25.2 Substitution: The Chain Rule in Reverse 791 u = x 2 du = 2xdx so xdx= 1 2 du x cos(x 2 )dx= cos(x 2 )xdx = cos u · 1 2 du Transform the integral using substitution. = 1 2 cos udu This is one of our basic integrals. = 1 2 sin u + C Now return to the original variable. = 1 2 sin(x 2 ) + C We can verify that this answer is correct by differentiating it. 3 d dx 1 2 sin(x 2 ) + C = 1 2 cos(x 2 ) · 2x = x cos x 2 . Why Is the Substitution du = u (x) dx Legitimate? In the example above we went from du dx = u (x) to du = u (x) dx. Why is this valid? Suppose that f(x)dx =F(x)+C. That is, F (x) = f(x). Then f (u) du = F(u) + C. Is it true that f (u(x)) · u (x) dx = F (u(x)) + C? Sure! d dx F(u(x)) = F (u(x)) · u (x) = f (u(x)) · u (x) Therefore, it “works” to replace u(x) by u and u (x) dx by du. Replacing du dx dx by du is legitimate. This is part of the genius of Leibniz’ notation du dx for the derivative. One generally doesn’t run into trouble treating du dx as it appears, despite the fact that du dx is really not a fraction. Now we return to more examples. 3 We can always check an answer to an integration problem by differentiation. 792 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration Using Substitution to Reverse the Chain Rule The key idea when making a substitution in order to reverse the Chain Rule is to choose u so that the integral is of the form f (u(x)) ·u (x) dx. We must choose u so that u is already sitting by in the integrand, up to a constant multiple. ◆ EXAMPLE 25.4 Evaluate the following integrals. (a) 5 3x+2 dx (b) 1 √ xe √ x dx (c) 3x 1+x 2 dx (d) cos 2 x sin xdx Trythese problems on your own and then compare your answers with those below. (You can always check your answers by differentiating.) SOLUTIONS (a) 5 3x+2 dx =5 1 3x+2 dx. We hope we essentially have 1 u du. Let u = 3x + 2. Then du dx =3so 1 3 du =dx. Rewrite the original integral in terms of u. 5 1 3x + 2 dx =5 1 u · 1 3 du = 5 3 1 u du This is a familiar integral. = 5 3 ln |u|+C Return to the original variable. = 5 3 ln |3x + 2|+C (b) 1 √ xe √ x dx = e − √ x √ x dx. We hope we essentially have e u du. Let u =− √ x. du dx =− 1 2 1 √ x −2du = 1 √ x dx Rewrite the original integral in terms of u. e − √ x 1 √ x dx = e u ·(−2)du =−2 e u du This is a familiar integral. =−2e u +C Return to the original variable. =−2e − √ x + C 25.2 Substitution: The Chain Rule in Reverse 793 (c) 3x 1+x 2 dx = 3 x 1+x 2 dx Depending upon how we look at this, we might make the substitution u = 1 + x 2 ,or we might let u = x 2 . We can do the problem either way, although the latter choice is less efficient. Option 1: Let u = 1 + x 2 .(We’re hoping we essentially have 1 u du.) du dx = 2x 1 2 du = xdx Rewrite the original integral in terms of u. 3 x 1 + x 2 dx = 3 1 1 + x 2 xdx =3 1 u · 1 2 du = 3 2 1 u du = 3 2 ln |u|+C = 3 2 ln(1 + x 2 ) + C Option 2: Let u = x 2 .(We’re trying to simplify.) du dx = 2x 1 2 du = xdx Rewrite the original integral in terms of u. 3 x 1 + x 2 dx = 3 1 1 + x 2 · xdx =3 1 1 + u · 1 2 du = 3 2 1 1 + u du You may be able to guess an antiderivative at this point. Alternatively, you can use substitution again. Let v = 1 + u. dv du = 1 dv = du Rewrite 3 2 1 1+u du in terms of v. 794 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration 3 2 1 1 + u du = 3 2 1 v dv = 3 2 ln |v|+C But v = 1 + u. = 3 2 ln |1 + u|+Cu=x 2 = 3 2 ln(1 + x 2 ) + C (d) cos 2 x sin xdx.We’re hoping we essentially have u 2 du. Let u = cos x. du dx =−sin x −du = sin xdx cos 2 x sin xdx= u 2 (−du) =− u 2 du = −u 3 3 + C =− 1 3 cos 3 x + C ◆ Substitution in Definite Integrals When using substitution to evaluate a definite integral b a f(x)dx,wehavetwooptions. Option 1: Change the limits of integration to the new variable and never look back. Option 2: Leave the limits in terms of the original variable throughout, but rewrite the integral as x=b x=a g(u) du to make it clear that we do not mean u = a and u = b. 4 After antidifferentiating with respect to u, replace u by its original expression in terms of x. Then evaluate at x = b and x = a, subtracting the latter from the former as usual. ◆ EXAMPLE 25.5 Evaluate π/3 0 cos 2 x sin xdx. SOLUTION This is the same integrand as in Example 25.3(d). We use the substitution u = cos x as before. u = cos x du dx =−sin x −du = sin xdx Option 1: When x = 0, u = cos 0 = 1; when x = π 3 , u = cos π 3 = 1 2 . 4 For instance, π 0 sin 2xdx=0. If we let u = 2x, then du = 2dx and dx = 1 2 du. π 0 1 2 sin udu=1=0, whereas 2π 0 1 2 sin ududoes equal zero. 25.2 Substitution: The Chain Rule in Reverse 795 π/3 0 cos 2 x sin xdx=− 1/2 1 u 2 du =− u 3 3 1/2 1 =− 1 3 1 8 − 1 =− 1 3 − 7 8 = 7 24 Option 2: π/3 0 cos 2 x sin xdx=− x=π/3 x=0 u 2 du =− 1 3 u 3 x=π/3 x=0 =− 1 3 cos 3 x π/3 0 =− 1 3 cos π 3 3 − ( cos 0 ) 3 =− 1 3 1 8 − 1 = 7 24 CAUTION Writing π/3 0 u 2 du is not an option! Unless you explicitly specify that the endpoints of integration are in terms of a different variable, it is understood that they are in terms of u if all else is in terms of u. ◆ Working many problems will help you become proficient at using substitution. You will see that integrands that at first glance appear similar may in fact yield vastly different antiderivatives. Try the exercises below. Answers are provided, although you can always check your answers on your own by differentiating. Fundamental goals of substitution: i. Simplify the integrand. ii. Obtain f (u(x)) · u (x) dx, which we write as f (u) du. Therefore, when choosing a “u,” be sure that you already have duup to a constant multiple. If you don’t, chances are you’ve bitten off too much to call “u,” so try a smaller bite. If you’re stuck on any of the problems that follow, look at the suggested substitutions given below the answers. 796 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration EXERCISE 25.2 Evaluate the following integrals. 5 (a) dx 1+2x (b) dx (1+2x) 2 (c) ln 2 0 e x 1+e x dx (d) e x 1+e 2x dx (e) xdx √ 2+3x 2 (f) tan xdx Answers (a) 1 2 ln |1 + 2x|+C (b) −1 2(1+2x) +C (c) ln 3 2 (d) arctan(e x ) + C (e) 1 3 √ 2 +3x 2 + C (f) − ln | cos x|+C Suggested substitutions (a) u = 1 + 2x (b) u = 1 + 2x (c) u =1 + e x or u =e x (d) u = e x (e) u = 2 +3x 2 (f) Express tan x as sin x cos x . Then let u = cos x. (Letting u = sin x won’t work because du will be cos xdx,not 1 cos x dx.) PROBLEMS FOR SECTION 25.2 1. Evaluate the following indefinite integrals. (Let your mind be limber. Integrals that look very similar may require very different mindsets!) (a) 1 x dx (b) 1 x+1 dx (c) 1 (x+1) 2 dx (d) 1 x 2 +1 dx (e) x x 2 +1 dx (f) x 2 +1 x dx (g) (1 + x) 5 dx (h) 1 (1+x) 5 dx (i) (1 + x 2 ) 2 dx 2. Find the following indefinite integrals. Check your answers. (a) 3 sin(5t) dt (b) π cos(πt ) dt (c) √ 3x + 5 dx (d) π e x dx (e) e −3t dt (f) √ e t dt (g) 6 √ t 3 dt (h) 1 3t+8 dt In Problems 3 through 8, find the indefinite integrals. 3. (a) (2x + 1) 3 dx (b) 1 (2x+1) 2 dx (c) 1 (2x+1) dx (d) 1 √ 2x+1 dx 4. (a) x √ 2x 2 + 1 dx (b) x √ 2x 2 +1 dx (c) cos √ x √ x dx (d) √ cos x sin xdx 5. (a) 5 1+9x 2 dx (b) ln x x dx (c) e x e −x dx (d) (ln w) 2 w dw 6. (a) t 2 sin(t 3 )dt (b) xe −x 2 dt (c) 1 x+5 dx (d) t 2t 2 +7 dt 7. (a) t (t 2 +1) 2 dt (b) tan 2tdt (c) e w e w +1 dw (d) e w e 2w +1 dw 5 dx 1+2x and 1 1+2x dx are two ways of writing the same thing. 25.2 Substitution: The Chain Rule in Reverse 797 8. (a) (x 2 + 3) 3 xdx (b) x √ x 2 + 4 dx (c) (2x +1) √ x 2 + xdx 9. Evaluate. (a) 1 0 3 1+4w 2 dw (b) 1 0 1+4w 2 3 dw (c) 3π π/2 cos t 2 dt (d) 3 1 4 3x+2 dx (e) 4 1 1 (2x+1) 2 dx 10. Find the following definite integrals. (a) 3 1 1 (2x+4) 2 + 1 x+4 dx (b) e 1 (ln x) 2 x dx In Problems 11 through 23, compute the following integrals. 11. cos(e −x ) e x dx 12. (e x + x) √ 2e x + x 2 dx 13. sec 2 (ln x) x dx 14. sin(x) √ cos(x) dx 15. π/4 0 tan xdx 16. e 2x √ e x dx 17. π/2 0 cos(x) 1+sin 2 x dx 18. ln 5 0 3e x √ e x +4 dx 19. ln 3 0 e 2x e 2x +1 dx 20. ln 2 0 e x e 2x +1 dx 21. 3 x 3 x +1 dx 22. sec 2 x tan 2 xdx 23. sec 2 √ x tan 2 √ x √ x dx 24. It is 10:00 a.m. and five ants have found their way into a picnic basket. Ants are notorious followers, so ants from all over the vicinity follow their five brethren into the basket. The culinary treat awaiting them is unsurpassed elsewhere, so once the ants find their way into the basket they choose not to leave. If the rate at which ants are climbing into the basket is well modeled by 100e −0.2t ants per hour, where t = 0 is the benchmark hour of 10:00 a.m., how many ants will be in the basket x hours after 10:00 a.m.? How many ants are in the basket at 1:00 p.m., when the picnic is supposed to begin? Give your answer to the nearest ant. 798 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration 25. Consider the integral cos x sin xdx. (a) Using the substitution u = sin x, show that cos x sin xdx= 1 2 sin 2 x + C. (b) Using the substitution u = cos x, show that cos x sin xdx=− 1 2 cos 2 x + C. (c) Explain why although these answers look different they are both correct. 26. It is April and, with the arrival of warm weather a pile of snow has turned into a puddle and the puddle is drying up. At time t = 0 there are 4 gallons of water in the puddle. The rate of change of water in the puddle is given by f(t)=−e 0.2t gallons per hour. How much water is in the puddle t hours from now? When will the puddle dry up? 27. A little rock rolls off a little cliff. It experiences an acceleration of −32 ft/sec 2 . (a) The derivative of the velocity function v(t) is acceleration. Therefore, the an- tiderivative (indefinite integral) of acceleration is velocity. v(t) = a(t) dt. Find v(t) assuming that the rock’s initial vertical velocity is zero. (You’ll use this initial condition to find the constant of integration.) (b) Let s(t) give the rock’s height at any time t.We’ll call s the position function. The derivative of the position function s(t) is velocity. Therefore, the antiderivative (the indefinite integral) of velocity is the distance function. s(t) = v(t) dt. Find s(t) assuming that the cliff has a height of 25 feet. (You’ll use this initial condition to find the constant of integration.) 25.3 SUBSTITUTION TO ALTER THE FORM OF AN INTEGRAL There are many ways of altering the form of an integral to make it more tractable. In this section we will look at some of them. We saw that one strategy for approaching an integral is to manipulate the integrand algebraically so it can be expressed as a sum and the integral can be pulled apart into the sum of simpler integrals. Sometimes substitution is useful in evaluating the resultant integrals and other times in actually altering the form of the integrand so that it can be expressed as a sum. Both uses of substitution are illustrated in the examples below. Your goal in each case is to write the integral as a sum of simpler integrals. Don’t allow your eagerness to express an integrand as a sum interfere with your algebraic judgement; naturally, the standard rules of algebra apply! ◆ EXAMPLE 25.6 Evaluate the following. (a) 1 0 3x+7 1+x 2 dx (b) x x+3 dx (c) 3x √ x + 5 dx Try these on your own before reading the detailed solutions. If you need help, first look at the hints supplied. Hints (a) Split this into the sum of two integrals. Then use substitution on one of them. (b) Let u = x + 3 so that you can split this integral. (c) Let u = x + 5 so that you can split this integral. 25.3 Substitution to Alter the Form of an Integral 799 SOLUTIONS (a) 1 0 3x+7 1+x 2 dx = 3 1 0 x 1+x 2 dx + 7 1 0 1 1+x 2 dx We’ ll evaluate each integral independently and then sum the results. 3 1 0 x 1+x 2 dx looks essentially like du u . Let u = 1 + x 2 . du = 2xdx so xdx= 1 2 du when x = 0, u = 1 + 0 2 = 1 when x = 1, u = 1 + 1 2 = 2 3 1 0 x 1 + x 2 dx = 3 2 1 1 u · 1 2 du = 3 2 2 1 1 u du = 3 2 ln |u| 2 1 = 3 2 [ln 2 − ln 1] = 3 2 ln 2 7 1 0 1 1 + x 2 dx = 7 tan −1 (x) 1 0 = 7 tan −1 (1) − tan −1 (0) = 7 π 4 − 0 = 7π 4 so 1 0 3x + 7 1 + x 2 dx = 3 2 ln 2 + 7π 4 (b) x x+3 dx CAUTION x x+3 = x x + x 3 . In order to express this integrand as a sum we need the numerator expressed as a sum, not the denominator. The substitution u = x + 3 transforms the integral so that we can break it up. Let u = x + 3. du = dx and x = u − 3 800 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration x x + 3 dx = u −3 u du = 1 − 3 u du = 1du−3 1 u du =u −3ln|u|+C Now return to the original variable. = x + 3 − 3ln|x+3|+C or x − 3ln|x+3|+C 1 (where C 1 = 3 + C) (c) 3x √ x + 5 dx CAUTION √ x + 5 = √ x + √ 5. 6 In order to express this integrand as a sum we need the quantity outside the square root to be a sum, not the quantity under the square root. The substitution u = x + 5 takes care of this. Let u = x + 5. du =dx and x = u − 5 3x √ x + 5 dx = 3(u −5) √ udu =3 u √ udu−15 √ udu =3 u 3/2 du −15 u 1/2 du =3 · 2 5 u 5/2 − 15 · 2 3 · u 3/2 + C Now return to the original variable. = 6 5 (x + 5) 5/2 − 10(x + 5) 3/2 + C ◆ Sometimes a substitution is used as a first step in simplifying a problem. It might be followed by a second substitution. ◆ EXAMPLE 25.7 Evaluate the following. (a) 3 dx 4+9x 2 (b) tan √ x √ x dx (c) e 2x e 2x +2e x +1 dx SOLUTIONS (a) 3 dx 4+9x 2 = 3 1 4+9x 2 dx This looks vaguely like a variation on 1 1+u 2 du.Totry to get 1 +u 2 in the denominator we’ll divide the numerator and denominator of the integrand by 4. 3 1 4 4 4 + 9x 2 4 dx = 3 4 1 1 + 3x 2 2 dx 6 We know 5 = √ 25 = √ 16 +9 = √ 16 + √ 9 =4 + 3 = 7. . some of them. We saw that one strategy for approaching an integral is to manipulate the integrand algebraically so it can be expressed as a sum and the integral can be pulled apart into the sum of. tan 2 xdx 23. sec 2 √ x tan 2 √ x √ x dx 24. It is 10:00 a.m. and five ants have found their way into a picnic basket. Ants are notorious followers, so ants from all over the vicinity follow their five brethren into. initial condition to find the constant of integration.) 25.3 SUBSTITUTION TO ALTER THE FORM OF AN INTEGRAL There are many ways of altering the form of an integral to make it more tractable. In this section