Techniques of Integration

Transcription

1 0 Techniques of Integration ½¼º½ ÈÓÛ Ö Ó Ò Ò Ó Ò Functions consisting of products of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. Some examples will suffice to explain the approach. EXAMPLE 0. Evaluate sin 5 xdx. Rewrite the function: sin 5 xdx = sinxsin 4 xdx = sinx(sin x) dx = sinx( cos x) dx. Now use u = cosx, du = sinxdx: sinx( cos x) dx = = ( u ) du ( u +u 4 )du = u+ u 5 u5 +C = cosx+ cos x 5 cos5 x+c. 89

2 90 Chapter 0 Techniques of Integration EXAMPLE 0. Evaluate sin 6 xdx. Use sin x = ( cos(x))/ to rewrite the function: sin 6 xdx = ( cosx) (sin x) dx = dx 8 = cosx+cos x cos xdx. 8 Now we have four integrals to evaluate: and dx = x cosxdx = sinx are easy. The cos x integral is like the previous example: cos xdx = cosxcos xdx = = = = cosx( sin x)dx ( u )du ) (u u ( ) sinx sin x. And finally we use another trigonometric identity, cos x = (+cos(x))/: +cos4x cos xdx = dx = ( x+ sin4x ). 4 So at long last we get sin 6 xdx = x 8 6 sinx ( ) sinx sin x + ( x+ sin4x ) +C EXAMPLE 0. Evaluate sin xcos xdx. Usetheformulassin x = ( cos(x))/ and cos x = (+cos(x))/ to get: cos(x) sin xcos xdx = +cos(x) dx. The remainder is left as an exercise.

3 0. Trigonometric Substitutions 9 Exercises 0.. Find the antiderivatives.. sin xdx.. sin 4 xdx cos xdx cos xsin xdx sec xcsc xdx 0. sin xdx cos xsin xdx sin xcos xdx sinx(cosx) / dx tan xsecxdx ½¼º¾ ÌÖ ÓÒÓÑ ØÖ ËÙ Ø ØÙØ ÓÒ So far we have seen that it sometimes helps to replace a subexpression of a function by a single variable. Occasionally it can help to replace the original variable by something more complicated. This seems like a reverse substitution, but it is really no different in principle than ordinary substitution. x EXAMPLE 0.4 Evaluate dx. Let x = sinu so dx = cosudu. Then x cos dx = sin ucosudu = ucosudu. We would like to replace cos u by cosu, but this is valid only if cosu is positive, since cos u is positive. Consider again the substitution x = sinu. We could just as well think of this as u = arcsinx. If we do, then by the definition of the arcsine, π/ u π/, so cosu 0. Then we continue: cos ucosudu = +cosu cos udu = = arcsinx + sin(arcsinx) 4 du = u + sinu 4 +C. This is a perfectly good answer, though the term sin(arcsinx) is a bit unpleasant. It is possible to simplify this. Using the identity sinx = sinxcosx, we can write sinu = sinucosu = sin(arcsinx) sin u = x sin (arcsinx) = x x. Then the full antiderivative is arcsin x + x x 4 = arcsinx + x x +C. +C

4 9 Chapter 0 Techniques of Integration This type of substitution is usually indicated when the function you wish to integrate contains a polynomial expression that might allow you to use the fundamental identity sin x+cos x = in one of three forms: cos x = sin x sec x = +tan x tan x = sec x. If your function contains x, as in the example above, try x = sinu; if it contains +x try x = tanu; and if it contains x, try x = secu. Sometimes you will need to try something a bit different to handle constants other than one. 4 9x EXAMPLE 0.5 Evaluate dx. We start by rewriting this so that it looks more like the previous example: 4 9x 4( (x/) dx = )dx = (x/) dx. Now let x/ = sinu so (/)dx = cosudu or dx = (/)cosudu. Then (x/) dx = sin u(/)cosudu = 4 cos udu = 4u 6 + 4sinu = arcsin(x/) = arcsin(x/) = arcsin(x/) +C + sinucosu +C + sin(arcsin(x/))cos(arcsin(x/)) + (x/) (x/) = arcsin(x/) + x 4 9x +C, using some of the work from example C +C +x EXAMPLE 0.6 Evaluate dx. Let x = tanu, dx = sec udu, so +x sec dx = +tan usec udu = usec udu. Since u = arctan(x), π/ u π/ and secu 0, so sec u = secu. Then sec usec udu = sec udu. In problems of this type, two integrals come up frequently: Both have relatively nice expressions but they are a bit tricky to discover. sec udu and secudu.

5 0. Trigonometric Substitutions 9 First we do secudu, which we will need to compute sec udu: secudu = secu secu+tanu secu+tanu du sec u+secutanu = du. secu+tanu Now let w = secu + tanu, dw = secutanu + sec udu, exactly the numerator of the function we are integrating. Thus sec u+secutanu secudu = du = dw = ln w +C secu+tanu w Now for sec udu: sec u = sec u = sec u + sec u = sec u + secutan u + secu = ln secu+tanu +C. + (tan u+)secu = sec u+secutan u + secu. We already know how to integrate secu, so we just need the first quotient. This is simply a matter of recognizing the product rule in action: sec u+secutan udu = secutanu. So putting these together we get sec udu = secutanu + ln secu+tanu +C, and reverting to the original variable x: +x dx = secutanu + ln secu+tanu = sec(arctanx)tan(arctanx) +C + ln sec(arctanx)+tan(arctanx) = x +x + ln +x +x +C, using tan(arctanx) = x and sec(arctanx) = +tan (arctanx) = +x. +C

6 94 Chapter 0 Techniques of Integration Exercises 0.. Find the antiderivatives.. cscxdx. csc xdx x dx 4. x x dx 6. +x dx dx 0. x (+x ) x dx. x 9+4x dx x x dx 8. x +xdx x 4 x dx x 4x dx ½¼º ÁÒØ Ö Ø ÓÒ Ý È ÖØ We have already seen that recognizing the product rule can be useful, when we noticed that sec u+secutan udu = secutanu. As with substitution, we do not have to rely on insight or cleverness to discover such antiderivatives; there is a technique that will often help to uncover the product rule. Start with the product rule: d dx f(x)g(x) = f (x)g(x)+f(x)g (x). We can rewrite this as f(x)g(x) = f (x)g(x)dx+ f(x)g (x)dx, and then f(x)g (x)dx = f(x)g(x) f (x)g(x)dx. This may not seem particularly useful at first glance, but it turns out that in many cases we have an integral of the form f(x)g (x)dx but that f (x)g(x)dx is easier. This technique for turning one integral into another is called integration by parts, and is usually written in more compact form. If we let u = f(x) and v = g(x) then

7 0. Integration by Parts 95 du = f (x)dx and dv = g (x)dx and udv = uv vdu. To use this technique we need to identify likely candidates for u = f(x) and dv = g (x)dx. EXAMPLE 0.7 Evaluate xlnxdx. Let u = lnx so du = /xdx. Then we must let dv = xdx so v = x / and xlnxdx = x lnx x x dx = x lnx x dx = x lnx x 4 +C. EXAMPLE 0.8 Evaluate dv = sinxdx so v = cosx and xsinxdx = xcosx cosxdx = xcosx+ EXAMPLE 0.9 Evaluate xsinxdx. Let u = x so du = dx. Then we must let cosxdx = xcosx+sinx+c. sec xdx. Of course we already know the answer to this, but we needed to be clever to discover it. Here we ll use the new technique to discover the antiderivative. Let u = secx and dv = sec xdx. Then du = secxtanx and v = tanx and sec xdx = secxtanx tan xsecxdx = secxtanx (sec x )secxdx = secxtanx At first this looks useless we re right back to sec xdx+ sec xdx = secxtanx sec xdx = secxtanx+ sec xdx = secxtanx+ sec xdx = secxtanx = secxtanx sec xdx+ secxdx. sec xdx. But looking more closely: + sec xdx+ secxdx secxdx secxdx + ln secx+tanx secxdx +C.

8 96 Chapter 0 Techniques of Integration EXAMPLE 0.0 Evaluate x sinxdx. Let u = x, dv = sinxdx; then du = xdx and v = cosx. Now x sinxdx = x cosx + xcosxdx. This is better than the original integral, but we need to do integration by parts again. Let u = x, dv = cosxdx; then du = and v = sinx, and x sinxdx = x cosx+ xcosxdx = x cosx+xsinx sinxdx = x cosx+xsinx+cosx+c. Suchrepeateduseofintegrationbypartsisfairlycommon, butitcanbeabittediousto accomplish, and it is easy to make errors, especially sign errors involving the subtraction in the formula. There is a nice tabular method to accomplish the calculation that minimizes the chance for error and speeds up the whole process. We illustrate with the previous example. Here is the table: sign u dv x sinx x cosx sinx 0 cos x or u dv x x sinx cosx sinx 0 cosx To form the first table, we start with u at the top of the second column and repeatedly compute the derivative; starting with dv at the top of the third column, we repeatedly compute the antiderivative. In the first column, we place a in every second row. To form the second table we combine the first and second columns by ignoring the boundary; if you do this by hand, you may simply start with two columns and add a to every second row. To compute with this second table we begin at the top. Multiply the first entry in column u by the second entry in column dv to get x cosx, and add this to the integral of the product of the second entry in column u and second entry in column dv. This gives: x cosx+ xcosxdx, or exactly the result of the first application of integration by parts. Since this integral is not yet easy, we return to the table. Now we multiply twice on the diagonal, (x )( cosx)

9 0.4 Rational Functions 97 and ( x)( sinx) and then once straight across, ()( sinx), and combine these as x cosx+xsinx sinxdx, giving the same result as the second application of integration by parts. While this integral iseasy, wemayreturnyetoncemoretothetable. Nowmultiplythreetimesonthediagonal to get (x )( cosx), ( x)( sinx), and ()(cosx), and once straight across, (0)(cosx). We combine these as before to get x cosx+xsinx+cosx+ 0dx = x cosx+xsinx+cosx+c. Typically we would fill in the table one line at a time, until the straight across multiplication gives an easy integral. If we can see that the u column will eventually become zero, we can instead fill in the whole table; computing the products as indicated will then give the entire integral, including the +C, as above. Exercises 0.. Find the antiderivatives.. xcosxdx.. xe x dx sin xdx xarctanxdx xsin xdx 0.. arctan( x)dx.. sec xcsc xdx x cosxdx xe x dx lnxdx x sinxdx xsinxcosxdx sin( x)dx ½¼º Ê Ø ÓÒ Ð ÙÒØ ÓÒ A rational function is a fraction with polynomials in the numerator and denominator. For example, x x +x 6, x + (x ), x, are all rational functions of x. There is a general technique called partial fractions that, in principle, allows us to integrate any rational function. The algebraic steps in the

10 98 Chapter 0 Techniques of Integration technique are rather cumbersome if the polynomial in the denominator has degree more than, and the technique requires that we factor the denominator, something that is not always possible. However, in practice one does not often run across rational functions with high degree polynomials in the denominator for which one has to find the antiderivative function. So we shall explain how to find the antiderivative of a rational function only when the denominator is a quadratic polynomial ax +bx+c. We should mention a special type of rational function that we already know how to integrate: If the denominator has the form (ax + b) n, the substitution u = ax + b will always work. The denominator becomes u n, and each x in the numerator is replaced by (u b)/a, and dx = du/a. While it may be tedious to complete the integration if the numerator has high degree, it is merely a matter of algebra. EXAMPLE 0. Find x dx. Using the substitution u = x we get ( x) 5 x ( x) dx = 5 = 6 = 6 = 6 ( ) u u 5 du = 6 u 9u +7u 7 du u 5 u 9u +7u 4 7u 5 du ( u 9u + 7u ( ( x) 9( x) ) 7u 4 +C 4 + 7( x) ) 7( x) 4 +C 4 = 6( x) + 9 ( x) 9 6( x) ( x) +C 4 We now proceed to the case in which the denominator is a quadratic polynomial. We can always factor out the coefficient of x and put it outside the integral, so we can assume that the denominator has the form x +bx+c. There are three possible cases, depending on how the quadratic factors: either x +bx+c = (x r)(x s), x +bx+c = (x r), or it doesn t factor. We can use the quadratic formula to decide which of these we have, and to factor the quadratic if it is possible. EXAMPLE 0. Determine whether x +x+ factors, and factor it if possible. The quadratic formula tells us that x +x+ = 0 when x = ± 4. Since there is no square root of, this quadratic does not factor.

11 0.4 Rational Functions 99 EXAMPLE 0. Determine whether x x factors, and factor it if possible. The quadratic formula tells us that x x = 0 when x = ± +4 = ± 5. Therefore x x = ( x + 5 )( x 5 ). If x +bx+c = (x r) then we have the special case we have already seen, that can be handled with a substitution. The other two cases require different approaches. If x +bx+c = (x r)(x s), we have an integral of the form p(x) (x r)(x s) dx where p(x) is a polynomial. The first step is to make sure that p(x) has degree less than. x EXAMPLE 0.4 Rewrite dx in terms of an integral with a numerator that has degree less than. To do this we use long division of polynomials to discover (x )(x+) that so x (x )(x+) = x x +x 6 x (x )(x+) dx = = x + 7x 6 x +x 6 = x + x dx+ The first integral is easy, so only the second requires some work. Now consider the following simple algebra of fractions: A x r + B x s = A(x s)+b(x r) (x r)(x s) 7x 6 (x )(x+) dx. 7x 6 (x )(x+), = (A+B)x As Br. (x r)(x s) That is, adding two fractions with constant numerator and denominators (x r) and (x s) produces a fraction with denominator (x r)(x s) and a polynomial of degree less than for the numerator. We want to reverse this process: starting with a single fraction, we want to write it as a sum of two simpler fractions. An example should make it clear how to proceed.

12 00 Chapter 0 Techniques of Integration x 7x 6 EXAMPLE 0.5 Evaluate dx. We start by writing (x )(x+) as the sum of two fractions. We want to end up with 7x 6 (x )(x+) = A x + B x+. If we go ahead and add the fractions on the right hand side we get 7x 6 (x )(x+) = (A+B)x+A B. (x )(x+) (x )(x+) So all we need to do is find A and B so that 7x 6 = (A+B)x+A B, which is to say, we need 7 = A+B and 6 = A B. This is a problem you ve seen before: solve a system of two equations in two unknowns. There are many ways to proceed; here s one: If 7 = A+B then B = 7 A and so 6 = A B = A (7 A) = A 4+A = 5A 4. This is easy to solve for A: A = 8/5, and then B = 7 A = 7 8/5 = 7/5. Thus 7x 6 8 (x )(x+) dx = 5x x+ dx = 8 7 ln x ln x+ +C. The answer to the original problem is now x (x )(x+) dx = x dx+ 7x 6 (x )(x+) dx = x x+ 8 7 ln x ln x+ +C. Now suppose that x +bx+c doesn t factor. Again we can use long division to ensure that the numerator has degree less than. Now we can complete the square to turn the integral into a trigonometric substitution problem. x EXAMPLE 0.6 Evaluate dx. We have seen that this quadratic does x +x+ not factor. We complete the square: x +x+ = x +x = x +x+ 4 + ( 4 = x+ ) + 4. Now factor out /4: 4 ( ( ) ( 4 x+ ( + = ) x+ ) ) +. 4 Now let tanu = x+

13 0.4 Rational Functions 0 sec udu = dx dx = sec udu. Now we can substitute in the original integral, using x = ( /)(tanu / ) in the numerator. x x +x+ dx = 4 ( /)(tanu / ) tan u+ = 4 ( /)(tanu / ) sec u ( tanu ) du = 4 = tanu du = ln cosu u +C. = ln secu u +C. sec udu sec udu Finally, we can substitute and to get secu = u = arctan( x+ ) tan u+ = ( x+ ) + ( ln x+ ) + arctan( x+ ) +C. The details here are admittedly a bit unpleasant, but the whole process is fairly mechanical and easy in principle. Exercises 0.4. Find the antiderivatives.. dx. 4 x x 4 4 x dx

14 0 Chapter 0 Techniques of Integration dx 4. x +0x+5 x 4 dx 6. 4+x x dx 8. 4+x dx 0. x x x 4 x dx x +0x+9 dx x +0x+ dx x +x dx ½¼º Ø ÓÒ Ð Ü Ö These problems require the techniques of this chapter, and are in no particular order. Some problems may be done in more than one way (t+4) dt. (e t +6)te t dt 4. tantsec tdt 6. dt 8. t(t 4) cost dt 0. sint e t et + dt. dt 4. t +t sec t dt 6. (+tant) e t sintdt 8. t dt 0. ( t ) 5/ arctant dt. +4t sin tcos 4 tdt 4. dt 6. t(lnt) t e t dt 8. t(t 9) / dt sintcostdt t+ t +t+ dt dt (5 t ) / tsec tdt cos 4 tdt t +t dt t t +dt (t / +47) tdt t(9+4t ) dt t t +t dt t 6t+9 dt t(lnt) dt t+ t +t dt