Textbook Formulas Summary

Textbook Formulas Summary - 392 |||| CHAPTER 5 INTEGRALS...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 392 |||| CHAPTER 5 INTEGRALS TABLE OF INDEFINITE INTEGRALS 1 y cf c y f x dx x dx y k dx kx n dx xn 1 n1 ye x dx ex 2 1 2 1 y sinh x d x y 1 1 dx x ya tan x dx n C y sec x tan x d x yx C cos x y sec x d x fx yf t x dx x dx ytx dx C yx y sin x d x y C sin x y csc x d x 2 C sec x cosh x ax ln a dx C y cos x d x C tan 1x x ln x C cot x y csc x cot x d x C y C 1 s1 x2 csc x dx y cosh x d x C C sin 1x sinh x C C C Recall from Theorem 4.9.1 that the most general antiderivative on a given interval is obtained by adding a constant to a particular antiderivative. We adopt the convention that when a formula for a general indefinite integral is given, it is valid only on an interval. Thus we write 1 1 C y x 2 dx x with the understanding that it is valid on the interval 0, or on the interval , 0 . This is true despite the fact that the general antiderivative of the function f x 1 x 2, x 0, is 1 x 1 x Fx The indefinite integral in Example 1 is graphed in Figure 1 for several values of C. The value of C is the y-intercept. C1 if x 0 C2 if x 0 N EXAMPLE 1 Find the general indefinite integral y 4 10 x 4 2 sec 2x d x S O L U T I O N Using our convention and Table 1, we have _1.5 1.5 _4 FIGURE 1 y 10 x 4 2 sec2x d x 10 y x 4 d x 10 x5 5 2 y sec2x d x 2 tan x You should check this answer by differentiating it. C 2x 5 2 tan x C M 392 |||| CHAPTER 5 INTEGRALS TABLE OF INDEFINITE INTEGRALS 1 y cf c y f x dx x dx y k dx kx n dx xn 1 n1 ye x dx ex 2 1 2 1 y sinh x d x y 1 1 dx x ya tan x dx n C y sec x tan x d x yx C cos x y sec x d x fx yf t x dx x dx ytx dx C yx y sin x d x y C sin x y csc x d x 2 C sec x cosh x ax ln a dx C y cos x d x C tan 1x x ln x C cot x y csc x cot x d x C y C 1 s1 x2 csc x dx y cosh x d x C C sin 1x sinh x C C C Recall from Theorem 4.9.1 that the most general antiderivative on a given interval is obtained by adding a constant to a particular antiderivative. We adopt the convention that when a formula for a general indefinite integral is given, it is valid only on an interval. Thus we write 1 1 C y x 2 dx x with the understanding that it is valid on the interval 0, or on the interval , 0 . This is true despite the fact that the general antiderivative of the function f x 1 x 2, x 0, is 1 x 1 x Fx The indefinite integral in Example 1 is graphed in Figure 1 for several values of C. The value of C is the y-intercept. C1 if x 0 C2 if x 0 N EXAMPLE 1 Find the general indefinite integral y 4 10 x 4 2 sec 2x d x S O L U T I O N Using our convention and Table 1, we have _1.5 1.5 _4 FIGURE 1 y 10 x 4 2 sec2x d x 10 y x 4 d x 10 x5 5 2 y sec2x d x 2 tan x You should check this answer by differentiating it. C 2x 5 2 tan x C M S E C T I O N 5 . 5 T H E S U B S T I T U T I O N RU L E |||| 401 because, by the Chain Rule, d Ftx dx F tx t x t x , then from Equation 3 If we make the “change of variable” or “substitution” u we have yF or, writing F t x t x dx Ftx C Fu C yF u du f , we get yf yf t x t x dx u du Thus we have proved the following rule. 4 THE SUBSTITUTION RULE If u t x is a differentiable function whose range is an interval I and f is continuous on I , then yf yf t x t x dx u du Notice that the Substitution Rule for integration was proved using the Chain Rule for differentiation. Notice also that if u t x , then du t x d x, so a way to remember the Substitution Rule is to think of d x and du in (4) as differentials. Thus the Substitution Rule says: It is permissible to operate with d x and du after integral signs as if they were differentials. EXAMPLE 1 Find yx 3 cos x 4 2 d x. x 4 2 because its differential is du 4 x 3 d x, which, apart from the constant factor 4, occurs in the integral. Thus, using x 3 d x du 4 and the Substitution Rule, we have S O L U T I O N We make the substitution u yx cos x 4 2 dx y cos u 1 4 1 4 3 C 1 4 N Check the answer by differentiating it. sin u sin x 4 du 2 1 4 y cos u du C Notice that at the final stage we had to return to the original variable x. M The idea behind the Substitution Rule is to replace a relatively complicated integral by a simpler integral. This is accomplished by changing from the original variable x to a new variable u that is a function of x. Thus, in Example 1, we replaced the integral x x 3 cos x 4 2 d x by the simpler integral 1 x cos u du. 4 The main challenge in using the Substitution Rule is to think of an appropriate substitution. You should try to choose u to be some function in the integrand whose differential also occurs (except for a constant factor). This was the case in Example 1. If that is not 404 |||| CHAPTER 5 INTEGRALS This rule says that when using a substitution in a definite integral, we must put everything in terms of the new variable u, not only x and d x but also the limits of integration. The new limits of integration are the values of u that correspond to x a and x b. THE SUBSTITUTION RULE FOR DEFINITE INTEGRALS If t is continuous on 6 N t x , then a, b and f is continuous on the range of u y b a y f t x t x dx tb ta f u du P R O O F Let F be an antiderivative of f . Then, by (3), F t x is an antiderivative of f t x t x , so by Part 2 of the Fundamental Theorem, we have y b a f t x t x dx Ftx b a F tb F ta But, applying FTC2 a second time, we also have y tb ta EXAMPLE 7 Evaluate y 4 0 f u du s2 x Fu tb ta F tb F ta M 1 d x using (6). S O L U T I O N Using the substitution from Solution 1 of Example 2, we have u dx when x 0, u 20 y Therefore The geometric interpretation of Example 7 is shown in Figure 2. The substitution u 2 x 1 stretches the interval 0, 4 by a factor of 2 and translates it to the right by 1 unit. The Substitution Rule shows that the two areas are equal. 4 0 1 s2 x 1 1 dx 2 1 su du 93 2 13 2 1 2 2 3 24 1 9 u3 2 9 1 26 3 M œ„ u 2 1 0 4 The integral given in Example 8 is an abbreviation for N 1 3 91 2 1 4, u 3 y=œ„„„„„ 2x+1 2 2 y when x y 3 1 1 and Observe that when using (6) we do not return to the variable x after integrating. We simply evaluate the expression in u between the appropriate values of u. y y and 1 3 N FIGURE 2 2x du 2. To find the new limits of integration we note that 5x 2 dx y= 0 x 1 EXAMPLE 8 Evaluate y S O L U T I O N Let u 5x. Then du 3 2 1 3 9 u dx . 5x 2 5 d x, so d x du 5. When x 1, u 2 and 7.1 I NTEGRATION BY PARTS Every differentiation rule has a corresponding integration rule. For instance, the Substitution Rule for integration corresponds to the Chain Rule for differentiation. The rule that corresponds to the Product Rule for differentiation is called the rule for integration by parts. The Product Rule states that if f and t are differentiable functions, then d f xtx dx f xt x txf x In the notation for indefinite integrals this equation becomes y t x f x dx yf or f xt x x t x dx ytxf f xtx x dx f xtx We can rearrange this equation as yf 1 x t x dx f xtx ytxf x dx Formula 1 is called the formula for integration by parts. It is perhaps easier to remember in the following notation. Let u f x and v t x . Then the differentials are du f x d x and dv t x d x, so, by the Substitution Rule, the formula for integration by parts becomes y u dv 2 EXAMPLE 1 Find uv y v du y x sin x d x. x and t x sin x. Then f x 1 cos x. (For t we can choose any antiderivative of t .) Thus, using Formula SOLUTION USING FORMULA 1 Suppose we choose f x and t x 1, we have y x sin x d x f xtx x cos x ytxf x dx y cos x d x x cos x y cos x d x x cos x sin x C It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as expected. 453 462 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION STRATEGY FOR EVALUATING y sin m x cos n x dx (a) If the power of cosine is odd n 2 k 1 , save one cosine factor and use cos 2x 1 sin 2x to express the remaining factors in terms of sine: y sin y sin x cos 2 k 1x d x Then substitute u m x cos 2x k cos x d x y sin m m x1 sin 2x k cos x d x sin x. (b) If the power of sine is odd m 2 k 1 , save one sine factor and use sin 2x 1 cos 2x to express the remaining factors in terms of cosine: y sin y x cos n x d x sin 2x k cos n x sin x d x y 2k 1 1 cos 2x k cos n x sin x d x Then substitute u cos x. [Note that if the powers of both sine and cosine are odd, either (a) or (b) can be used.] (c) If the powers of both sine and cosine are even, use the half-angle identities sin 2x 1 2 1 1 2 cos 2x cos 2 x 1 cos 2 x It is sometimes helpful to use the identity 1 2 sin x cos x sin 2 x We can use a similar strategy to evaluate integrals of the form x tan mx sec nx dx. Since d dx tan x sec 2x, we can separate a sec 2x factor and convert the remaining (even) power of secant to an expression involving tangent using the identity sec 2x 1 tan 2x. Or, since d dx sec x sec x tan x, we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant. V EXAMPLE 5 Evaluate y tan 6x sec 4x d x. SOLUTION If we separate one sec 2x factor, we can express the remaining sec 2x factor in terms of tangent using the identity sec 2x 1 tan 2x. We can then evaluate the integral by substituting u tan x so that du sec 2x d x : y tan x sec x d x y tan x sec x sec x d x 6 4 6 2 y tan x 6 yu 6 1 7 tan 7x tan 2x sec 2x d x 1 u 2 du 1 u7 7 2 u9 9 y u6 u 8 du C 1 9 tan 9x C M S ECTION 7.2 TRIGONOMETRIC INTEGRALS EXAMPLE 6 Find y tan 5 |||| 463 sec 7 d . SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left with a sec 5 factor, which isn’t easily converted to tangent. However, if we separate a sec tan factor, we can convert the remaining power of tangent to an expression sec 2 1. We can then evaluate the involving only secant using the identity tan 2 integral by substituting u sec , so du sec tan d : y tan 5 y tan sec 7 d 4 sec 6 sec tan d y sec 2 y u2 1 2 u 6 du y u 10 2u 8 u 6 du u9 9 u7 7 u 11 11 1 11 1 2 sec 6 sec 2 2 9 sec 11 tan d C 1 7 sec 9 sec 7 C M The preceding examples demonstrate strategies for evaluating integrals of the form x tan mx sec nx d x for two cases, which we summarize here. STRATEGY FOR EVALUATING y tan m x sec nx dx (a) If the power of secant is even n 2 k, k 2 , save a factor of sec 2x and use sec 2x 1 tan 2x to express the remaining factors in terms of tan x : y tan Then substitute u y tan x sec 2 kx d x m x sec 2x y tan m m x1 k1 sec 2x d x tan 2x k1 sec 2x d x tan x. (b) If the power of tangent is odd m 2 k 1 , save a factor of sec x tan x and use tan 2x sec 2x 1 to express the remaining factors in terms of sec x : y tan Then substitute u x sec n x d x y tan 2x k sec n 1x sec x tan x d x y 2k 1 sec 2x 1 k sec n 1x sec x tan x d x sec x. For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. We will sometimes need to be able to 464 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION integrate tan x by using the formula established in (5.5.5): y tan x d x ln sec x C We will also need the indefinite integral of secant: y sec x d x 1 ln sec x tan x C We could verify Formula 1 by differentiating the right side, or as follows. First we multiply numerator and denominator by sec x tan x : y sec x d x y sec x sec x sec x y If we substitute u becomes x 1 u du sec x ln u sec 2x sec x tan x dx sec x tan x sec 2x d x, so the integral tan x, then du sec x tan x C. Thus we have y sec x d x EXAMPLE 7 Find tan x dx tan x ln sec x tan x C y tan x d x. 3 SOLUTION Here only tan x occurs, so we use tan 2x sec 2x 1 to rewrite a tan 2x factor in 2 terms of sec x : y tan x d x y tan x tan x d x y tan x 3 2 sec 2x 1 dx y tan x sec x d x y tan x d x 2 tan 2x 2 ln sec x In the first integral we mentally substituted u C tan x so that du sec 2x d x. M If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of sec x. Powers of sec x may require integration by parts, as shown in the following example. EXAMPLE 8 Find y sec x d x. 3 SOLUTION Here we integrate by parts with u du sec x sec x tan x d x dv v sec 2x d x tan x SECTION 7.2 TRIGONOMETRIC INTEGRALS y sec x d x sec x tan x y sec x sec x tan x 465 y sec x tan x d x sec x tan x 3 Then |||| y sec x d x y sec x d x 2 sec 2x 1 dx 3 Using Formula 1 and solving for the required integral, we get y sec x d x 1 2 3 (sec x tan x ln sec x tan x ) C M Integrals such as the one in the preceding example may seem very special but they occur frequently in applications of integration, as we will see in Chapter 8. Integrals of the form x cot m x csc n x d x can be found by similar methods because of the identity 1 cot 2x csc 2x. Finally, we can make use of another set of trigonometric identities: 2 To evaluate the integrals (a) x sin m x cos n x d x, (b) x sin m x sin n x d x, or (c) x cos m x cos n x d x, use the corresponding identity: (a) sin A cos B 1 2 sin A B sin A (b) sin A sin B 1 2 cos A B cos A B (c) cos A cos B These product identities are discussed in Appendix D. N 1 2 cos A B cos A B EXAMPLE 9 Evaluate B y sin 4 x cos 5x d x. SOLUTION This integral could be evaluated using integration by parts, but it’s easier to use the identity in Equation 2(a) as follows: y sin 4 x cos 5x d x y 1 2 y 1 2 7.2 1 2 (cos x x sin 9x d x sin x sin 9x d x 1 9 cos 9x C M EXERCISES 1– 49 Evaluate the integral. y sin x cos x d x 2. y sin x cos x d x 3. y 4. y 3 4 2 sin 5x cos 3x d x 6 2 0 5. y sin 2 7. y 2 2 x cos 5 cos2 d x dx 6. y 8. y 0 y sin 4 3 t dt y 1 13. y 15. y ssin 3 cos 5x d x sin3 (sx ) dx sx 2 9. 11. 1. 3 0 sin sin 2 2 d 0 0 2 cos 10. 2 d sin 2x cos 2x d x cos 5 d y 12. y x cos x d x 14. y 16. y cos 0 cos6 d 2 0 sin 2 t cos 4 t dt cos5 sin d S ECTION 7.3 TRIGONOMETRIC SUBSTITUTION y sin m x cos n x d x 68. y sin m x sin n x d x 69. y cos m x cos n x d x 467 70. A finite Fourier series is given by the sum 67–69 Prove the formula, where m and n are positive integers. 67. |||| N 0 a n sin n x fx n1 0 0 if m if m n n a 1 sin x a 2 sin 2 x a N sin Nx Show that the mth coefficient a m is given by the formula if m if m n n 7.3 TRIGONOMETRIC SUBSTITUTION am 1 y f x sin m x d x In finding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 d x arises, where a 0. If it were x x sa 2 x 2 d x, the substitution u a 2 x 2 would be effective but, as it stands, x sa 2 x 2 d x is more difficult. If we change the variable from x to by the substitution x a sin , then the identity 1 sin 2 cos 2 allows us to get rid of the root sign because sa 2 x2 sa 2 sa 2 1 a 2 sin 2 sa 2 cos 2 sin 2 a cos Notice the difference between the substitution u a 2 x 2 (in which the new variable is a function of the old one) and the substitution x a sin (the old variable is a function of the new one). In general we can make a substitution of the form x t t by using the Substitution Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is, t is one-to-one. In this case, if we replace u by x and x by t in the Substitution Rule (Equation 5.5.4), we obtain yf x dx yf t t t t dt This kind of substitution is called inverse substitution. We can make the inverse substitution x a sin provided that it defines a one-to-one function. This can be accomplished by restricting to lie in the interval 2, 2 . In the following table we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities. In each case the restriction on is imposed to ensure that the function that defines the substitution is one-to-one. (These are the same intervals used in Section 1.6 in defining the inverse functions.) TABLE OF TRIGONOMETRIC SUBSTITUTIONS Expression Substitution sa 2 x2 x a sin , sa 2 x2 x a tan , sx 2 a2 x a sec , 2 0 1 2 2 or 3 2 sin 2 cos 2 1 2 2 Identity tan 2 sec 2 sec 2 1 tan 2 484 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION memorized since they are easily derived. Formula 19 can be avoided by using partial fractions, and trigonometric substitutions can be used in place of Formula 20. TABLE OF INTEGRATION FORMULAS Constants of integration have been omitted. 1. y x n dx xn 1 n1 3. y e x dx ex 5. y sin x d x 7. y sec x d x 9. y sec x tan x d x 2 2. tan x ln x y a x dx ax ln a y cos x d x 8. cos x 1 dx x 6. 1 y 4. n y csc x d x sin x 2 cot x 10. sec x y csc x cot x d x 12. y csc x d x ln csc x ln sin x 11. y sec x d x ln sec x 13. y tan x d x ln sec x 14. y cot x d x 15. y sinh x d x cosh x 16. y cosh x d x 17. yx 1 tan a 18. y sa *19. yx * 20. y sx dx 2 a 2 dx 2 a2 1 tan x x a 1 x ln 2a x a a x 2 dx 2 cot x sinh x dx 2 csc x a2 sin 1 ln x x a sx 2 a2 Once you are armed with these basic integration formulas, if you don’t immediately see how to attack a given integral, you might try the following four-step strategy. 1. Simplify the Integrand if Possible Sometimes the use of algebraic manipulation or trigonometric identities will simplify the integrand and make the method of integration obvious. Here are some examples: y s x (1 y y (s x sx ) dx tan d sec2 y cos sin cos2 d y sin y sin x cos x 2 d x x) d x cos d y sin 2x y 1 1 2 y sin 2 2 sin x cos x 2 sin x cos x d x d cos 2x d x S ECTION 7.5 STRATEGY FOR INTEGRATION |||| 485 t x in the t x d x also occurs, apart from a constant fac- 2. Look for an Obvious Substitution Try to find some function u integrand whose differential du tor. For instance, in the integral y x x 2 1 dx we notice that if u x 2 1, then du 2 x d x. Therefore we use the substitution u x 2 1 instead of the method of partial fractions. 3. Classify the Integrand According to Its Form If Steps 1 and 2 have not led to the solution, then we take a look at the form of the integrand f x . (a) Trigonometric functions. If f x is a product of powers of sin x and cos x, of tan x and sec x, or of cot x and csc x, then we use the substitutions recommended in Section 7.2. (b) Rational functions. If f is a rational function, we use the procedure of Section 7.4 involving partial fractions. (c) Integration by parts. If f x is a product of a power of x (or a polynomial) and a transcendental function (such as a trigonometric, exponential, or logarithmic function), then we try integration by parts, choosing u and dv according to the advice given in Section 7.1. If you look at the functions in Exercises 7.1, you will see that most of them are the type just described. (d) Radicals. Particular kinds of substitutions are recommended when certain radicals appear. (i) If s x 2 a 2 occurs, we use a trigonometric substitution according to the table in Section 7.3. n n (ii) If sax b occurs, we use the rationalizing substitution u sax b . n More generally, this sometimes works for st x . 4. Try Again If the first three steps have not produced the answer, remember that there are basically only two methods of integration: substitution and parts. (a) Try substitution. Even if no substitution is obvious (Step 2), some inspiration or ingenuity (or even desperation) may suggest an appropriate substitution. (b) Try parts. Although integration by parts is used most of the time on products of the form described in Step 3(c), it is sometimes effective on single functions. Looking at Section 7.1, we see that it works on tan 1x, sin 1x, and ln x, and these are all inverse functions. (c) Manipulate the integrand. Algebraic manipulations (perhaps rationalizing the denominator or using trigonometric identities) may be useful in transforming the integral into an easier form. These manipulations may be more substantial than in Step 1 and may involve some ingenuity. Here is an example: y1 dx cos x y y 1 1 1 cos x 1 1 cos x dx sin 2x cos x dx cos x y csc 2x y 1 1 cos x sin 2x cos x dx cos 2x dx (d) Relate the problem to previous problems. When you have built up some experience in integration, you may be able to use a method on a given integral that is similar to a method you have already used on a previous integral. Or you may even be able to express the given integral in terms of a previous one. For 486 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION instance, x tan 2x sec x d x is a challenging integral, but if we make use of the identity tan 2x sec 2x 1, we can write y tan x sec x d x y sec x d x y sec x d x 2 3 and if x sec 3x d x has previously been evaluated (see Example 8 in Section 7.2), then that calculation can be used in the present problem. (e) Use several methods. Sometimes two or three methods are required to evaluate an integral. The evaluation could involve several successive substitutions of different types, or it might combine integration by parts with one or more substitutions. In the following examples we indicate a method of attack but do not fully work out the integral. EXAMPLE 1 y tan 3x dx cos 3x In Step 1 we rewrite the integral: y tan 3x dx cos 3x y tan x sec x d x 3 3 The integral is now of the form x tan m x sec n x d x with m odd, so we can use the advice in Section 7.2. Alternatively, if in Step 1 we had written tan 3x dx cos 3x y sin 3x y cos x 3 1 dx cos 3x sin 3x y cos x d x 6 then we could have continued as follows with the substitution u sin 3x y cos x d x y 1 6 y V EXAMPLE 2 ye sx cos 2x sin x d x cos 6x u2 1 u 6 y du u y 4 cos x : u2 1 u6 u 6 du du M dx According to (ii) in Step 3(d), we substitute u ye sx dx s x . Then x u 2, so dx 2u du and 2 y u e u du The integrand is now a product of u and the transcendental function e u so it can be integrated by parts. M ...
View Full Document

This note was uploaded on 01/26/2012 for the course ECON 401 taught by Professor Burbidge,john during the Fall '08 term at Waterloo.

Ask a homework question - tutors are online