MATH111-200530-PS05-Solutions

MATH111-200530-PS05-Solutions - Math 111 Problem Set 5...

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Unformatted text preview: Math 111 Problem Set 5 Solutions Edward Doolittle November 21, 2005 1. (a) Make the substitution x 4 sec θ, dx Now making the transformation u 1 64 sin θ, du u2 du cos θ dθ, C By the right triangle with sides A O H 4 x2 16 x , sin θ x2 16 dx x4 x2 16 3 192x3 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 dx 9 9 sin2 θ Using the identity sin2 θ cos2 θ 1 or the appropriate right triangle, C 3 1 x 3 2 x x2 9 dx 3 1 sin2 θ 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 ¥ ¦ ¤ ¤ ¦© § ¤  ¤ ¦ C 9 3 2 1 1 a2 tan2 θ sec2 θ dθ ¥ ¦ 3 cos θ dθ 3 sin θ dθ ¤ x 3 sin θ  ¤ ¢ ¥ ¦ sin2 θ cos θ dθ 1 64 1 u3 64 3 1 sin3 θ 192 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 44 sec3 θ 1 64 ¡ ¡ ¥ ¦ © ¤ § © ¨ ¤ ¢¨ § © ¨ ¨ § ¦ ¡ ¡ ¡ ¤ ¤  ¦¤ ¤  ¡ ¢ £¡ 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 tanθ Reversing the substitution, tan θ where (b) Here the algebraic substitution may work better than a trig substitution. Let u Then 4 9x2, du x 4 9x2 dx u1 2 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 4 90 2 0 (c) Make the trig substitution 2x 4x2 1 dx tan θ, 2dx sec2 θ dθ, we have By example 8 on page 523 of the textbook, 4x2 1 dx 1 sec x tan x 2 Check by differentiating. Now the definite integral is 0 (d) Let ax b sec θ, a dx dx a 2 x2 b 2 5 2 2 ¥ b a sec θ tan θ dθ b5 tan5 θ 1 ab4 cos3 θ dθ sin4 θ 1 ab4 ¡ ¡ © ¤ § ¡ ¡ ¦ 4x2 1 dx 1 5 1 ln 2 5 2 0 1 ln 2 1 0 5 1 ln 2 2 b sec θ tan θ dθ. Then cot3 θ csc θ dθ !¥© ¢ ¦ § ¦  ¦  ¢ ¢ ¤ ¤  ¦ ¢ ¦ ¢ 1 ¦ ¦ ln sec x tanx C x 4x2 1 4x2 5 ¥ ¦ ¦ ¦ 1 ln 2  ¦  © ¥ sec2 θ sec2 θ dθ sec3 θ dθ ¥ ¤   ¤  2 3 1 0 27 1 2 27 3 4 9 2 3 2 0 0 0 1 ¤ ¥ ¦  © ¤ § ¤ ¦  ¤ ¤  ¤  ¡ ¤ © § ¦ ¤  ¤  ¦   ¤  ¤   ¤ ¢  ¤ ¢ ¤ ¢ ¦ ¤ ¢ ¤ ¢ ¡ ¡ sec2 θ 16x2 1 16x2 C 9 3 and dx 16x2 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 9 sec2 θ 9 1 4 ¦ ¥ ¦ © § ¦  ¦ § ¦  ¡ ¦ ¢ ¡  ¡ ¡ ¤ ©§ ¤  ¤   ¡ ¦ © ¦ § x2 x2 1 sin3 θ a2 3 1 x3 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 Since the power of cot is odd we should convert as many cot to csc as possible and make the substitution u csc θ, du cot θ csc θ dθ: Reversing the substitutions using a A O H dx a 2 x2 b 2 5 2 Check by differentiating. 3. These problems require completing the square. By the hint, 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 6 x2 2x 2 3 5 x 1 24 x2 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 t2 3 dt 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 Check by differentiating. (d) Adding the previous two results, t2 6t 1 t2 6t 1 As usual, check by differentiating. ¦ ¤ ¢ ¦ ¦ ¤ ¢ t dt t 3 dt 3 t2 6t 1 dt t2 6t 1 3 ln t 3 ¥ ¦ ¦ ¤  ¦ ¤  ¦ ¦ ¤  ¡ ¦ ¤ ¢ ¡ ¤ ¦ ¤ ¢ t t2 3 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: 6t du 2u1 2 1 u1 2 21 2 C t2 6t 1 ¤ © ¤§ ¦ ¢ ¤ ¦ ¦ ¤  ¦ © ¢ & ¦ ¤ ¢ ¤  © § ¥ ¦ ¦  3 sec θ dθ ¤ ¤§ ¤ 6 2 sec θ tan θ dθ 2 2 tan θ ¤ © ¤ ¤ ¦ ¦ ¦ © ¥ ¦© ¦ § $ ¦ ¦ ¦ ¦ #© ¦ § ¢ ¥ ¦ ¦ ¦ ¦ (a) Write x2 2x sec2 θ dθ, ¥ ¦© ¤ § ¤ © ¤ §  ¤ ¦ ¤  ¤ © ¤ § ¡ © ¨  © ¤ "§¨ § © ¨ ¨ § © ¤ © ¤ § ¡ ¤ ¤ § ¡ 5 2 csc2 θ 1 cot θ csc θ dθ b a 2 x2 C u2 1 du b2 1 2 ax right triangle, 3 2 1 ab4 csc3 θ 3 cscθ 1 ab4 a 3 x3 3 a 2 x2 b 2 a 2 x2 ax b2 2 dx 2x x2 2x 1 1 x 1 2 1. Then making the substitution x 1 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 2 15 x 1 48 x2 2x 2 15 tan 48 3 2 8. Making the substitution t 3 ln sec θ 2 tan θ C 8 1 t2 6t 1 2 2, t2 6t 1 C ln 2 2 into C. (Check by different 3 2 1 8, but this time make C 3 t2 ¥ ¦ ¤  u C 1 2 tan θ, dx 2x 2, cos θ 1 x 1 C 3 6t 1 C ¦ ¤  '  ' ¦ '' '' ¦ '' ¦ ¢ ¤ ¢ ¦ ¢¤ '' §  © ¢ %#© ¤ § ¦ ¢ ¢ ¡ ¡ ¢ § ¤ ¦ ¤ ¦ ¤ ¦ ¦ ¦ ¦ ¦ ¦ § ¦ © ¦ § © © ¦ ¦ ¢¨ ¦ ¨ § ¦ ¦ ¦ © © ¦ ¥ ¡ ¡ ¦ © ¦ § ¦ ¦ ¦ © ¤ § ¡ ¤ dx a 2 x2 b 2 1 ab4 1 ab4 1 ab4 u3 3 C ¤ ¤ ¢ ¤ ¢ ¦ § ¦ § ¦ ¦¡  ¢ ¦ § ¡ ¡ ¡ ¢ ¡ ¡ ¡ 4. (a) For large x, the integrand is approximately x x2 1 x, and we know that a∞ 1 x dx is divergent, so the given integral is likely divergent. To make sure we use the comparison test, bounding the integrand from below by some multiple of 1 x: 1 x x2 2x2 for x large enough (specifically for x bigger than the largest root of the quadratic equation 1 so for x large enough. However, 1 2 ∞ 1 1 dx x diverges, so it follows that ∞ 1 diverges as well by the comparison test. Changing the lower bound of integration from 1 to 0 doesn’t make any difference in the divergence of an improper integral, so the given integral diverges. Alternatively, the integral may be evaluated explicitly using techniques similar to those of 3(d) above. (b) The integrand is approximately 1 t 2 for large t so we should expect it to converge. However, we also need to check whether there is any type II “improperness” since the denominator of the integrand may vanish at t 1 t 3 , so the denominator vanishes at t 1 some point. Factoring the denominator, t 2 4t 3 and t 3; those points are outside of the interval over which we are integrating, namely 0 ∞ , so we don’t need to worry about the integrand going infinte in the domain of integration. Since t 2 t 2 4t 3 for t 0, t2 1 4t 3 1 t2 by the hint. Reversing the substitution by the A O H 1 t2 4t 3t 1 t2 3 dt ln t2 4t 3 Check by differentiating. To evaluate the definite improper integral, ∞ 0 M M 3 where the limit was obtained by dividing through by M in the fraction inside the logarithm. Since the limit exists, the comparison test wasn’t really necessary, but in general it’s a good idea to quickly check the integral using the comparison test before putting too much of an effort into evaluating it. 4 '' t2 3 lim ∞ 0 t2 3 lim ln ∞ M2 ln 0 0 3 ' ' ' '' ¦ ¦ ¦ ''' ¤' dt 4t M dt 4t ' ¢ ¤ 1 4t t 2 1 t2 4t 3 C ln t M 1 4M ¥ ¦ ¦ ¦  ¤ © ¦¨ § ¦ ¦ 2 ¦ ¦ ¡ ¡ ¦ ¦ '' ¦ ¦ ¦ ¢ '' ¡ ¦ ¦ t2 3 dt ¦ csc θ dθ ¦ ¦ '' ¦ ¢ '' 2 ' ¦  ¦ ''' ¦ ¦ ¦ ¦ ¨ § © ¨ ¨ ¢  ln csc θ ¤ 1 4t sec θ tan θ dθ tan2 θ cot θ C 2 right triangle, 4t 3 1 ln t 2 2 C 0 1 ln 3 ¦ for t large enough; since 1∞ 1 t 2 dt converges, the given integral converges by the comparison test. In order to evaluate the integral, complete the square in the denominator and make the substitution t 2 sec θ, dt sec θ tan θ dθ to obtain ¤ © ¨0 © ¦ !§© ¦ § ¦ ¦ ¡ © § (  1 ¡ ¦ ¦ ) ¦ ¦ ) ¦ ¦ 1 ¦ ¦ ) 1 2x 1 x x x2 x dx x x2 ¦ ¦ x x2 © § (    ) ¦ ¦ ¡ 2x2 ), ¤ ¡ ¡ (c) The integrand blows up as x approaches 2 from below, so the integral is improper of type II. In this particular case it’s probably easier to evaluate the integral, but further down I’ll explain how to use the comparison test to check for convergence in this case. To evaluate the integral, let x 2 sinθ, dx 2 cos θ dθ to obtain Therefore 2 and the integral is convergent to the value π 2. It is possible to check for convergence using the comparison test, but it is a little more difficult than in the other cases we’ve studied. The idea is to factor out the part that goes to infinity from the integrand and show that the rest is approximately constant: Near the singular point x 2 the first factor is approximately 1 2 x 1 4 1 2, while the second factor is of the order u 1 2 where u 2 x. Let’s make that change of variables so that the improper integral becomes 2 4 u 1 2 u 1 2 2 du u 1 2 du 0 0 because 4 u 1 2 1 on the interval 0 2 . The latter integral is convergent by the “p-test” because the power of u is greater than 1; by the comparison test the given integral is convergent. This still leaves us with the problem of evaluating the integral, which we then take care of as in the first part of this solutions. (d) In this case it isn’t hard to check using the comparison test. The integrand blows up near z integral is improper of type II at z 2; the integral really means 3 z 2 1 3 a dz lim z 2 1 3 3 dz lim z 2 1 3 dz 0 a 2 0 b 2 b For the latter integral we let u z 2 which leads to comparison with 01 u 1 3 which is convergent by the p-test. Similar reasoning applies to 0a z 2 1 3 dz with the change of variables v 2 z. Now that we know that the integral is convergent, we can evaluate it as if it were not improper. Making the change of variables u z 2, z 2 1 3 dz u 1 3 du u2 3 2 3 C 3 z 2 2 2 3 C so the definite integral is 3 z 2 1 3 dz 0 3 3 2 2 2 3 3 0 2 2 2 3 3 1 2 3 3 4 2 Alternatively, we could have evaluated the limits in (1) to obtain the same result. 5. Taking the limit throughout the given formula, lim ∞ 0 t 0 5 $ xn eax dx 2 ¤8 ¤ 2 t 1 lim t n eat t ∞a 1 n a0 0 e a n lim at ∞ t xn 1 eax dx; ¤  ¢ ¦ ¢  ©$ ¤ § $ ( ¥  ©$ ¤ § 7 2 ¦  ©$ ¤ ¡  © § $ ¤4 3 $ 5 2 ¤ ¢ ¡5 2 ¤ ¢ ¡ dx a 0 4 x2 lim dx a 2 0 4 ¥ ¢ ¤ © §  © ¤ § ¤  © ¤ §  ©$ ¤ § ¡ ¦  © ¤ § ¦    $ ¡  ©$ ¤ § ¡ ¤ ¡ ¤ 6¨0 )  ©$ ¤ §  $ ¡ )  $  #$© ¤ § ¡ ¤  $ ¦ ¥ ¤ ¢ ¢ ¤ ¢ 1 4 x2 1 2 x 1 2 x ¥ ¦4 3 $ ¦ θ C sin 1 dx 4 x2 2 cos θ dθ 2 cos θ x 2 C x2 lim sin 1 a 2 a 2 sin 1 0 π 2 2, so the (1) ( ¤ § 5 2  ©$ ¤ § ¡ ¡ ¤ ¢ ¡ ¡ ¡ 2 the second limit above is 0 by the hint so the formula becomes 0 0 which is actually simpler than the corresponding formula on a finite interval. It follows that dx dx 0 0 0 (Question: this improper integral can give us a generalization of the factorial function to any nonnegative real number; how?) 6. Make the substitution u cost, du sint dt to obtain At this point a second trig substitution is called for. Let u du sec2 θ dθ sec θ tan θ, du Reversing the substitutions, Check by differentiating. To evaluate the definite integral, ¦ 0 sint dt 1 cos2 t 9¥© ¢ ¦ §  ¦ ¢ ¦  ¦  ¦ ¢ ¦  ¤ π 2 ln 0 1 0 ln 1 1 1 ln 1 6 ¦ ln u 1 u2 C ln cost 1 cos2 t 2 ¥ ¦ sint dt 1 cos2 t  ¦  ¤ ¦ ¦  ¦  ¤ ¦ 1 u2 tanθ ¥ ¦ ¦ sec θ dθ ln sec θ  ¤ ¡¤ ¥ ¦ sint dt 1 cos2 t du 1 u2 $ ¤ $ ¢ ¤ ¡ ¤ ¡ $ x2 e 3x x1 e 3x x0 e 3x dx '' ¥ '' $ ¤ 3x 0 ∞ 2 3 ∞ 21 33 ¡ $ ∞ xn eax dx ¡ ¤ n a ¡ ¦ ¦ ¢ ¢ ¤ ¡  ¡ ¢ ¢ ¡ ¡ ¡ ¡ ∞ xn 1 eax dx ∞ 2 1 e 9 3 ∞ 2 27 sec2 θ dθ, to obtain C C ...
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