Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 91 pptx

29.1 Integration by Parts—The Product Rule in Reverse 881 The new integral is no simpler than the original, but it is no more difficult either. We do integration by parts again. Let u = e 2x , dv =cos 3xdx du = 2e 2x dx, v = 1 3 sin 3x  e 2x sin 3xdx=− 1 3 e 2x cos 3x + 2 3  1 3 e 2x sin 3x − 1 3  sin 3x · 2e 2x dx  Then  e 2x sin 3xdx    I =− 1 3 e 2x cos 3x + 2 9 e 2x sin 3x − 4 9  e 2x sin 3xdx    I . The original integral and the integral on the right are identical. We can solve algebraically for it. If you like you can denote the original integral by I and solve for I. 13 9  e 2x sin 3xdx=− 1 3 e 2x cos 3x + 2 9 e 2x sin 3x (29.1)  e 2x sin 3xdx=− 3 13 e 2x cos 3x + 2 13 e 2x sin 3x + C REMARKS 1. Notice that in Equation(29.1) we are missing a “+ C.”This is because we never actually computed  v du.We’ve got to insert a + C. 2. One of the key ingredients that makes this method work is that if sin x either is differentiated or integrated twice then we obtain sin x again, and similarly with e x . 3. This integral could have been solved similarly by letting u = sin 3x and du =e 2x dx and repeating this. It will not, however, be productive to let u = sin 3x the first time and then u = e 2x the second time. ◆ EXERCISE 29.3 Find  e 2x sin 3xdxby letting u be the trigonometric function. Then try using integration by parts twice but letting u = e 2x the first time and u = sin 3x the second time in order to verify that this is unproductive. EXERCISE 29.4 The integral  sin 2 xdx can be computed most easily using the trigonometric identity sin 2 x = 1 2 (1 − cos 2x). However, it can also be solved using the techniques of Example 29.5. Do the latter. Check your answer by differentiating. In Example 29.4 and Exercise 29.2 we saw that integration by parts can be used to reduce the complexity of an integral. There are numerous reduction formulas that can be derived by integration by parts. Exercise 29.5 below leads you through one such derivation. EXERCISE 29.5 Derive the reduction formula  sin n xdx=− 1 n cos x sin n−1 x + n − 1 n  sin n−2 xdx, where n is an integer, n ≥ 2 as follows. 882 CHAPTER 29 Computing Integrals (a) Use integration by parts, letting u = sin n−1 x and dv =sin xdx. (b) In the resulting new integral, replace cos 2 x by 1 − sin 2 x to obtain an equation with  sin n xdxon both sides. (c) Solve algebraically for  sin n xdx, as was done in Example 29.5. Using Integration by Parts in Definite Integrals To figure out how to adapt the integration by parts formula  udv= uv −  v du to definite integrals, let’s return to functions of x. Recall that u = f(x)and v = g(x). Then du = f  (x) dx and dv =g  (x) dx and the formula becomes  f(x)g  (x) dx = f (x)g(x) −  g(x)f  (x) dx. Suppose we want to compute the definite integral  b a f(x)g  (x) dx. Assuming f  and g  are continuous, we simply evaluate between x = a and x = b.  b a f(x)g  (x) dx = f (x)g(x)  b a −  b a g(x)f  (x) dx In other words,  x=b x=a udv= uv  x=b x=a −  x=b x=a v du. This is equivalent to doing the indefinite integral and taking the difference between the antiderivative at x = b and the antiderivative at x = a. ◆ EXAMPLE 29.6 Evaluate  1 0 tan −1 xdx. SOLUTION Let u = tan −1 x, dv = dx du = 1 1 + x 2 dx, v=x.  1 0 tan −1 xdx= x tan −1 x  1 0 −  1 0 x 1 + x 2 dx You may be able to do the latter integral in your head once you realize that the numerator is a constant multiple of the derivative of the denominator. Otherwise, use substitution. Let w = 1 + x 2 dw =2xdx xdx= 1 2 dw. 29.1 Integration by Parts—The Product Rule in Reverse 883  x 1 + x 2 dx = 1 2  1 w dw = 1 2 ln |w|+C = 1 2 ln(1 + x 2 ) + C  1 0 tan −1 x = x tan −1 x  1 0 − 1 2 ln(1 + x 2 )     1 0 = tan −1 1 − 0 −  1 2 ln(1 + 1) − 1 2 ln(1)  = π 4 − 1 2 ln 2 ◆ The choice of u and dv sometimes requires a fair amount of thinking ahead. This is illustrated in the next example. ◆ EXAMPLE 29.7 Evaluate  1 0 x 3 e x 2 dx. SOLUTION Choosing u = x 3 won’t work because we can’t integrate e x 2 . Choosing u = e x 2 is unpro- ductive; the complexity of the integral will increase. Instead, we choose u = x 2 ; we can integrate xe x 2 . Let u = x 2 , dv =xe x 2 dx du = 2xdx, v =  xe x 2 dx = 1 2  e x 2 · 2xdx= 1 2 e x 2 .  1 0 x 3 e x 2 dx = x 2 · 1 2 e x 2  1 0 −  1 0 1 2 e x 2 2xdx = 1 2 x 2 e x 2  1 0 − 1 2  1 0 e x 2 2xdx = 1 2 e − 0 − 1 2  e x 2  1 0 = 1 2 e − 1 2 [e − 1] = 1 2 ◆ PROBLEMS FOR SECTION 29.1 In Problems 1 through 5, evaluate the integrals. 1.  x sin xdx 2.  x cos xdx 3.  3xe −2x dx 884 CHAPTER 29 Computing Integrals 4.  e 1 ln xdx 5.  1 0 cos −1 xdx 6.  sin −1  x 2  dx 7.  e −t sin(2t) dt 8.  e −x cos xdx 9.  x 2 cos 3xdx 10.  x ln 1 x dx (You may want to use substitution first.) 11.  x sec 2 xdx 12.  1 0 t 3 e −t dt 13.  √ x ln xdx 14.  cos(ln x) dx 15.  (ln x) 2 dx 16. Find  cos 2 xdxin two ways. (a) Use the trigonometric identity cos 2 x = 1 2 (1 + cos 2x). This is the most efficient way to do the problem. (b) Use integration by parts. You will solve algebraically for  cos 2 xdx. (c) Check that your answers to parts (a) and (b) are correct by differentiating them. (d) Your answers to parts (a) and (b) are both antiderivatives of cos 2 x; therefore they must differ by a constant (where the constant is possibly zero). What is the constant? In Problems 17 through 29, evaluate the integral. In many cases it will be advantageous to begin by doing a substitution. For example, in Problem 19, let w = √ x, w 2 =x; then 2wdw = dx. This eliminates √ x by replacing x with a perfect square. Not every integral in Problems 17 through 29 requires the technique of integration by parts. 17.  x 5 cos x 3 dx 18.  (ln x) 3 dx 19.  √ xe √ x dx 20.  e 1 ln x x dx 21.  ln x √ x dx 29.1 Integration by Parts—The Product Rule in Reverse 885 22.  π 4 0 sec 2 x tan xdx 23.  e 1 ln √ wdw 24.  √ x ln xdx 25.  sin(ln x) dx 26.  1 0 cos √ xdx 27.  x 3 ln xdx 28.  π 2 0 x sin x cos xdx(Hint: Use a trigonometric identity.) 29.  2 0 e √ x dx Problems 30 through 34 involve reduction formulas. 30. (a) Show that  cos n xdx= 1 n cos n−1 x sin x + n−1 n  cos n−2 xdx, for n an integer, n ≥ 2. If you need some hints, see Exercise 29.5. (b) Use the results of part (a) to find  cos 2 xdx. Verify your answer using differen- tiation. (c) Use the results of parts (a) and (b) to find  cos 6 xdx. 31. (a) Use the reduction formula given in Exercise 29.5 to find  π 2 0 sin 3 xdx. (b) Find  π 2 0 sin 5 xdx. (c) Use the reduction formula given in Exercise 29.5 to show that for n a positive odd integer,  π 2 0 sin n xdx− 2 ·4 ·6 · ·(n −1) 3 · 5 · 7 · · n . This is called a Wallis product. 32. (a) Use the reduction formula given in Exercise 29.5 to find  π 2 0 sin 4 xdx. (b) Find  π 2 0 sin 6 xdx. (c) Show that for n a positive even integer,  π 2 0 sin n xdx− 1 · 3 · 5 · · (n − 1) 2 ·4 ·6 · ·n π 2 . Like the formula in part (c) of the previous problem, this is called a Wallis product. 33. Show that  (ln x) n dx = x(ln x) n − n  (ln x) n−1 dx. 34. Show that  x n e x dx = x n e x − n  x n−1 e x dx. 886 CHAPTER 29 Computing Integrals 29.2 TRIGONOMETRIC INTEGRALS AND TRIGONOMETRIC SUBSTITUTION The two fundamental techniques of integration involving calculus are integration by substitution—coming from the Chain Rule—and integration by parts—coming from the Product Rule. The techniques discussed in this section are simply methods of transforming an integrand into something more manageable by exploiting trigonometric identities. Trigonometric Integrals Suppose we want to integrate the product of integer powers of trigonometric functions. We can convert the problem into one of the form  sin p x cos q xdxwhere p and q are integers. Integrals of this form can be evaluated. We’ll demonstrate some methods below. Integrals Involving sin x and cos x When dealing with trigonometric integrals, trigonometric identities can come in handy. Recall: cos(A + B) = cos A cos B − sin A sin B so cos(2A) =cos 2 A −sin 2 A  cos 2A = 2 cos 2 A − 1  cos 2A = 1 −2 sin 2 A ⇒ cos 2 A = 1 2 (1 + cos 2A) sin 2 A = 1 2 (1 − cos 2A) These two identities, together with sin 2A = 2 sin A cos A and the Pythagorean identities, can be useful. We illustrate by example. ◆ EXAMPLE 29.8 Evaluate  2π 0 sin 2 xdx. SOLUTION As mentioned in Section 29.1, integration by parts can beapplied to this integral. It’s simpler, however, to use the trigonometric identity sin 2 x = 1 2 (1 − cos 2x).  2π 0 sin 2 xdx= 1 2  2π 0 (1 − cos 2x) dx Integrate cos 2x by letting u = 2x. = 1 2  x − 1 2 sin 2x      2π 0 = 1 2  2π − 1 2 sin(4π)  − 1 2  0 − 1 2 sin 0  = π ◆ The trigonometric identities obtained from the formula for cos 2x sin 2 x = 1 2 (1 − cos 2x) and cos 2 x = 1 2 (1 + cos 2x) are useful if the powers of sine and cosine are both positive and even. They are used to reduce powers. On the other hand, if the power of either sine or cosine is odd, the Pythagorean identities are useful. 29.2 Trigonometric Integrals and Trigonometric Substitution 887 EXERCISE 29.6 Evaluate  sin 4 xdx by writing sin 4 x as [sin 2 x] 2 and using the trigonometric identity sin 2 x = 1 2 (1 − cos 2x). Along the way you’ll have to use cos 2 u = 1 2 (1 + cos 2u) as well. Answer: 3 8 x − 1 4 sin 2x + 1 32 sin 4x + C Sometimes when integrating trigonometric functions we can use substitution to trans- form the integral into the form  u n du. This is the strategy we employ when facing  sin p x cos q xdx when at least one of p and q is odd. The Pythagorean identities can be useful in making this transformation. ◆ EXAMPLE 29.9 Evaluate  cos 2 x sin 3 xdx. SOLUTION We’d like to convert sin 2 x to cosines using the Pythagorean identity and save the spare sin x;it’s just what we need when we make the substitution u =cos x and du =−sin xdx.  cos 2 x sin 3 xdx=  cos 2 x(sin 2 x) sin xdx (sin 2 x = 1 − cos 2 x) =  cos 2 x(1 − cos 2 x) sin xdx =  (cos 2 x − cos 4 x) sin xdx Let u = cos x, du =−sin xdx − du = sin xdx. =  (u 2 − u 4 )(−du) =  (−u 2 + u 4 )du =− u 3 3 + u 5 5 + C =− cos 3 x 3 + cos 5 x 5 + C ◆ The method used in Example 29.9 will work for integrals of the form  sin p x cos q xdx, where p and q are positive and at least one of p and q is odd. For instance, to evaluate  cos 7 x sin 2 xdx, convert cos 6 x = [cos 2 x] 3 to sines using the Pythagorean identity and save the other cos x to use with dxafter making the substitution u = sin x. To evaluate  cos 9 x sin 3 xdx, convert sin 2 x to cosines and save a sin x as part of du after making the substitution u = cos x. This is much more efficient than converting [cos 2 x] 4 to sines and letting u = sin x. EXERCISE 29.7 Find  cos 3 xdx. Convert cos 2 x to sines and save a cos x for the subsequent substitution. Answer: sin x − sin 3 x 3 + C 888 CHAPTER 29 Computing Integrals NOTE Integrals of the form  sin n xdxor  cos n xdx, where n is a positive integer, can be treated using a reduction formulas. (This is not always the most efficient strategy.)  sin n xdx=− sin n−1 x cos x n + n − 1 n  sin n−2 xdx, n = 0  cos n xdx= cos n−1 x sin x n + n − 1 n  cos n−2 xdx, n = 0. See Exercise 29.5 for an outline of the derivation of these reduction formulas. If the powers of sin x and cos x are not positive, it can be useful to express the integrand using tan x and sec x. Integrals Involving tan x and sec x Recall that d dx tan x = sec 2 x, d dx sec x = sec x tan x, and tan 2 x + 1 = sec 2 x. (The Pythagorean identity is obtained from dividing sin 2 x + cos 2 x = 1bycos 2 x.) Again, we will proceed by example. Where possible, we attempt to transform the integral into the form  u n du. ◆ EXAMPLE 29.10 Evaluate  tan 2 x sec 4 xdx. SOLUTION Convert sec 2 x to tangents, saving a sec 2 xdxfor du when we let u =tan x.  tan 2 x sec 4 xdx=  tan 2 x(tan 2 x + 1) sec 2 xdx =  (tan 4 x + tan 2 x) sec 2 xdx Let u = tan x, du = sec 2 xdx. =  (u 4 + u 2 )du = u 5 5 + u 3 3 + C = tan 5 x 5 + tan 3 x 3 + C ◆ EXERCISE 29.8 Evaluate  tan x sec 3 xdx. Let u = sec x. Use tan x sec xdxas du. Answer sec 3 x 3 + C ◆ EXAMPLE 29.11 Find  tan 3 xdx. 29.2 Trigonometric Integrals and Trigonometric Substitution 889 SOLUTION  tan 3 xdx=  tan 2 x tan xdx =  (sec 2 x − 1) tan xdx =  tan x sec 2 xdx−  tan xdx =  tan x(sec 2 xdx)+  1 cos x sin xdx In the first integral we let u = tan x, du = sec 2 xdx. In the second, we let w = cos x, dw =−sin xdx.  tan 3 xdx=  udu−  1 w dw = u 2 2 − ln |w|+C = 1 2 tan 2 x − ln | cos x|+C ◆ The next integral is hard to figure out unless you’ve seen the method. It might be worth remembering the answer or programming it into your calculator (if it is not already there). ◆ EXAMPLE 29.12 Evaluate  sec xdx. SOLUTION  sec xdx=  sec x · sec x + tan x sec x + tan x dx =  sec 2 x + sec x tan x sec x + tan x dx Now the derivative of the denominator is sitting in the numerator. Let u = sec x + tan x, du = (sec x tan x + sec 2 x) dx.  sec xdx=  1 u du = ln |u|+C = ln | sec x + tan x|+C ◆ Integrals of the form  tan n xdxor  sec n xdx, where n is a positive integer, can be treated using reduction formulas.  tan n xdx= tan n−1 x n − 1 −  tan n−2 xdx, n ≥ 2  sec n xdx= sec n−2 x tan x n − 1 + n − 2 n − 1  sec n−2 xdx, n ≥ 2. The derivation of the first of these reduction formulas is given as Problem 46 at the end of this section. 890 CHAPTER 29 Computing Integrals EXERCISE 29.9 Use the reduction formula given above to find  sec 3 xdx. (Problem 45 at the end of this section leads the reader through this problem without reference to the reduction formula.) Answer  sec 3 xdx= 1 2 (sec x tan x +ln | sec x + tan x|) + C Other Trigonometric Integrals If you are faced with integrals of the form  sin Ax cos Bx dx,  sin Ax sin Bx dx,or  cos Ax cos Bx dx, you can use the appropriate product formula: i. sin Ax cos Bx = 1 2 [sin(A − B)x +sin(A +B)x] ii. sin Ax sin Bx = 1 2 [cos(A − B)x − cos(A + B)x] iii. cos Ax cos Bx = 1 2 [cos(A − B)x + cos(A + B)x] An alternative is to use integration by parts twice and solve algebraically for the original integral. These types of integrals will come up when you study Fourier series, a powerful tool in modern physics and engineering. We derive formula (i) below. The other product formulas can be derived similarly. sin(A + B)x =sin Ax cos Bx + sin Bx cos Ax + sin(A − B)x =sin Ax cos Bx − sin Bx cos Ax sin(A + B)x +sin(A −B)x = 2 sin Ax cos Bx 1 2 [sin(A + B)x +sin(A −B)x] =sin Ax cos Bx This is not meant to be an exhaustive discussion of trigonometric integrals. If you run into something you can’t handle, see what a computer algebra system or a calculator can do with it. Trigonometric Substitution Trigonometric substitutions can be useful in transforming integrals whose integrands con- tain √ c 2 − x 2 , √ x 2 − c 2 , and √ x 2 + c 2 .We’ll first illustrate why it can be useful to insert trigonometric functions into problems where they don’t appear to belong, and then specify which substitutions to use. ◆ EXAMPLE 29.13 Find  1 √ 9−x 2 dx. SOLUTION Method 1: Recall that  1 √ 1−u 2 du =arcsin u +C. We can transform the given integral into this form by dividing both numerator and denominator by 3, because 1 3 = 1 √ 9 . . Exercise 29.5 for an outline of the derivation of these reduction formulas. If the powers of sin x and cos x are not positive, it can be useful to express the integrand using tan x and sec x. Integrals. methods of transforming an integrand into something more manageable by exploiting trigonometric identities. Trigonometric Integrals Suppose we want to integrate the product of integer powers of trigonometric. = 1 2 (1 − cos 2x) and cos 2 x = 1 2 (1 + cos 2x) are useful if the powers of sine and cosine are both positive and even. They are used to reduce powers. On the other hand, if the power of either sine