MATH111-200630-PS04-Solutions

MATH111-200630-PS04-Solutions - MATH 111 Problem Set 4...

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Unformatted text preview: MATH 111 Problem Set 4 Solutions Edward Doolittle October 24, 2006 You should check the answer by differentiating. Again, you should check the answer by differentiating. (c) Following the usual trick for evaluating the integrals of inverse trig functions, introduce a factor of 1 and cos 1 x, g x 1, f x 1 1 x2 , g x x: then apply integration by parts with f x 1 x2 As usual, check by differentiating. (d) This is a case where integration by parts doesn’t seem to go anywhere; we need to aply it twice and solve sin 2θ , g θ e3θ . Then f θ 2 cos 2θ , g θ 1 3 e3θ, for the required integral. First, let f θ and Solving for the required integral, As usual, you should check the result by differentiation. 1   ¡ ¦ #¡ ¢ e3θ sin 2θ dθ 9 13 1 3θ e sin 2θ 3 2 3θ e cos 2θ 9 C  ¡ ¦ ¡ ¦ ¡ ¢ e3θ sin 2θ dθ e3θ sin 2θ dθ 1 3θ e sin 2θ 3 2 3θ e cos 2θ 9 4 9 © Substituting (2) into (1),  ¡  "¡ ¢ e3θ cos 2θ dθ e3θ sin 2θ dθ 1 3θ e cos 2θ 3 2 3 ¡ ¨ ¢ ¡ ¡ ¢ ¡ Let’s try again, this time with f θ cos 2θ , g θ e3θ , f θ 2 sin 2θ , g θ ¦ !¢ ¡ ¤  ¡ ¢ ¡ ¤ ¡ © ¦ ¡ ¢ e3θ sin 2θ dθ e3θ cos 2θ dθ 1 3θ e sin 2θ 3 2 3 1 3 e3θ: (2) ¡ ¨  ¢ ¡ ¡ ¦ ¦  ¢ ¡ ¤ ¢  ¨ ¢ ¡ ¤ ¦  ¡ ¢  © ¢ ¥¡ ¦  ¢ ¡ ¡ ¡ ¡  cos x dx x cos x x cos x C x cos x 1   ¡ 1 1 1 1 1 2 du u1 2 1 u1 2 21 2  ¢ ¦ ¢ ¦© ¢ Now, making the substitution u 1 x2, du 2x dx, x dx du ¡ ¦ ¨  ¦    ¢  cos 1 x dx x cos 1 x x dx 2, x2 1 2 ¢ ¡ ¦  ¦ ¢ ¨ ¡ ¤  ¢ ¡ ¤  ¦  ¢ ¡ ¢ © ¦ ¢ reπr dr 1 πr re π 1 π eπr dr 1 πr re π 1 πr e π2 C ¡ ¨ ¢ ¡ ¢ ¡ (b) Similar to the previous answer, let f r tion by parts formula gives r, g r eπr . Then f r 1, g r  1 π eπr , and the integraC (1)  ¢ ¡ ¤  ¦ ¢ ¢ ¡ ¤  © ¦ ¢ x sin 7x dx 1 x cos 7x 7 1 7 cos 7x dx 1 x cos 7x 7 ¡ ¨ ¦ ¢ §¥¡ ¢ £¡ 1. (a) Let f x gives x, g x sin 7x. Then f x 1, g x © ¢ ¥¡ ¤ ¢ ¥¡ ¤ © © © © © © © © 1 7 cos 7x, and the integration by parts formula 1 sin 7x 49 C Check by differentiating and applying trig identities. Reversing the substitution, Check by differentiating and applying trig identities. (c) Let I be the integral in question. The powers of sin x and cos x are both even, so pair off sines and cosines and use double angle identities: The first term can be evaluated by another double angle formula: Assembling the above results, As usual, you should check by differentiating and applying trig identities. (d) The power of sect is even so convert sec2 to tan (leaving sec2 in the integrand) and make the substitution u tant, du sec2 t dt: As usual, check by differentiating and applying trig identities. 3. It’s probably best to evaluate the indefinite integrals first, because then you can check your results by differentiating. However, just for the sake of variety and for the sake of applying a technique that we have learned that may be useful in some circumstances, I have solved the problems in this section using the integration by parts rule for definite integrals. You should check these results by evaluating the indefinite integrals first and then evaluating the definite integrals. 2    ¢   ¢ ¡  ¢ ¡  ¢ tan2 t sec4 t dt tan2 t 1 tan2 t sec2 t dt u2 1 u2 du   "¡ © ¦ ¡ ¦ ¢ sin4 θ cos2 θ dθ 1 θ 16 1 sin 4θ 64 1 sin3 2θ 48 C u3 3 u5 5 C 1 3 tan t 3   "¡ ¦ ¢  ¦ ¢ ¦ ¢ cos 2θ sin2 2θ dθ u2 du C ¡ © ¡ 1 8 1 16 1 3 u 48 1 sin3 2θ 48 C 1 5 tan t C 5 ¡ ¢ ¨ ¡ ¢ ¡ ¢ The second term can be evaluated by the substitution u sin 2θ , du 2 cos 2θ dθ, du 2 cos 2θ dθ:   "¡ ¦ ¢ ¡ © ¦ ¢ sin2 2θ dθ 1 cos 4θ dθ ¡ 1 8 1 16 1 θ 16 1 sin 4θ 64 C  ¡ %¡ ¡ ¦ $¡ ¢ " ¢ © ¡ ¢ © © ¢ I sin2 θ sin θ cos θ 2 dθ sin2 2θ cos 2θ sin2 2θ dθ ¡ ¡ ¦ 1 cos 2θ sin2 2θ dθ 2 4 1 8 ©   ¦ ©  ¦ !¢ sin5 x cos4 x dx 1 cos5 x 5 2 cos7 x 7 1 cos9 x 9 C   ¦  ¦ ¢ ¡  ¦ ¦ ¢ ¡  ¦ ¦ !¢ sin5 x cos4 x dx 1 2u2 u4 u4 du © Expanding the polynomial and integrating, u4 2u6 u8 du  ¡ ¦ ¦ ¢ ¡ ¦ © ¢ sin5 x cos4 x dx 1 cos2 x 2 cos4 x sin x dx 1 u2 2 u4 du u5 5 2u7 7 u9 9 ¢ (b) The power of sin x is odd, so convert as many sin x to cos x as possible and make the substitution u du sin x dx: cos x, C   ¦ ¢  ¦ ¢ ¡ © ¦ ¢ ¡ ¦ © ¢ sin2 x cos3 x dx sin2 x 1 sin2 x cos x dx u2 u4 du u3 3 u5 5 C 1 3 sin x 3 1 5 sin x C 5 ¢ © © © ¦ ©¢ © © © ¢© © 2. (a) The power of cos x is odd, so convert as many cos x to sin x as possible and make the substitution u sin x: ¦ π 0 0 0 0 2 1 1 1 1 1 1 1 1 4 1 4 1 1 Substituting (4) into (3) gives 1 8 0 0 0 0 (Check by differentiating.) Now, evaluating the definite integral, π 2 (Check by differentiating.) Evaluating the definite integral, 9 9 4 4 (c) Let I be the desired integral. The power of tan is odd so change as many tan to sec as possible and substitute u sec θ: As usual, check by differentiating and applying trig identities. 3   ¦ ¢  ¦ ¢ ¦ ¢ ¡ ¦ ¢ I sec2 θ 1 sec2 θ sec θ tan θ dθ u4 u2 du 1 5 u 5 1 3 u 3  ¦ ¢  ¦ © ¦ ¢ ¦  ¢ e x dx 2 xe x 2e x 6e3 2e3 4e2 2e2 4e3   ¦  ¢  ¦ ) ¢ ) ¢ e x dx 2 ueu du 2ueu 2eu C 2 xe x 2e ) ) ¢ ©  ¢ (b) As above, let u x, 2u du ¢ cos x dt dx and integrate by parts:  2 cos π1 4 2π1 4 sin π1 4 2 cos π 2 1 4 2π 2 1 4 sin π 2 x C 2e2 C 1 sec5 θ 5  2¡  ¡ ¨ %  ¡ ¨  ¡  ¡ ¨ 1 ¦ ¡  ¦ ¡  π     ¦  ¢   ¦ ¢ ¢ cos x dx 2u cos u du 2u sinu 2 cosu C 2 cos x 2 x sin x C 1 4 1 sec3 θ 3 ¡  ¨ ¢  ¢ 4. (a) First evaluate the indefinite integral. The obvious thing to try is the substitution u 2u du dx. After the substitution we integrate by parts: x, du dx 2 x , ¡ 2 2   ¡ ¦ ¢ ¡ ¦ ¢ ¦ ©  ¢ t2t dt 1 t t2 ln 2 8 1 ln 2 8 2t dt 211 ln 2 ¡ ¨ ¢ ¡  )   © ) ¢ ¡ © (d) Let f t t, g t 2t ; then f t 1, g t 1 ln 2 2t , and 1 2t ln 2 2 8  ¦ ¢ ¡ ¦ ¢© ¡ ¤ ¦ ¢ ¢ ¡ ¤ ¡  x2 4 ex dx 20e4 5e 2 3e4 4 14e4 5e  ¢  ¦ ¦ ¢ ¦ ¢ xex dx xex 4 ex dx 4e4 e e4 e 3e4 211 ln 2 28 ln 2 1 ln 2 ¢ ¡ ¢ ¡ We need to integrate by parts again. Letting f x x, g x ex , we have f x 1, g x ¢ ¡ ¤  ¦ ¢ ¡ ¤ ¦ ¢ ¦ ¡  © ¢ ¡  x2 4 ex dx 4 4 x2 4 ex 4 2xex dx 20e4 5e 2 © ¢ ¡  ¢ ¡ © (c) Let f x x2 4, g x ex ; then f x 2x, g x ex , and 4 xex dx   ¦ ¦ !¢  ¦ ¦ ¢ ¢© ¡ ¤   ¢ ¡ ¤ ¦ (¢ ln x dx x3 ln x 2x2 2 1 2 2 x 3 dx 1 ln 2 8 1 x 4 2 ¡ ¢ '¡ ¨  ¢ ¡ © (b) Let f x ln x, g t x 3 ; then f x 1 x, g x x 2 2 , and the integration by parts formula gives 2 1 ln 2 8  1 16 1 4 (3) ex , and (4) C ¢ ¦ ¨   ¢ ¢ &¡ ¤ © ¦ ¢ &¡ ¤ ¢ t cos 4t dt 1 t sin 4t 4 π 1 4 π sin 4t dt 0 1 cos 4t 16 ¡ ¨ ¢ ¡ ¢ ¡ © ) 0© © © ¢© © © (a) Let f t t, g t cos 4t. Then f t 1 and g t ¢ ©¡ ¤ ¢ ¡ ¤ 1 4 sin4t and π 0 ¢ The second integral in the right side of (5) can be evaluated by again applying the Pythagorean identity: Assembling our results, (Check by differentiation and application of trig identities.) Evaluating the definite integral, 0 0 The integral can now be written in terms of sec x and tan x: Alternatively, you can write the answer as 1 Check by differentiating. (b) The power of tan θ is even, and the power of sec θ is odd, so there is no simple substitution. Integration by parts is necessary in this case, but you can reduce the amount of work required by changing all the tan θ to sec θ, and applying a known reduction formula, namely formula 77 from the integral tables at the back of the textbook: Recall from the lectures that By reduction formula (6) and formula (7), 4   87  7  ¢  ¢ sec3 θ dθ 1 tan θ sec θ 2 1 2 sec θ dθ  ©  87  7 ¢ sec θ dθ ln tan θ secθ C ¦ ¦ n 1 1 tan θ 2    ¢ secn u du tan u secn 2 u ¦ 1 n n 2 1 secn 2 u du 1 ln tan θ 2 secθ C 6  ¦ ¢ © ¡ ¦ ¢ tan4 θ sec θ dθ sec2 θ 1 2 sec θ dθ ©   ©  ¢ dx sinx 1 sinx cos x C sec5 θ    ¢  ¢ dx 1 sinx sec2 x secx tan x dx tan x ¦  5 ¦ © ¦ 1 1 secx C 2 sec3 θ secθ dθ   ¢  ¢  ¢ dx sinx 1 1 sinx 1 sinx dx sinx 1 sinx dx 1 sin2 x 1 sinx dx cos2 x ¡  © © ¦ ¦ © 5. (a) Multiply the integrand by the ‘conjugate trig function’ 1 sinx 1 sinx : ¦ ¦ 3¢  ¦ ¦   © 4¡ ¨ ¦ ¢  ¦ ¢ tan4 x dx 1 3 tan x 3 tanx x  π 4 π 4    ¦ ¢ tan4 x dx 1 3 tan x 3 tan x x C 1 3 π tan 3 4    ¦ ¢  ¦ ¢ tan2 x dx sec2 x 1 dx tan x x C   ¢  ¢ ¢ tan2 x sec2 x dx u2 du u3 3 C 1 3 tan x 3 C tan π 4 π 4 1 3 tan 0 3 tan0 ¢ ¢ The first integral in the right side of (5) can be evaluated by the substitution u  tan x, du sec2 x dx: 0 2 3 ¦ ¢ © © ¢ tan4 x dx tan2 x tan2 x dx tan2 x sec2 x tan2 x dx ¢  (d) The strategy is to change tan2 into terms of csc2 using the Pythagorean identity tan2 x 1 sec2 x: (5) © © ©  ¦ © © © © © © © © © © π 4 (6) (7) (8) By reduction formula (6) and formula (8), Assembling the above results, Evaluating the definite integral, 6. We’re going to end up in some difficulty with notation further down the line, so I recommend changing f to g in the statement of the problem; now we need to prove that ga a This is one case where it’s very helpful to use the definite integral forms of the substitution rule ga a a a ga a a a a ga a as required. Without the definite integral forms of the substitution and integration by parts rules, the problem is rather messy. The result is illustrated in Figures 1 and 2, which also provide an alternate argument for the result. Figure 2 is just Figure 1 flipped along the line y x. We calculate the areas of various regions in the diagrams. Region A is just a rectangle, so its area is a f a . Region B is the region under the curve y f x , so its area is ab f x dx. f b Region C in Figure 2 is just the region under the curve x f 1 y so its area is f a f 1 y dy. The sum of the three areas A, B, and C is a rectangle the area of which is b f b . Therefore a f a which is equivalent to the required result. (Question: what happens if f x is negative for some values of x? What happens if a is negative?) 5 ¡ ¡ ¢ ¡   ¡  ¡ af a f x dx f 1 y dy bf b @ 9 @ 9 b f b ¡ A ¡  @@ 99 A ¡ ¢ ¡ ¡  ¢ 6 ¡ ¦ #¡ ¢ ¡ ¦ ¡ © ¢ ¡ ©  g 1 y dy bg b ag a g x dx gb © Putting it all together,  ¡ ¦ ¡ ¢ xg x dx xg x g x dx ¡ ¤ b b b b ¢ ¡ Now applying the integration by parts rule (11) with f x x, f x ¢ ¡ ¤ ¢ © ¢ ¡  g 1 y dy g 1 g x g x dx xg x dx a ¡ ¤ ¡ ¤ %¡ ¡  gb b b 1,  ¢ Let’s work on the left hand side of (9). First use the substitution rule (10) with f cancellation equation g 1 g x x:  ¡ © ¦ ¡ ¢ ¡ %¡ ¡ ©  ¢ f x g x dx f xgx f x g x dx ¡ ¤ ¡ ¤ ¡ b b © and the integration by parts rule ¢ f y dy f gx a ¡ ¤ 1¡ ¡ ¡ @ 9 @ 9 gb b g x dx b  ¡  ¡ ¦ ¡ © ¢ g 1 y dy bg b ag a g x dx @ 9 ¡  @ 9 gb b  0 g 1 , followed by the  4¡    ¦   "¡ ¦ ¡ 2 ln 1 ¢ ¡  © ¢ tan4 θ sec θ dθ 3 21 2 ln 1  ¡ π 4 1 1 4 5 1 8 1 2   87  7  ¦ ¢ tan4 θ sec θ dθ 1 tan θ sec3 θ 4 5 tan θ 8 ln tan θ secθ C 5 8 2  (9) (10) (11)  87  7   ¢  ¢ sec5 θ dθ 1 tan θ sec3 θ 4 3 4 ©  © © © sec3 θ dθ 1 tan θ sec3 θ 4 3 tan θ 8 3 ln tan θ 8 secθ C @ 9 @ 9 @ 9 @ 9 © © © © © © 0 0 a b x b B a A 0 0 C 6 ¡ B B ¡ ¡  Figure 2: Graph of f 1 y, f a y f b ¡ ¡ f a f b ¡  f 1 x y B B Figure 1: Graph of f x , a ¡ ¡ f a ¡ f b C B A ¡ x b y f x y ...
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This note was uploaded on 01/12/2010 for the course MATH 111 taught by Professor Doolittle during the Fall '06 term at Berkeley.

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