Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 92 ppsx

29.2 Trigonometric Integrals and Trigonometric Substitution 891 1 √ 9 − x 2 = 1 3  9 9 − x 2 9 = 1 3 1  1 −  x 3  2  1 √ 9 − x 2 dx =  1  1 −  x 3  2 1 3 dx Let u = x 3 , du = 1 3 dx, = arcsin  x 3  + C Method 2: We’d love to get rid of that square root in the denominator. To do this, let’s replace 9 − x 2 by a perfect square. We can exploit the trigonometric identity 1 − sin 2 θ = cos 2 θ or 9 − 9 sin 2 θ = 9 cos 2 θ. Let x =3 sin θ, where θ ∈  − π 2 , π 2  , dx = 3 cos θdθ. Then √ 9 − x 2 =  9 − 9 sin 2 θ = √ 9 cos 2 θ = 3 √ cos 2 θ  1 √ 9 − x 2 dx =  1 3 √ cos 2 θ 3 cos θdθ 1 =  1 dθ = θ + C But x =3 sin θ,so x 3 = sin θ and θ = arcsin  x 3   1 √ 9 − x 2 dx = arcsin  x 3  + C This second method will come in very handy, particularly when the first method cannot be applied. Notice that although the integrand involves no trigonometry, the antiderivative does. ◆ ◆ EXAMPLE 29.14 Find  1 0 √ 4 − x 2 dx. SOLUTION The strategy is to replace 4 − x 2 by a perfect square in order to eliminate the square root. Let x =2 sin θ, where θ ∈  −π 2 , π 2  , dx = 2 cos θdθ.  4 − x 2 =  4 − 4 sin 2 θ =  4 cos 2 θ = 2  cos 2 θ What are the new endpoints of integration?  When x =0, 0 = 2 sin θ ⇒ sin θ = 0 ⇒ θ = arcsin 0 = 0. When x =1, 1 =2 sin θ ⇒ sin θ = 1 2 ⇒ θ = arcsin 1 2 = π 6 . 1 √ cos 2 θ is actually | cos θ| as opposed to cos θ. In the context of this problem, however, √ cos 2 θ can be replaced by cos θ because θ has been restricted to  − π 2 , π 2  and cos θ ≥0 on this interval. 892 CHAPTER 29 Computing Integrals  1 0  4 − x 2 dx =  π 6 0 2  cos 2 θ 2 cos θdθ = 4  π 6 0 cos 2 θdθ Use cos 2 θ = 1 2 [1 + cos 2θ]. = 2  π 6 0 (1 + cos 2θ) dθ = 2  θ + 1 2 sin 2θ      π 6 0 = 2  π 6 + 1 2 sin π 3 − 0  = π 3 + 1 2 √ 3 2 Clarifications and Justifications Notice that when we let x = 2 sin θ, we are restricting x to lie between −2 and 2. That’s fine, because √ 4 − x 2 only makes sense for x in this interval. In both of the previous examples we’ve said √ cos 2 θ = cos θ. Actually, √ cos 2 θ = |cos θ|, which is equal to cos θ only ifcos θ is nonnegative. It is because we’ve restricted θ to [− π 2 π 2 ] that we can write | cos θ|=cos θ. In fact, the substitution x =g(θ) is only valid if g(θ) is 1-to-1. In other words, when we say x = 2 sin θ we must restrict θ . In this case we choose θ ∈  −π 2 , π 2  , the same restrictions used for sin −1 x. This type of restriction will hold in all the examples that follow. Notice that once we’ve established that θ ∈  − π 2 , π 2  we can say, as above, if sin θ = 0, then θ = 0. ◆ Generalizing the Method of Trigonometric Substitution: k is a constant. Original integrand The square contains Substitution root transformed Restrictions on θ  k 2 − u 2 u =k sin θ  k 2 − u 2 = C cos θθ∈  − π 2 , π 2   k 2 + u 2 u =k tan θ  k 2 + u 2 = C sec θθ∈  − π 2 , π 2   u 2 − k 2 u =k sec θ  u 2 − k 2 = C tan θθ∈  0, π 2  ,  π, 3π 2  The restriction on θ for the substitution u = k sec θ is designed to guarantee that | tan θ|=tan θ. 2 Restrictions on θ specified above will hold in the examples that follow. CAUTION As always, try to make your work as simple as possible. The table supplied above does not mean you ought to use trigonometric substitution on  x √ x 2 −4 dx, for instance. Use ordinary substitution, u = x 2 − 4, instead. 2 In this particular case, care must be taken in converting an expression involving θ back to one involving u if sec θ is negative. 29.2 Trigonometric Integrals and Trigonometric Substitution 893 ◆ EXAMPLE 29.15 Find  1 x 2 √ 9+x 2 dx. SOLUTION Let x =3 tan θ, dx = 3 sec 2 θdθ, θ ∈  − π 2 , π 2  .  9 + x 2 =  9 + 9 tan 2 θ = 3  1 + tan 2 θ = 3  sec 2 θ = 3 sec θ  1 x 2 √ 9 + x 2 dx =  1 9 tan 2 θ 3 sec θ · 3 sec 2 θdθ = 1 9  sec θ tan 2 θ dθ = 1 9  cos 2 θ cos θ sin 2 θ dθ = 1 9  cos θ sin 2 θ dθ Let u = sin θ, du = cos θdθ. = 1 9  1 u 2 du = 1 9  u −2 du =− 1 9 u −1 + C =− 1 9 sin θ + C Now we must change from θ back to x. It is not appealing to write sin θ = sin  tan −1 x 3  without simplifying. Instead, we can use a triangle to sort out the conversion to x as follows: x =3 tan θ ⇒tan θ = x 3 . Draw a right triangle, label θ, and label the sides so that tan θ = x 3 . We determine the hypotenuse using the Pythagorean Theorem. (See Figure 29.1) 3 xx θ 3 θ ⇒ √9 + x 2 Figure 29.1 sin θ = x √ 9 + x 2 894 CHAPTER 29 Computing Integrals  1 x 2 √ 9 + x 2 dx =− 1 9 1 sin θ + C =− 1 9 √ 9 + x 2 x + C =− √ 9 + x 2 9x + C Check this by differentiating. ◆ ◆ EXAMPLE 29.16 Find  √ 4 − x 2 dx. SOLUTION This is the same integrand as in Example 29.14, so we use the same methods. We let x =2 sin θ and find   4 − x 2 dx = 2  θ + 1 2 sin 2θ  + C = 2θ + sin 2θ + C. The task now is to change from θ back to x.We’d like to use a triangle, as in the previous example, but we must first apply the trigonometric identity sin 2θ = 2 sin θ cos θ.   4 − x 2 dx = 2θ + 2 sin θ cos θ + C x = 2 sin θ ⇒ sin θ = x 2 . Draw a right triangle, label θ, and label the sides so that sin θ = x 2 . (See Figure 29.2.)   4 − x 2 dx = 2θ + 2 sin θ cos θ + C = 2 arcsin  x 2  + 2  x 2   √ 4 − x 2 2  + C = 2 arcsin  x 2  + x √ 4 − x 2 2 + C 22 xx θ θ ⇒ √4 – x 2 Figure 29.2 ◆ Can problems become more involved than this? Of course they can. Consider, for instance,  √ 4x −x 2 dx. √ 4x −x 2 can be expressed as  4 − (x −2) 2 by completing the square.   4 − (x −2) 2 dx is of the form  √ 4 − u 2 du and we continue as in the previous example. At some point, when the level of complexity overpowers the level of intrigue, you’ll turn to a computer or powerful calculator. Regardless, it is useful to have this method of trigonometric substitution in your array of tools. 29.2 Trigonometric Integrals and Trigonometric Substitution 895 PROBLEMS FOR SECTION 29.2 In Problems 1 through 18, evaluate the integral. 1.  cos 2 xdx 2.  π 0 sin 3 xdx 3.  cos x sin 2 xdx 4.  cos 3 x sin 2 xdx 5.  cos 4 xdx 6.  cos 2 x sin 2 xdx 7.  cos 4 x sin 3 xdx 8.  cos 3 x sin 11 xdx 9.  cos 3 (3x) dx 10.  π 2 0 cos 5 x √ sin xdx 11.  sin x √ cos 3 x dx 12.  tan 3xdx 13.  tan 2x sec 2xdx 14.  π 4 0 √ tan x sec 2 xdx 15.  tan x sec 4 xdx 16.  tan 3 x sec 4 xdx 17.  tan 3 x sec 5 xdx 18.  tan 3 x sec xdx 19.  tan 3 x sec 4 x dx 20.  tan 8 x sec 4 xdx 21.  sin x cos 2 x dx Problems 22 through 24 refer to the information provided about Fourier series. A Fourier series expresses a function as a weighted infinite sum of terms of the form 896 CHAPTER 29 Computing Integrals sin nx and cos nx, where n is a nonnegative integer. Fourier series are a very powerful tool. In order to construct such a series we need the following results. m and n are positive integers. (a)  π −π sin mx cos nx dx =0 (b)  π −π sin mx sin nx dx =  π if m = n 0ifm = n (c)  π −π cos mx cos nx dx =  π if m = n 0ifm = n These results can be obtained using the product formulas given in this section. 22. Prove statement (a) above. 23. Prove statement (b) above. 24. Prove statement (c) above. 25. Find the volume generated by revolving the region under one arch of cos x about the x-axis. 26. Find the volume generated by revolving the region between the graphs of y = sin x and y =cos x from x = π 4 to x = 5π 4 around the horizontal line y =2. 27. Find  1 sin θ dθ. 28. Evaluate  sec 3 xdxusing a reduction formula. In Problems 29 through 43, evaluate the integrals. Not all require a trigonometric substitution. Choose the simplest method of integration. 29.  x 3 √ 4 − x 2 dx 30.  x 3 √ 9−x 2 dx 31.  x √ 4+x 2 dx 32.  x 3 √ 4+x 2 dx 33.  1 √ 9+x 2 dx 34.  x 2 √ x 2 − 9 dx 35.  3 4 0 √ 9 − 4x 2 dx 36.  √ 4 − 9x 2 dx 37.  x √ 4 − 9x 2 dx 29.2 Trigonometric Integrals and Trigonometric Substitution 897 38.  3 0 dx (4+x 2 ) 3 2 39.  √ x 2 −4 x dx 40.  x √ x 2 −1 dx 41.  2 2 √ 3 √ x 2 −1 x dx 42.  2x−3 √ 9−4x 2 dx 43.  x 4−x 2 dx 44. Find  √ k 2 − x 2 dx for any constant k. 45. Show that the area enclosed by the ellipse x 2 a 2 + y 2 b 2 = 1, where a and b are positive constants, is given by πab. y b –b –aa x x 2 a 2 y 2 b 2 += 1 Hint: Find the area in the first quadrant and multiply by 4. Express y as a function of x. Notice the similarity between the formula for the area inside a circle and the formula for the area inside an ellipse. 46. Find  sec 3 xdx as follows: Use integration by parts with u = sec x. The resulting new integral will contain tan 2 x. Replace tan 2 x by sec 2 x −1 and split the integral into the difference of two integrals,  sec 3 xdx−  sec xdx. Integrate the latter and solve algebraically for the former. 47. Derive the reduction formula  tan n xdx= tan n−1 x n − 1 − tan n−2 xdx, where n is an integer greater than or equal to 2. (To do this, rewrite the integrand as tan n−2 x ·tan 2 x and use a Pythagorean identity to convert tan 2 x into an expression involving sec x.) 48. Derive the formula sin Ax sin Bx = 1 2 [cos(A − B)x − cos(A + B)x] given in this section. (Begin with the addition formula for cosine.) 898 CHAPTER 29 Computing Integrals 29.3 INTEGRATION USING PARTIAL FRACTIONS Practical motivation: Models commonly used in epidemiology and population biology are often based on what is known as a logistic growth model. For instance, while population growth may initially look exponential, due to limited resources growth generally levels off at what is known as the carrying capacity for the population. In order to derive an expression for P(t), the population at time t, one computes  1 kP(P−L) dP, where L is the carrying capacity and k is a constant determined by the population. The methods of this section enable us to calculate this integral. Suppose an integral is of the form  P(x) Q(x) dx where P(x) and Q(x) are polynomials and Q(x) factors. The method of partial fractions is an algebraic technique enabling us to split up the integrand into the sum of simpler rational functions. For example, we know that 1 x +2 − 1 x +3 = x +3 −(x + 2) (x +2)(x +3) = 1 x 2 + 5x +6 . Suppose we are struggling to find  1 x 2 +5x+6 dx. If we replace the integrand by the equiv- alent expression 1 x+2 − 1 x+3 , our struggles are over.  1 x 2 + 5x +6 dx =   1 x +2 − 1 x +3  dx =  1 x +2 dx −  1 x +3 dx = ln |x + 2|−ln |x +3|+C = ln     x +2 x +3     + C The method of decomposition into partial fractions is a method of integrating rational functions, functions of the form P(x) Q(x) , where P(x) and Q(x) are polynomials by decom- posing P(x) Q(x) into the sum of simpler rational functions. The method of partial fraction decomposition works only if P(x) Q(x) is a proper fraction, that is, only if the degree of P is strictly less than the degree of Q(x):deg(P) < deg(Q). Suppose deg(P ) ≥deg(Q); the fraction is improper. Then we can do long division to split the fraction into a sum of a polynomial and a proper rational function. We can use the method of partial fractions on the proper part. We illustrate this below. ◆ EXAMPLE 29.17 Integrate  x 2 x 2 −1 dx. SOLUTION Doing long division we get x 2 x 2 − 1 = 1 + 1 x 2 − 1 or 1 + 1 (x −1)(x + 1) . 29.3 Integration Using Partial Fractions 899 1 (x−1)(x+1) can be decomposed into a sum of the form A x−1 + B x+1 where A and B are constants. 1 (x −1)(x + 1) = A x −1 + B x +1 Clear denominators. 1 = A(x +1) + B(x − 1) Gather x terms; gather constants. 0x +1 =(A +B)x +(A −B) This can be true only if A + B =0 and A − B =1. Solve these simultaneous equations to obtain A = 1 2 and B =− 1 2 . Then x 2 x 2 − 1 = 1 + 1 2 1 x −1 − 1 2 1 x +1 .  x 2 x 2 − 1 dx =   1 + 1 2 1 x −1 − 1 2 1 x +1  dx = x + 1 2 ln |x − 1|− 1 2 ln |x + 1|+C ◆ The first step in partial fraction decomposition of a proper rational function is factoring the denominator. It can be proven that any polynomial can be factored into the product of linear and irreducible quadratic factors. A quadratic ax 2 + bx + c is irreducible if it cannot be factored into the product of two linear factors, that is, if the discriminant, b 2 − 4ac,is negative. We’ll look at polynomials that are not hard to factor. Computer algebra systems can take care of more difficult factoring jobs. A proper rational function can be decomposed into the sum of proper rational functions. We’ll look at some examples before giving the general procedure. Suppose the denominator factors into the product of distinct linear factors. For example, consider −3 (x+1)(x−2) .Wecan express −3 (x+1)(x−2) as a sum of the form A x+1 + B x−2 . The task of determining A and B reduces to simultaneous equations. −3 (x +1)(x − 2) = A x +1 + B (x −2) Get a common denominator. −3 (x +1)(x − 2) = Ax −2A + Bx + B (x +1)(x − 2) Equate the numerators. (Equivalently, clear denominators.) −3 = Ax −2A + Bx + B Collect like powers of x. −3 = (A + B)x + (B −2A) 0x −3 =(A +B)x +(B − 2A) Equate the coefficients of x; equate the constant terms. Then  A + B =0 −3 = B −2A. simultaneous linear equations A =−B so − 3 = B −2(−B) − 3 = 3B − 1 = B and A = 1, Then −3 (x +1)(x − 2) = 1 x +1 − 1 x −2 . Check by adding. 900 CHAPTER 29 Computing Integrals Knowing the partial fraction decomposition of −3 x 2 −x−2 makes finding its antiderivative a snap.  −3 x 2 − x −2 dx =  1 x +1 − 1 x −2 dx = ln |x + 1|−ln |x −2|+C = ln     x +1 x −2     + C There’s a shortcut to finding A and B.Given −3 (x+1)(x−2) = A x+1 + B x−2 , clear denominators and then evaluate both sides at values of x that would have made the denominator zero. It can be shown that this is legal; certainly it is quick. − 3 = A(x −2) + B(x + 1) Evaluate at x =−1: − 3 = A(−1 − 2) ⇒ A = 1 Evaluate at x = 2: − 3 = B(2 + 1) ⇒ B =−1 We’ll apply this shortcut in the next example. ◆ EXAMPLE 29.18 Find  x+1 x 2 −3x 2 +2x dx. SOLUTION The integrand is a proper fraction, so we factor the denominator. x 2 + 1 x 3 − 3x 2 + 2x = x 2 + 1 x(x 2 − 3x +2) = x 2 + 1 x(x − 2)(x − 1) The denominator has distinct linear factors, so we set up partial fractions as follows. x 2 + 1 x(x − 2)(x − 1) = A x + B x −2 + C x −1 x 2 + 1 = A(x −2)(x −1) + Bx(x − 1) + Cx(x − 2) Evaluate at x = 0: 1 = A(−2)(−1) ⇒ A = 1 2 Evaluate at x = 2: 5 = B(2)(1) ⇒B = 5 2 Evaluate at x = 1: 2 =C(1)(−1) ⇒ C =−2 x 2 + 1 x 3 − 3x 2 + 2x =− 1 2 x + 5 2 x −2 − 2 x −1 Therefore  x 2 + 1 x 3 − 3x 2 + 2x dx = 1 2  1 x dx + 5 2  1 x −2 dx − 2  1 x −1 dx = 1 2 ln |x|+ 5 2 ln |x − 2|−2ln|x − 1|+C ◆ . and Q(x) are polynomials and Q(x) factors. The method of partial fractions is an algebraic technique enabling us to split up the integrand into the sum of simpler rational functions. For example,. first step in partial fraction decomposition of a proper rational function is factoring the denominator. It can be proven that any polynomial can be factored into the product of linear and irreducible. greater than or equal to 2. (To do this, rewrite the integrand as tan n−2 x ·tan 2 x and use a Pythagorean identity to convert tan 2 x into an expression involving sec x.) 48. Derive the formula