos x x sin 4 x cos x x (sin x) cos x x ( ) ( ) ( cos x) ( + cos x) x ( cos x + cos x)( + cos x) x 8 cos x cos x + cos 3 x x 8 Now x is no bother. cos x x is not much harer than cos x x sin x + C. If we look at sin x (cos x) x

30 S3 006 07 R. Timoney we can see that cos x x sin x+c. (This can also be one by a substitution u x but that is harly neee.) Next cos x x (using the same ieas as for cos x x). ( + cos 4x) x (x + 4 ) sin 4x + C For cos 3 x x we are in a situation where we have an o power of cos times a zeroth power of sin. So we can use the earlier metho (the fact that the angle is x oe snot make a big ifference) of substituting u sin x. Then u cos x x, x u cos x, cos 3 x x cos 3 u x cos x cos x u sin x x u u (u 3 ) u3 + C sin x 6 sin3 x + C Putting all the bits together sin 4 x cos x x cos x cos x + cos 3 x x 8 ( x 8 sin x x 8 sin 4x + sin x ) 6 sin3 x 6 x 64 sin 4x 48 sin3 x + C + C (iii) Powers of sin x an cos x Metho: Use the previous two methos treating sin n x x sin n x(cos x) 0 x an similarly for cos m x x (that is treat the secon power as the zeroth power). Example:

Chapter 6 Techniques of Integration 3 cos 3 x x Solution: cos 3 x x (sin x) 0 cos 3 x x. Power of cos o. Substitute u sin x, u cos x x, x u cos x cos 3 x x cos 3 x u cos x cos x u u u u 3 u3 + C sin x 3 sin3 x + C sin 4 x x Solution: Use even powers metho. sin 4 x x (sin x) x ( ) ( cos x) x cos x + cos x x 4 Note: still have one even power cos x + ( + cos 4x) x 4 3 4 cos x + cos 4x x ( 3 4 x sin x + ) sin 4x + C 8 3 8 x 4 sin x + sin 4x + C 3 There is another way to o these problems, an it is given in the book by Anton. Possibly also some of you know it alreay. Alternative metho: Use the reuction formulae sin n x x n sinn x cos x + n n cos m x x m cosm x sin x + m m sin n x x cos m x x

3 S3 006 07 R. Timoney (vali for n an m ) to express the integrals of a power in terms of integrals where the power is reuce by. Using the formula enough times, will get own to power or zero eventually. Example: sin 4 x x 4 sin3 x cos x + 3 sin x x 4 (using the case n 4) 4 sin3 x cos x + 3 ( 4 sin x cos x + ) (sin x) 0 x (using the case n ) 4 sin3 x cos x 3 8 sin x cos x 3 8 x 4 sin3 x cos x 3 8 sin x cos x 3 8 x + C Note that the answer looks ifferent to the answer we got by the other metho. They can be reconcile using trigonometry: sin x sin x cos x sin 4x sin x cos x ( sin x cos x)( sin x) 4 sin x cos x 8 sin 3 x cos x 3 8 x 4 sin x + 3 sin 4x 3 8 x sin x cos x + 8 sin x cos x 4 sin3 x cos x 3 8 x 3 8 sin x cos x 4 sin3 x cos x

Chapter 6 Techniques of Integration 33 The way to establish the reuction formulae is to use integration by parts. For example sin n x x sin n x sin x x u v where u sin n x v sin x x u (n ) sin n x cos x x v cos x uv v u sin n x( cos x) ( cos x)(n ) sin n x cos x x sin n x cos x + (n ) sin n x cos x x sin n x cos x + (n ) sin n x( cos x) x sin n x cos x + (n ) sin n x x (n ) sin n x x Though this looks ba, we have in fact foun an equation that is satisfie by sin n x x an if we solve the equation for this quantity in terms of the other things, we get n sin n x x sin n x cos x + (n ) sin n x x sin n x x n sinn x cos x + n sin n x x n an that is the reuction formula as claime. The one for powers of cos x works out in a similar way. Richar M. Timoney April 5, 007