MATH111-200630-PS05-Solutions

MATH111-200630-PS05-Solutions - MATH111-002 200630 Problem...

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Unformatted text preview: MATH111-002 200630 Problem Set 5 Solutions DRAFT Edward Doolittle October 30, 2006 1. (a) Make the substitution x 4 sec θ, dx Now making the change of variables u 1 16 sin θ, du 1 u3 16 3 4 cos θ dθ, C By the right triangle with sides A O H x2 16 x , sin θ x2 16 dx x4 x2 16 48x3 3 2 C Check by differentiating. (b) It’s probably easier to make an algebraic substition here; try it. However, the trig substitution also works. Let x 3 sin θ, dx 3 cos θ dθ. Then 9 x2 9 9 x2 dx 3 1 sin2 θ C 3 1 x 3 2 C 9 Check by differentiating. The definite integral is now x x2 dx 0 9 (c) Let x a sin θ, dx a cos θ dθ. Then dx a2 5 2 Now making the substitution u 1 a2 tan θ, du u2 du sec2 θ dθ, 1 u3 a2 3 C By using a right triangle with the appropriate sides, a2 5 2 dx Check by differentiating. ¥ ¦ © ¤ § © ¤ § x2 x2 1 x3 2 a2 3a x2 3 2 C 1 ¥ ¦ tan2 θ sec2 θ dθ ¦ 1 a2 tan3 θ 3a2 C ¥ ¡ © ¤ § x2 x2 a2 sin2 θ a cos θ dθ a5 cos5 θ ¡ ¥ ¤ ¤ ¢¤ ¢ 5 9 9 5 3 2 1 1 a2 tan2 θ sec2 θ dθ ¥ ¦ ¤ ¤ ¦© § ¤ ¤ ¦ ¤ ¡ ¤ x ¦ Using the identity sin2 θ cos2 θ ¤ dx 9 sin2 θ 1 or the appropriate right triangle, ¥ ¦ 3 cos θ dθ 3 sin θ dθ ¤ x 3 sin θ  ¤ ¢ ¥ ¦ sin2 θ cos θ dθ 1 16 u2 du 1 sin3 θ 48 C x2 16 x, so putting it all together 3 cos θ C x2 C ¥ x2 16 dx x4 16 sec2 θ 16 4 sec θ tan θ dθ 44 sec4 θ 4 tan θ tan θ dθ 43 sec3 θ 1 16 ¡ ¡ ¥ ¦ © ¤ § © ¨ ¤ ¢¨ § © ¨ ¨ § ¦ ¡ ¡ ¤ ¢ £¡ 4 sec θ tan θ dθ: sin2 θ cos θ dθ ¡ ¤ ¢ ¤ £¡ ¢ ¤ £¡ ¢ ¤ ¢ ¤ ¢ ¡  ¡ ¡ ¡ ¡ ¡ ¡ (d) Let x a tan θ, dx a2 5 2 dx Since the power of cos is odd, make the substitution u sin2 θ cos θ dθ u2 du sin θ, du cos θ dθ: Reversing the substitutions using the appropriate right triangle, dx C a2 5 2 Check by differentiating. 2. (a) Let 4x 3 sec θ, 4dx dx 16x2 9 3 sec θ tan θ dθ. Then 9 sec2 θ 9 tanθ Reversing the substitution, tan θ dx 16x2 where (b) Here the algebraic substitution may work better than a trig substitution. Let u Then x 4 9x2 dx u1 2 4 9x2, du du 18 1 u3 2 18 3 2 C 1 4 27 9x2 3 2 C (Check by differentiating.) Evaluating the definite integral, x 4 9x2 dx 90 2 3 2 4 0 To four decimal places I get the answer to the question is 0 2963. (c) Make the trig substitution 2x tan θ, 2dx sec2 θ dθ, we have By example 8 on page 523 of the textbook, Check by differentiating. Now the definite integral is 0 ¦ 4x2 1 dx 1 5 1 ln 2 5 2 0 1 ln 2 1 0 5 1 ln 2 2 2 ¥© ¢ ¦ § ¦ ¦ ¢  ¢ ¤ ¤  ¦ ¢ ¦ ¢ 1 ¦ ¦ ¦ ¦ 4x2 1 dx ln sec x tanx C x 4x2 1 4x2 5 ¥ ¦ ¦ ¦  ¦ ©  ¦ § ¦ 4x2 1 dx 1 sec x tan x 2 ¥ sec2 θ sec2 θ dθ sec3 θ dθ ¥ ¤ ¡ ¤ ©© § ¤ § ¢ ¡ ¤ 2 3 1 4 27 1 27 2 9 3 2 3 2 8 27 1 ln 2 1 ¤ ¥ ¦  © ¤ § ¤ ¦  ¤ ¤  ¤ ¡ ¡ ¤ © §¤ ¦ ¤ ¤  ¦  ¤ ¤   ¤ ¢ ¡  ¤ ¢ ¤ ¢ ¦ ¤ ¤ ¢ ¢ ¡ ¡ ¡ sec2 θ 16x2 1 16x2 C 9 3 and 9 1 ln 4x 3 4 9 3 1 ln 4x 4 16x2 9 1 4 ln3 has been absorbed into C. Check by differentiating. ¥ ¦ ¥   sec θdθ 3 4 sec θ tan θ dθ 1 4 ¦ ¥ ¦ © § ¦ © ¦ § x2 x2 1 sin3 θ a2 3 x3 1 2 a2 3a x2 ¦ 1 a2 1 a2 1 u3 a2 3 C 3 2 C 1 ln sec θ 4 ¥ ¡ ¡ © ¦ § x2 x2 a2 tan2 θ a sec2 θ dθ a5 sec5 θ ¡ 1 a2 ¡  ¡ ¡ ¡ ¡ ¡ ¡ a sec2 θ dθ. Then sin2 θ cos θ dθ C C 18x dx. 2x C (d) Let ax b sec θ, a dx dx a 2 x2 b 2 5 2 It’s possible to do the integral by changing the quotient of sines and cosines to a product of cotangent and cosecant, but since we are avoiding the use of those functions if possible, let’s try another way. Change cos2 in the numerator to a function involving sines and make the substitution u sin θ, du cos θ dθ: 5 2 u 4 u 2 du Reversing the substitutions using a A O H b a 2 x2 C b2 1 2 ax right triangle, 3 2 dx 2 x2 a b2 5 2 1 ab4 csc3 θ 3 cscθ 1 ab4 a 3 x3 3 a 2 x2 b 2 Check by differentiating. 3. These problems require completing the square. By repeated application of reduction formula 74 at the back of the textbook, x2 2 4 Reversing the substitution by a 1 x 1 1 x2 2x 2, and x2 dx 2x 2 4 x2 2x 2 right triangle, sin θ 1 x 1 2 6 x 2x 2 3 5 x 1 2 24 x 2x 2 9 8 t (b) Completing the square, t 2 6t 1 2 2 sec θ, dt 2 2 sec θ tan θ dθ, t2 3 dt 6t t2 6t 1 Reversing the substitution, sec θ t 3 2 2 , tan θ 1 t 3 2 8 1 t2 3 dt 2 t 6t 1 3 ln t 3 2 2 t2 6t 2 2 C 3 ln t 3 by the law of logarithms ln a b tiating.) ln a lnb, absorbing the constant 6t ln 2 2 into C. (Check by different 3 2 Check by differentiating. ¦ ¤ ¢ t 3 t2 6t 1 dt ¥ ¦ ¦ ¤ ¦   ¡ © ¤§ ¤© ¤§ (c) As with the previous problem, complete the square to obtain t 2 the algebraic substitution u t 3 2 8, du 2 t 3 dt: 1 du 2u1 2 1 u1 2 21 2 C t2 6t 1 C 3 ¤ © ¤§ ¢ ¦ ¦ ¤ © ¢ ! ¦ ¤ ¢ § ¥ ¦ ¦ ¦ ¤ ¤ ¦ ¤ ¤ © ¤§  3 sec θ dθ 6 2 sec θ tan θ dθ 2 2 tan θ ¤ © ¤ ¤ ¦ ¦ ¦ © ¥ ¦© ¦ §  ¦ ¦ ¦ ¦ © ¦ § ¢ ¥ ¦ ¦ ¦ ¦ (a) Write x2 2x sec2 θ dθ, 2 dx 2x x2 2x 1 1 x 1 2 1. Then making the substitution x 1 tan θ, dx x2 2 4 sec2 θ dθ sec8 θ cos6 θ dθ dx 2x 1 sin θ cos5 θ 6 5 sin θ cos3 θ 24 15 sin θ cos θ 48 15 θ 48 C x 1 x2 2x 2, cos θ 2 15 x 1 48 x2 2x 2 15 tan 48 1 x 1 C 3 2 8. Making the substitution t 3 3 ln sec θ tan θ C t2 6t 1 2 2, 6t 1 C 8, but this time make ¦ dx 2 x2 a b2 1 ab4 ¦ ¥ ¦ © ¤ § ¤ © ¤ § ¤ ¤ ¦ © ¨  © ¤ §¨ § © ¨ ¨ §  ¤  © ¤ § ¡ ¡ 1 1 ab4 1 ab4 a 2 x2 ¥ ¦ ¤ ¤ ¤ sin2 θ cos θ dθ sin4 θ u3 3 ax b2 ¥ ¤    ¦ ¦ ¦ ¢ ¤ ¢ ¦ ¢¤ © ¢ § © ¤ § ¦ ¢ ¢ ¡ ¡ ¢ § ¤ ¦ ¤ ¦ ¤ ¦ ¦ ¦ ¦ ¦ ¦ § ¦ © ¦ § © ¦ © ¦ ¦ ¢¨ ¨ § ¦ ¦ ¦ © © ¦ ¥ ¡ ¡ ¦ © ¦ § ¦ ¦ ¦ b a sec θ tan θ dθ b5 tan5 θ 1 ab4 ¡ © ¤ § ¡ ¡ © ¤ § ¡ ¤ ¤ ¢ ¦ § ¡ ¡ ¡ ¡ © ¤ § ¡ ¤ ¢ ¡ ¢ ¦ § ¦ ¦¡  ¢ ¦ § b sec θ tan θ dθ. Then cos3 θ dθ sin4 θ u 1 1 C 1 2 C (d) Adding the previous two results, t2 dt t2 dt t2 dt t2 6t 1 3 ln t 3 t2 6t 1 1 1 As usual, check by differentiating. 4. (a) Since the degree of the denominator is not greater than the degree of the numerator, we must first simplify the integrand by polynomial long division: r r r 4 r 4 r 4 r 4 r 4 r Now the integral can be evaluated directly: r2 r 4 16 dr r 4 r2 2 r 4 4r 16 ln r 4 C Check by differentiating. (b) Again, the degree of the denominator is not greater than the degree of the numerator, so we must apply long division: x x 1 x2 x 6 Check by evaluating x 1 x2 x 6 and apply partial fractions to obtain 9x 16 x 3 x 2 A B 9x 16 . Now factor the denominator x2 B x 2A 3B x 3 x 2 x 6 x 3 x 2 A x 3 x 2 Equating the coefficients of similar powers of x in the numerators, we obtain the system A B 9 2A 3B 16 Adding 3 times the first equation to the second we get A 43 5; adding 2 times the first equation to the second we get B 2 5. You should check those results by substituting into the above equations. Altogether, the integrand is x3 4x x2 x 10 6 x 1 43 5 x 3 2 5 x 2 which you should check by placing the terms on the left hand side over a common denominator. It follows that the indefinite integral is x 1 x 3 (which you should check by integration) so the definite integral is 1 0 x2 Note the importance of the absolute value signs in the indefinite integral. 4 ¥ ¦ ¦ ¤ ¦ ¦ ¤ 1 x3 4x x 10 6 1 2 43 ln 2 5 2 ln 3 5 43 ln 3 5 2 ln 2 5 3 2 41 ln 2 5 41 ln 3 5 ¦ ¦  ¤ ¤  ¤ ¦ x3 4x x2 x 10 dx 6 43 5 x 3 2 5 dx x 2 x2 2 43 ln x 5 2 ln x 5 2 C 3 2 41 3 ln 5 2 ¥ x3 4x x2 x 10 6 x x2 x 6 ¦ ¥ © © ¤ § "§© ¦ © ¤ § ¦ § © ¦ §"© ¤ § ¦ ¤ © ¦ § ¤© ¦ ¦ ¦¤ ¤ ¦ ¤ ¤© ¦ ¤ § ¦ ¤ ¤ x2 10x 10 1 x2 x 6 9x ¥ r2 r r 4 4r 4r 4r 4 16 16 4 16 9x 16 x2 x 6 ¥ ¦ ¦ ¤ 6t 1 C ¦ ¤ ¤ ¦ ¦ ¤ ¦ ¦ ¦ ¥ ¦ ¦  ¦ ¤ ¤ ¡ ¡ ¦ ¦ ¦ ¦ ¦ ¦ ¤ ¤© ¦ § ¤ ¤ ¤© ¦ §  ¦ ¦  ¤ ¤ ¤ ¡ ¦ ¦  ¤  ¤ ¤ ¤¤ ¤ ¤  ¤ ¦ ¦ ¦ ¤ © ¦ "§©¦ ¤ § ¦ ¤ "§© ¦ § ¦ ¤© ¦ ¤ § ¤¤ ¤ ¤ ¤ ¦ ¤¤ ¤ ¤ ¤ ¤ ¤ ¤ ¡ ¦ ¤ ¢ ¡ t t 3 6t ¦ ¤ ¢ ¦ ¡ 3 6t ¦ ¤ ¢ ¡ ¤ ¡ (c) The partial fractions decomposition of the integrand is s2 s 1 2 s 1 2 Expanding the numerator, A C s3 2A B C D s2 A 2B s B 0s3 0s2 0s 1 from which we obtain B ds s2 s 1 2 1, A 2 s 2, C 2 s 2, and D 1 s 1 2 1. (Check.) Therefore our integral is 2 ln s 1 s 2 ln s 1 1 s 1 1 s2 1 ds As usual, check by differentiating. (d) Again, long division gives x3 x3 1 x3 1 1 x3 1 1 x3 1 To get any further we need to know how to factor x3 1. Guessing a root, we have know it has a factor of the form x 1. By long division we have x3 1 x 1 x2 x 1 and the latter factor is irreducible because the discriminant b2 4ac 0. So the partial fractions decomposition of the second term is x3 1 giving the system Adding the second and third equations above gives B (Check.) Altogether our integral is x3 x3 1 dx 1 The first integral on the right hand side above can be handled by the substitution u 2x 1 dx. The second integral must be handled by completing the square to obtain x2 x so the appropriate substitution is x dx x2 1 Putting it all together, Guess what you should do at this point. ¦ x3 1 dx x ln x 1 x 1 5 ¢  1 ¤ ¢ ¤ ¦ ¤  ¤ ¦  ¦ x3 1 ln x2 2  ¦ dx dθ 10 tan 3 2 x 3 1 2 ¢  1 ¤ ¦ ¤ 1 x 1 3 4 tan2 θ 3 sec2 θ dθ 3 4 2 ¢ ¢ ¡ ¢ ¥ ¢ © § ¡ ¨ 1 3 tan θ 2 2 Carrying this through, ¢ ¢ ¤ ¦ ¦ ¤ ¦ ¦ ¤ A A A 1 x3 1 B B 1 1 ¦ ¤ ¦ ¦ ¤ ¤ ¦ ¤ ¦ ¦ ¦ ¦¤ ¤ ¥  ¦ ¦ ¡ ¤ ¤ ¡ ¤ ¡ ¤ ¤ ¤ ¦ ¦ ¦ ¦ ¦ © ¦ § ¦ © ¦ ¦¤ ¤§ ¦ © ¦ § ¦ ¤ ¤ © "§© § ¤ © ¤ § ¤ 1 2 41 1 A Bx C A B x2 x 1 x2 x 1 A B C x x2 x 1 A C C C 0 0 1 1 from which we obtain A 1 2 2x 1 dx x 1 x 2 x 1 x2 x 1 dx x ln x 1 x2 x 1 dx x 1 2 2 3 4 3 2 sec θ dθ 2 2 3 2 θ 3 2 tan 3 ¦ © ¤§ ¦ ¥ 1 1 3 1 and C 5 x 5 2 x2 x2 ¥ 1 0 so we 3 is less than 2. 1 dx x 1, du 2 x 3 1 2 © ¤§ ¦ ¤ ¤ ¦ ¦¤ © ¦ ¤ ¤§ ¦ ¤ ¦ ¤ § © ¦ ¦ ¦ ¦ © ¤ "§© § ¦ ¦ ¦ ¤ ¤ ¦ © ¤§ ¡ ¡ ¦ ¤ § ¦ © ¦ § ¦ © ¤§ ¦ ¡ 1 A s B s2 C D s 1 As s ¥ ¤ ¤ ¤  ¦ ¤ ¦ ¦ ¦ ¦ © ¦ © ¤ § © ¦ ©¤ § ¤ § ¦ © ¤ § 1 2 B s 1 2 Cs2 s s2 s 1 2 1 Ds2 ¤ ¡ ¡ © ¤ § ¦ ¡ 1 1 x 2 dx 2u 1 u du 21 u 1 u 2 du 2u 2 ln 1 u C 2 x 2 2 ln 1 Check by differentiating. 3 (b) Let u 1 dx x. Then 3u2 du dx and 1 3 x 3u 3 1 u du 3u 3 ln 1 u C 3 3x1 3 3 ln 1 x1 Check by differentiating. sin x, du (c) Let u cos x dx. Then the integral becomes 1 u2 u du 1 1 u 1 du u 2 ln 1 u ln u C ln 1 sinx Check by differentiating. (d) This is a tricky one. Try the substitution u 1 dx 12u11 du u4 u3 12 x, 12u11 du u8 du dx, 3 x u4 , 4 x u3 : 3 x 4 x 12 u 1 The long division is rather tedious but is not difficult: u8 u7 u6 u5 u4 u3 u2 u 1 1 1 u 1 u (check) so the integral is 1 3 x 4 x Reversing the substitution gives 3 x 4 x dx 12 12 12 12 12 12 12 Isn’t that wonderful? Check by differentiating. 6. Since t tan x 2 we have 1 sec2 x 2 By the double angle formulas we have cos x 2 cos2 x 2 1 2 We also have sin2 x 1 cos2 x which implies that sin x 2t ¦ 1 t2 6 § © ¦ § § ¤ © ¦ § 1 1 t2 t2 ¦ t2 1 1 2 2 1 1 ¦ ©© ¦ § ©© ¤ ¦ ¥ ¤ 1 1 t2 t2 ¦ t2 1 t2 t2 2 2 ¥ t2 cos2 x 2 1 4t 2 1 t2 2 ¦ ¦   ¦  ¤  ¦  ¤  ¦  ©  § ¤  ¤ ¦  ¤©  § # $ ©  § ¦ © § ¤  ¤ ¢¦ ¢ 1 12 8 x 8 12 7 x 7 12 6 x 6 12 5 x 5 12 4 x 4 12 3 x 3 12 2 x 2 ¦ dx 12 1 x 1 12 ¥ ¦ ¦  ¦ ¤ ¤ 12 8 u 8 12 7 u 7 12 6 u 6 12 5 u 5 12 4 u 4 12 3 u 3 12 2 u 2 12 u 1 ln u ln x1 ¥ ¦  ¤ ¢ ¦  ¦  cos x dx sin x sinx ¥ ¦  ¦  ¦  ¤  3u2 du 1 u 3 3 2 u 2 3 2 x 2 ¥ ¦ ¦ ¢ ¤  x 2 C 3 ¤ ¦ ¢ ¤ ¦ ¦ ¢¦ ¢ ¥ ¡ ¡ ¡ ¢ ¢ ¦ ¦ ¦ ¤ ¦  ¤ ¡ ¡ ¡ ¦ ¦ ¦ ¢¦ ¦ ¦  ¦ ¤ ¤ ¡ ¡ ¡ ¢ ¦ ¢ ¤ ¦ ¤  ¤ ¤ ¦ © ¤ ¤ § ¤ ¤ ¡ ¡ ¡  © ¦ "§©  § ¦ ¢ 5. (a) Let u x 2, du 1 2 x 2 1 2 dx, 2u du ¦ ¦ ¤ ¦ ¤ ¢ ¢ ¡ ¦ ¦ ¤ ¦ ¤ ¦ ¤ ¦ ¤ ¦ dx. Then the integral becomes C ln sin x C 1 C ¡ 12 1 C (at least for positive values of t; what could we do with negative values of t?). Differentiating, dt dx 1 t2 dt Armed with the above, we can integrate any rational trig function. If another method is available, we should use the other method, but in some cases the half-angle substitution is the only method available. For example, 3 3 3t 2 3 dt Factoring the denominator, 3t 2 so 3t 2 10t 3 3t 1 t 3 2 10t A B A 3 3t 1 t 3 3B t 3A 3t 1 t 3 B from which we obtain B dx 5 sinx 1 4, A 3 4, and the integral is 1 1 ln t 4 3 3 3t C 1 Check by differentiating. 7 ¥ ¦ ¤©  §  ¦ ¤©  §  3 4 1 1 4 dt t 3 1 ln 3t 4 1 ln 3 tan x 2 4 ¥ dx 5 sinx ¤ ¦  ¤  ¦  ¤  ¤ ¤ ¦  ¤  ¤  ¦ ¦ © %§© § © ¤ ¤ ¤ § ¦ ©¤ ¦ § ¤ ¤ © ¤ %§© ¤ § ¦ ¦ ¤ ¦ © ¦ §! ¤ ¡ 1 10t 1 2 dt t2 1 t2 2 10t ¡ ¥ ¦ # $ ©  § 1 sec2 x 2 dx 2 2 ¤ ¡ ¤ ¤ ¤ ¡ ¤ ¡ 1 ln tan x 2 4 3 C ...
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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 University of California, Berkeley.

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