The 2014 Integration Bee Solutions and comments. Mike Hirschhorn. u 4 du = 1 5 u5 +C = 1 5 (x3 1) 5 +C cosx dx = 1 2 x 1 2 sinx+c.

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1 The Integration Bee Solutions and comments Qualifying Round Mike Hirschhorn. x x dx u du 5 u5 +C 5 x 5 +C.. 5 x ] x dx 5 x.. sin x dx cosx dx x sinx+c.. a +x dx a tan x +C. a 5. x x+ dx 7 x+ dx x 7 log x+ +C. 6. +sinθ dθ + t +t +t dt +t dt +t ] x dx x5 5 +C. 8. x dx x x+ dx log x +C. x+ Typeset by AMS-TEX

2 9. e x cosx+sinx dx Ae x cosx+be x sinx+c for some A, B. Differentiating with respect to x gives e x cosx+sinx A+Be x cosx+ A+Be x sinx. Comparing coefficients gives A+B, A+B, A, B, so e x cosx+sinx dx e x sinx+c. Alternatively, you can do two integrations by parts.. xlogx dx u du log u +C log logx +C... x dx e xlog dx log exlog +C log x +C. x dx x +C x +C... x x dx x ]. x + x dz x + x z +C. We must assume x is a constant function of z. 5. x 5 log x dx x6 x 5 6 log x 6 dx x6 6 log x x6 6 +C by integration by parts with u log x, dv x 5 dx.

3 6. Let It follows that I xsinx +cos x dx I. If we make the substitution x u, we find I usinu +cos u du. sinx +cos x dx ] +u du tan u, so xsinx +cos x dx I. 7. +sinxdx cosx+sinx dx cosx+sinxdx cosx+sinx+c. Not exactly legitimate, I m afraid. It is not always the case that v v. This is one reason why definite integrals make better quiz questions than indefinite integrals. 8. sinx +sin x dx sinx cos x dx +u ] log u u du log + u +u + log+. du cosecx sinx dx sinx dx sinx 9 7 cosx dx ]9 sinx sinx 7 sin x sinx dx sin 9 sin 7.

4 . cotx dx cosx sinx dx log sinx +C x dx 5 x dx+ Alternatively, from a graph, 5 x dx x x] + 5 x x pi x dx ] 5. e 5logx dx x 5 dx x6 6 +C.. e ex +x dx e ex e x dx e u du e u +C e ex +C.. secxtanx dx secx+c. This is/used to be bookwork. 5. logx x dx u du log u +C log logx +C. 6. dx sin x +C. x This certainly is bookwork! 7. cosx+ dx t +t + +t dt dt t+c tan x +C.

5 5 8. cosec θ dθ cotθ+c. This used to be bookwork once. Differentiate the right side to check! 9. x e x dx 8 x 6xe x dx 8 8 ueu du 8 ueu e u +C x e x e x +C 8 ex x +C.. tan x dx usec u du utanu tanu du utanu+log cosu +C utanu log secu +C utanu log+tan u+c xtan x log+x +C. Alternatively, do an integration by parts with u tan x, dv dx.

6 6 Group Stage. sina+b sinacosb +cosasinb, so 8 6 sin5xcosx dx ] cos8x cosx 8 sin xcos x dx 5 5 Look at the graph!. 5 x dx 6 sin8x+sinx dx cos8x x dx x x ] 8 6 cosx ] sin x dx 8 cosx dx 8 x ] 6 sinx 5 x dx 5 x] x dx x u u du, because the integrand is odd and the interval of integration is balanced about the origin dx 5 5x 5 5 x dx 5 sin 5x ] cosx sin x dx u du +u + u log. du log +u u ]

7 7 7. e sin x e cos x dx e dx e. 8. x x+5 dx x + dx. tan x ] tan tan 9. x e x dx xx e x dx ueu du ueu e ] u e.. sinxsinx dx 6. sinx cosx cosx dx sinx ] 6 6. e x x dx ue u du ue u e u ] e e.. 5 cossinxcosx dx sin5 sin cosu du ] sin5 sinu sinsin5 sinsin. sin. e x x+ dx e x x dx e log x +C x e C x. x. e x tane x dx e tanu du ] e log secu log sece. sec

8 8 That s what a blind calculation gives, but it is not correct, would you believe! Sketch the graph of the integrand, and see what goes wrong! It makes you wonder about some of the other integrals! 5. xe x dx xe x e x ] e. The indefinite integral can be guessed, or you can do an integration by parts. 6. e x + e x dx e u+ du u+] e sinxcos x dx u du u ]. ] +x dx tan x tan. 9. tanx+cotx dx logcot. ] log secx +log sinx ] log tanx logtan. e x dx +e x e x e e x + dx e ] e u + du tan u e tan e tan e tan e e +e. b a x sec logx dx logb loga sec u du ] logb tanu tanlogb tanloga. loga

9 9. sin x+cos x dx dx. Draw the graph!. log ] log secx+tanx +, + secx dx log + log+ log + + which is negative, because of the unusual limits on the integral.. x x x dx x x ] dx 5 x5 x sin x dx sinu du cosu x ]. 6. +cosx dx +sinx dx. Done earlier! 7. e e x+9 x 9 dx + 98 ] e x 9 dx x+98log x 9 e +98log e e e +98log. 8 Draw the graph!

10 logx dx 7 5 x x log log+ dx log+ ] log+ 5 log+. log+ 9. secx+tanxsinx dx cosx +sin x dx ] +u du tan u.. e logx t x dx e t e t dt te t dt te t ] e... sec xtanx dx x x + dx u du u du. u 5 5 ] 5.. sin x dx cosx dx x ] sinx sin.. e cosx cos x cosx dx dx e. e You have to know that cosx cos x cosx. Try to prove it tan θ +cot θ dθ 8 dθ 8. Unusual. tan and cot are usually applied to x, not θ.

11 6. e tanx cos x dx e u du e. 7. xcosx dx 6 6 ] cosu du sinu log sec e x dx e x +e x dx u e x dx. du log log log.. x+ x +x+ dx u ] u + du logu + tan u log.. +x x dx +u u du u tan tan. +u du tan u ]. x x+ x dx does not exist, because of the discontinuity at x.. dx x sin x ].

12 . sin x dx x sinx ] x +x dx u +uu du u tan u ]. u u + du u + du 6. x+ dx ] x x dx u u du u u +u du u u +u ] x x +x dx x x + x] x

13 Quarter Finals. sin xcosx dx tan x/sec x dx tanx +C. sec x tanx dx tanx +C. tanx sinxcosx dx tanx tanx/sec x dx tanx+c. sec x tanx dx tanx +C. sin 5 x sinx cos cosx dx x sinx cos x+cos x dx dx cosx cosx tanx cosxsinx+cos xsinx dx log secx +cos x cos x+c.. cos x dx ucosu du usinu+cosu+c xsin x+cos x+c. 5. x x +x + dx x x + x + dx x x + x x + dx logx + tan x logx ++ tan x +C. 6. x x + dx u + du tan u+c tan x +C.

14 7. logcosx cotx dx logu u du v dv v +C log u +C log cosx +C. 8. sin 5 xcos 6 xdx cos 6 x cos x sinxdx 7 cos7 x+ 9 cos9 x cos x+c. 9. x +x dx x+ dx sec θ secθtanθ dθ u du secθ dθ log secθ+tanθ +C log u+ u +C log x++ x +x +C.. +xlogx dx +e t te t dt te t +te t dt te t e t +te t et +C xlog x x+x log x x +C x +xlog x x x+c.. x +x +6 dx x + x +6 dx tan x tan x +C.. If for the moment we pretend x x, we find x +x dx x +x dx u du u +C +x +C x x +x +C.

15 5 Luckily, since this function is odd apart from the C, it is correct.. x dx sin θcosθ dθ cos θ dθ + cosθ dθ + sinθ +C θ+ sinθcosθ +C sin x+ x x +C. Actually, a similar comment applies to this integral as to the previous one!. dx +x +tan θ cosθ dθ sinθ +C sec θ dθ x +x +C. 8sec θ sec θ dθ 5. e x dx secθtanθ dθ secθtanθ sec e x +C. e x e x e x dx u u du dθ θ+c sec u+c Alternatively, tan e x +C. 6. cos x dx + cosx dx +cosx+cos x dx +cosx+ + cosx dx 8 + cosx+ cosx dx 8 8 x+ sinx+ sinx+c 8 x+ sinxcosx+ 6 sinxcosx+c 8 x+ sinxcosx+ 8 sinxcosxcos x +C 8 x+ 8 sinxcosx+ sinxcos x+c.

16 6 7. dx 9+6x x 5 dx 9 +x x dx +C. x sin 5 x dx 5 x 8. sinlogx dx e u sinu du e u sinu cosu+c xsinlogx coslogx+c. See question 9 of the Qualifying Round. 9. xtan x dx x tan x x +x dx x tan x x tan x x+tan x+c +x tan x x+c. +x dx. x e x x + dx ue u u+ du v e v v dv v e v e v dv {v e v v e } v vev dv } { e v + ev e v + e v dv e v e v +C e e u+ u+ +C e u u+ +C e x x + +C e x x + +C.. x x+ dx u u du u u du u 6 u +C x+ 6 x+ +C 78 x+ 6x+ +C 78 x+ 6x 7+C..

17 7. +x dx +u u du v v dv v v + v dv v v +v log v +C u+ 6u+ +u+ log u+ +C u u +u log u+ +C x x +x log +x +C.. sin 5 x cosx x dx sin 5 ucosudu 6 sin6 u+c 6 sin6 x +C. cos x+ dx sec x+ dx tan x+ +C. 5. sin x dx u cosu du u sinu+ucosu sinu+c x sin x + x sin x x+c.

18 8 Semi Finals. sec x secx+tanx dx The notoriously ugly sec xsecx tanx dx sec x dx tan x+c. sec x dx can be calculated by integration by parts. sec θ dθ secθtanθ+ log secθ+tanθ +C. So our integral secθtanθ+ log secθ +tanθ tan θ +C tanθsecθ tanθ+ log secθ +tanθ +C.. e xx logx+x x dx ue u du ue u e u +C x x e xx e xx +C e xx x x +C.. x 9 x+9 u x 9u+9 u 8u +8u dx u du x u u u u u uu 6 du u u u u u du v dv 6 5 v 5 +C 6 5 u u 5 +C 6 5 x x 5 +C. du. x 5 dx u du u u +u +x u 5 u5 u +u +C 5 u u u+5+c 5 x + x ++5 +x +C 5 x x +8 +x +C. du

19 9 5. log+logx dx logu du ulogu u+c x +logxlog+logx logx+c. 6. x u 6 u u dx x +x u +u u du u+ du + du u+ u u +u u +u 9 u 8 +u 7 u 6 +u 5 u +u u +u + du u+ 6 7 u u +u u u u9 + u8 7 u7 +u 6 5 u5 +u u +6u u+log u+ +C Now set u x. 7. x +x +x x+ x +x +6x +x+ dx x+ dx log x+ + x+ +C. x+ x+ dx 8. secxcosecx dx log cosecx+cotx +C. cosx sinx dx sinx dx cosecx dx 9. x+x dx A xx+x x+ dx x + B x+ + Cx+D x x+ dx Alog x +Blog x+ + C logx x++ D + C sin x +K,

20 where A, B, C and D are given by Ax+x x++bxx x++cx+dxx+ for all x. We have A+B +C, B +C +D, B +D, A, so A, B, C, D, and the integral is log x log x+ logx x++k log x log x + +K log x x +K + log x +K. x+x. cosecx dx log cosecx+cotx +C. Pure bookwork.. Obvious! cossinsinxcossinxcosx dx sinsinsinx+c.. x x x dx x dx x+ u u du u sin u +C x x sin x+ +C.. x log5x dx 5 u logu du 5 vev dv 5 vev 6 ev +C 5 u logu 6 u +C x log5x 6 x +C.

21 . logx dx x u e u e u du u e u ue u e u +C logx x u e u du logx x x +C. 5. If n is an integer, n, x n +x dx xx n + dx dx x n x n x n + x xn x n + x n x n x n + dx dx log x n log xn + +C.

22 Finals. 8 sin xcos x dx sinx + cosx dx sin x+sin xcosx dx 8 cosx+sin xcosx dx 6 x 6 sinx+ 8 sin x+c. This solution was indicated to me by one of the competitors.. x 5x+9 x 5x x+ dx Alog x 5 + B logx x++ A x 5 + Bx+C x + dx B +C tan x +K, where A, B and C are given by Ax x++bx+cx 5 x 5x+9 for all x. We have A+B, A 5B +C 5, A 5C 9, so A, B, C and the integral is log x 5 + logx x++ tan x +K.. +sinxcotx +sinx +u dx cosx dx du sinx u v v v v dv v dv + v dv + v dv v + v +log v +C +u +C +u+log v + +u+ +sinx +C. +sinx+log +sinx+

23 . sinx +sin x dx sinx cosx dx u du log u +C log cosx +C log cosx+c. 5. x x dx x x dx x + x dx x x+ log +x +C x x + x dx x + x + +x x ++ x dx dx 6. x x dx sin θ cos θ sinθcosθ dθ sinθ dθ sinθ sinθ +sinθsinθ dθ sinθ + cosθ dθ cosθ+θ sinθ +C θ cosθ sinθcosθ +C sin x x x x+c. 7. e cotx+logcosecx dx e cotx cosec x dx e u du e u +C e cotx +C. 8. xe sin x x dx x x e sin x +C e θ sinθ dθ sinθ cosθeθ +C 9. sinx+cosx sinx+cosx dx 5 cosx sinx sinx+cosx sinx+cosx 5 cosx sinx sinx+cosx dx x 5 log sinx+cosx +C. dx

24 . sinx+sinx dx sinx+sinxcosx dx sinx+cosx dx cos sinx dx x+cosx u +u du +u 6 u +u du log +u + 6 log u log +u +C log +cosx + 6 log cosx log +cosx +C 6 log +cosx cosx +cosx log sinx+cosx +cosx +C +C 6 log sin x+cosx +cosx +C

25 5 Tie breaker questions. Let I Then I sinx sinx+ cosx dx. sin x sin x + cos dx x cosx cosx+ sinx dx, I.. e dx, so I. sin logx x dx ] + u 7 x 7 logx dx 7 sin u du +. x y dy dx y + dy logy + ] 7 7 log8. Clearly not suitable for first year students. ] usin u x y dx dy 7 u u du x y+ y + ] dy. 5. Let I Then I and I 6 sec θ dθ +tan x dx. +tan dx x dx, I. + log+. sec u du +cot x dx tan x tan x+ dx, secθtanθ+ log secθ +tanθ ]