NGUYN HM TCH PH N HM L×ÑNG GIC CHINH PHÖC OLYMPIC TON Nguy¹n Minh Tu§n ft Ph¤m Vi»t Anh TP CH V T× LIU TON HÅCCh÷ìng 1 C¡c d¤ng to¡n v ph÷ìng ph¡p 1 C¡c d¤ng to¡n cì b£n D¤ng 1 T½nh t½ch ph¥n têng qu¡t sau I 1 = Z (sinx) n dx;I 2 Z (cosx) n dx Ph÷ìng ph¡p Ta chó þ c¡c cæng thùc h¤ bªc sau sin 2 x = 1 cos2x 2 ;cos 2 x = 1+cos2x 2 ; sin 3 x = sin3x+3sinx 4 ;cos 3 x = cos3x+3cosx 4 N¸u n ch®n ho°c n = 3 th¼ ta s³ sû döng cæng thùc h¤ bªc tri»t º N¸u n l´ v lîn hìn 3 th¼ ta s³ sû döng ph²p bi¸n êi sau. Bi¸n êi 1. Ta câ I 1 = Z (sinx) n dx = Z (sinx) 2p+1 dx = Z (sinx) 2p sinxdx = Z 1 cos 2 x p d(cosx) = Z C 0 p C 1 p cos 2 x+:::+( 1) k C k p cos 2 x k +:::+( 1) p C p p cos 2 x p d(cosx) = C 0 p cosx 1 3 C 1 p cos 3 x+:::+ ( 1) k 2k+1 C k p (cosx) 2k+1 +:::+ ( 1) p 2p+1 C p p (cosx) 2p+1 ! +C Bi¸n êi 2. Ta câ I 2 = Z (cosx) n dx = Z (cosx) 2p+1 dx = Z (cosx) 2p cosxdx = Z 1 sin 2 x p d(sinx) 1Nguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC = Z C 0 p C 1 p sin 2 x+:::+( 1) k C k p sin 2 x k +:::+( 1) p C p p sin 2 x p d(sinx) = C 0 p sinx 1 3 C 1 p sin 3 x+:::+ ( 1) k 2k+1 C k p (sinx) 2k+1 +:::+ ( 1) p 2p+1 C p p (sinx) 2p+1 ! +C Nh¼n chung ¥y l mët d¤ng to¡n khæng khâ, c¡i khâ cõa nâ l ph²p bi¸n êi t÷ìng èi d i v cçng k·nh ,v m§u chèt l h¤ bªc d¦n d¦n º ÷a v· nguy¶n h m cì b£n. Sau ¥y ta s³ còng t¼m hiºu v½ dö v· ph¦n n y! T¼m c¡c nguy¶n h m sau I = Z cos 6 xdx. I = Z (sin5x) 9 dx. I = Z (cos2x) 13 dx. I = Z (3+cosx) 5 dx. B i 1 Líi gi£i 1. Bi¸n êi nguy¶n h m ta câ I = Z cos 6 xdx = Z cos 2 x 3 dx = Z 1+cos2x 2 3 dx = 1 4 Z (1+cos2x) 3 dx = 1 4 Z 1+3cos2x+3cos 2 2x+cos 3 2x dx = 1 4 Z 1+3cos2x+ 3(1+2cos4x) 2 + cos3x+3cosx 4 dx = 1 16 Z (7+12cos2x+12cos4x+cos3x+3cosx)dx = 1 16 7x+6sin2x+3sin4x+ 1 3 sin3x+3sinx +C 2. Bi¸n êi nguy¶n h m ta câ I = Z (sin5x) 9 dx = Z (sin5x) 8 (sin5x)dx = 1 5 Z 1 cos 2 5x 4 d(cos5x) = 1 5 Z 1 4cos 2 5x+6cos 4 5x 4cos 6 5x+cos 8 5x d(cos5x) Chinh phöc Olympic To¡n 2 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh = 1 5 cos5x 4 3 cos 3 5x+ 6 5 cos 5 5x 4 7 cos 7 5x+ 1 9 cos 9 5x +C 3. Bi¸n êi nguy¶n h m ta câ I = Z (cos2x) 13 dx = Z (cos2x) 12 cos2xdx = 1 2 Z 1 sin 2 2x 6 d(sin2x) 1 6sin 2 2x+15sin 4 2x 20sin 6 2x+15sin 8 2x 6sin 10 2x+sin 12 2x d(sin2x) = 1 2 sin2x 2sin 3 2x+3sin 5 2x 20 7 sin 7 2x+ 5 3 sin 9 2x 6 11 sin 11 2x+ 1 13 sin 13 2x +C 4. Bi¸n êi nguy¶n h m ta câ I = Z (3+cosx) 5 dx = Z 3 5 +5:3 4 cosx+10:3 3 cos 2 x+10:3 2 cos 3 x+5:3cos 4 x+cos 5 x dx = Z 243+405cosx+135(1+cos2x)+ 45 2 (cos3x+3cosx)+ 15 2 (1+cos2x) 2 +cos 5 x dx = Z 378+ 945 2 cosx+135cos2x+ 45 2 cos3x+ 15 2 1+2cos2x+ 1+cos4x 2 +cos 5 x dx = Z 1557 4 + 945 2 cosx+150cos2x+ 45 2 cos3x+ 15 4 cos4x dx+ Z cos 4 xcosxdx = 1 4 Z (1557+1890cosx+600cos2x+90cos3x+15cos4x)dx+ Z 1 sin 2 x 2 d(sinx) = 1 4 1557x+1890sinx+300sin2x+30sin3x+ 15 4 sin4x + Z 1 2sin 2 x+sin 4 x d(sinx) = 1 4 1557x+1894sinx+300sin2x+30sin3x+ 15 4 sin4x 8 3 sin 3 x+ 4 5 sin 5 x +C Tâm l¤i. Qua 4 v½ dö tr¶n ta ¢ ph¦n n o nm ÷ñc d¤ng to¡n n y, ri¶ng ð v½ dö 4 ta ¢ sû döng tîi cæng thùc khai triºn h» sè Newton º khai tr¶n biºu thùc trong d§u nguy¶n h m v c¡c b÷îc cán l¤i ch¿ l bi¸n êi thæng th÷íng. Chinh phöc Olympic To¡n 3 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC D¤ng 2 æi khi trong khi l m c¡c b i t½nh t½ch ph¥n ta bt g°p c¡c b i to¡n li¶n tuan tîi t½ch c¡c biºu thùc sinx;cosx khi â ta s³ sû döng c¡c cæng thùc bi¸n t½ch th nh têng º gi£i quy¸t c¡c b i to¡n n y. Sau ¥y l c¡c cæng thùc c¦n nhî I = Z (cosmx)(cosnx)dx = 1 2 Z (cos(m n)x+cos(m+n)x)dx I = Z (sinmx)(sinnx)dx = 1 2 Z (cos(m n)x cos(m+n)x)dx I = Z (sinmx)(cosnx)dx = 1 2 Z (sin(m+n)x+sin(m n)x)dx I = Z (cosmx)(sinnx)dx = 1 2 Z (sin(m+n)x sin(m n)x)dx Nh¼n chung ¥y l mët d¤ng to¡n cì b£n, sau ¥y ta s³ còng t¼m hiºu c¡c b i to¡n v· nâ. T¼m c¡c nguy¶n h m sau I = Z (cosx) 3 sin8xdx I = Z (cos2x) 13 dx B i 2 Líi gi£i 1. Bi¸n êi nguy¶n h m ta câ I = Z (cosx) 3 sin8xdx = Z (3cosx+cos3x) 4 sin8xdx = 1 4 Z (3cosxsin8x+cos3xsin8x)dx = 1 4 Z (3cosxsin8x+cos3xsin8x)dx = 1 4 Z 3 2 (sin9x+sin7x)+ 1 2 (sin11x+sin5x) dx = 1 8 3 9 cos9x+ 3 7 cos7x+ 1 11 cos11x+ 1 5 cos5x +C 2. Bi¸n êi nguy¶n h m ta câ I = Z (sinx) 4 (sin3x)(cos10x)dx = 1 8 Z (1 cos2x) 2 (sin13x+sin7x)dx Chinh phöc Olympic To¡n 4 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh = 1 8 Z 1 2cos2x+cos 2 2x (sin13x+sin7x)dx = 1 8 Z 1 2cos2x+ 1+cos4x 2 (sin13x+sin7x)dx = 1 16 Z (3 4cos2x+cos4x)(sin13x+sin7x)dx = 1 6 Z (3(sin13x+sin7x) 4cos2x(sin13x+sin7x)+cos4x(sin13x+sin7x))dx = 1 6 Z (3(sin13x+sin7x) 2(sin15x+sin11x+sin9x+sin5x)+ D¤ng 3 T½nh t½ch ph¥n têng qu¡tI = Z sin m xcos n xdx Ph÷ìng ph¡p Tr÷íng hñp 1. N¸u m;n l c¡c sè nguy¶n. N¸u m v n ch®n th¼ dòng cæng thùc h¤ bªc bi¸n t½ch th nh têng. N¸u m ch®n v n l´ th¼ ta bi¸n êi I = Z (sinx) m (cosx) 2p+1 dx = Z (sinx) n (cosx) 2p cosxdx = Z (sinx) m 1 sin 2 x p d(sinx) = Z (sinx) m C 0 p C 1 p sin 2 x+:::+( 1) k C k p sin 2 x k +:::+( 1) p C p p sin 2 x p d(sinx) =C 0 p (sinx) m 1 m+1 C 1 p (sinx) m+3 m+3 +:::+( 1) k C k p (sinx) 2k+1+m 2k+1+m +:::+( 1) p C p p (sinx) 2p+1+m 2p+1+m +C N¸u m l´ v n ch®n th¼ ta công bi¸n êi t÷ìng tü nh÷ tr÷íng hñp tr¶n. N¸u m l´ v n l´ th¼ dòng ta s³ t¡ch ra 1 biºu thùc sinx ho°c cosx º ÷a v o trong d§u vi ph¥n. Tr÷íng hñp 2. N¸u m;n l c¡c sè húu t. Trong tr÷íng hñp n y ta s³ °t u = sinx v tòy theo tr÷íng hñp ta s³ bi¸n êi nâ º ÷a v· b i to¡n cì b£n. Ta s³ t¼m hiºu kÿ thuªt n y qua c¡c b i to¡n d÷îi. Chinh phöc Olympic To¡n 5 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC T¼m c¡c nguy¶n h m sau I = Z (sinx) 2 (cosx) 4 dx I = Z (sin3x) 10 (cos3x) 5 dx I = Z (sin5x) 9 (cos5x) 111 dx I = Z (sin3x) 7 5 p cos 4 3x dx B i 3 Líi gi£i 1. Bi¸n êi nguy¶n h m ta câ I = Z (sinx) 2 (cosx) 4 dx = 1 4 Z (sin2x) 2 (cosx) 2 dx = 1 16 Z (1 cos4x)(1+cos2x)dx = 1 16 Z (1+cos2x cos4x cos2xcos4x)dx = 1 16 Z 1+cos2x cos4x 1 2 (cos6x+cos2x) dx = 1 32 Z (2+cos2x 2cos4x cos6x)dx = 1 32 2x+ sin2x 2 sin4x 2 sin6x 6 +C 2. Bi¸n êi nguy¶n h m ta câ I = Z (sin3x) 10 (cos3x) 5 dx = Z (sin3x) 10 (cos3x) 4 cos3xdx = 1 3 Z (sin3x) 10 1 sin 2 3x 2 d(sin3x) = 1 3 Z (sin3x) 10 1 2sin 2 3x+sin 4 3x d(sin3x) = 1 3 Z 1 0 (sin3x) 10 2(sin3x) 12 +(sin3x) 14 d(sin3x) = 1 3 (sin3x) 11 11 2(sin3x) 13 13 + (sin3x) 15 15 ! +C 3. Bi¸n êi nguy¶n h m ta câ I = Z (sin5x) 9 (cos5x) 111 dx = Z (cos5x) 111 (sin5x) 8 sin5xdx = 1 5 Z (cos5x) 111 1 cos 2 5x 4 d(cos5x) Chinh phöc Olympic To¡n 6 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh = 1 5 Z (cos5x) 111 1 4cos 2 5x+6cos 4 5x 4cos 6 5x+cos 8 5x d(cos5x) = 1 5 (cos5x) 112 112 4(cos5x) 114 114 + 6(cos5x) 116 116 4(cos5x) 118 118 + (cos5x) 120 120 ! +C 4. Bi¸n êi nguy¶n h m ta câ I = Z (sin3x) 7 5 p cos 4 3x dx = Z (cos3x) 1 5 (sin3x) 6 sin3xdx = 1 3 Z (cos3x) 4 5 1 cos 2 3x 3 d(cos3x) = 1 3 Z (cos3x) 4 5 1 3cos 2 3x+3cos 4 3x cos 6 3x d(cos3x) = 1 3 5(cos3x) 1 5 15 11 (cos3x) 11 5 + 15 21 (cos3x) 21 5 5 31 (cos3x) 31 5 +C D¤ng 4 T½nh t½ch ph¥n têng qu¡t I 1 = Z (tanx) n dx;I 2 = Z (cotx) n dx(n2N) Ph÷ìng ph¡p Trong c¡c b i to¡n nh÷ th¸ n y ta c¦n chó þ tîi c¡c cæng thùc sau Z tanxdx = Z sinx cosx dx = Z d(cosx) cosx = lnjcosxj+C; Z cotxdx = Z cosx sinx dx = Z d(sinx) sinx = lnjsinxj+c; Z 1+tan 2 x dx = Z dx cos 2 x = Z d(tanx) = tanx+C; Z 1+cot 2 x dx = Z dx sin 2 x = Z d(cotx) = cotx+C: º l m c¡c b i to¡n t½nh Z (tanx) n dx ta s³ c¦n cè gng t¡ch v· d¤ng tan m x(tan 2 x+1) ¸n cuèi còng º ÷a v· b i to¡n cì b£n. Sau ¥y chóng ta s³ còng t¼m hiºu c¡c v½ dö minh håa º hiºu rã hìn c¡c b i to¡n n y. Chinh phöc Olympic To¡n 7 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC T¼m c¡c nguy¶n h m sau I = Z (tanx) 8 dx I = Z (tan2x) 13 dx I = Z (cotx) 12 dx I = Z (cot4x) 9 dx I = Z (tanx+cotx) 5 dx B i 4 Líi gi£i 1. Bi¸n êi nguy¶n h m ta câ I = Z (tanx) 8 dx = Z (tanx) 6 1+tan 2 x (tanx) 4 1+tan 2 x +(tanx) 2 1+tan 2 x (tanx) 0 1+tan 2 x +1 dx = Z (tanx) 6 (tanx) 4 +(tanx) 2 (tanx) 0 d(tanx)+ Z dx = (tanx) 7 7 (tanx) 5 5 + (tanx) 3 3 tanx 1 +x+C 2. Bi¸n êi nguy¶n h m ta câ I = Z (cotx) 12 dx = Z (cotx) 10 1+cot 2 x (cotx) 8 1+cot 2 x +(cotx) 6 1+cot 2 x (cotx) 4 1+cot 2 x +(cotx) 2 1+cot 2 x (cotx) 0 1+cot 2 x +1 = Z (cotx) 10 (cotx) 8 +(cotx) 6 (cotx) 4 +(cotx) 2 (cotx) 0 d(cotx)+ Z dx = (cotx) 11 11 (cotx) 9 9 + (cotx) 7 7 (cotx) 5 5 + (cotx) 3 5 cotx 1 ! +x+C 3. Bi¸n êi nguy¶n h m ta câ I = Z (tan2x) 13 dx = Z (tan2x) 11 1+tan 2 2x (tan2x) 9 1+tan 2 2x +(tan2x) 7 1+tan 2 2x Chinh phöc Olympic To¡n 8 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh (tan2x) 5 1+tan 2 2x +(tan2x) 3 1+tan 2 2x tan2x 1+tan 2 2x +tan2x = 1 2 Z (tan2x) 11 (tan2x) 9 +(tan2x) 7 (tan2x) 5 +(tan2x) 3 tan2x d(tan2x)+ Z tan2xdx = 1 2 (tan2x) 12 12 (tan2x) 10 10 + (tan2x) 8 8 (tan2x) 6 6 + (tan2x) 4 4 (tan2x) 2 2 lnjcos2xj ! +C 4. Bi¸n êi nguy¶n h m ta câ I = Z (cot4x) 9 dx = Z (cot4x) 7 1+cot 2 4x (cot4x) 5 1+cot 2 4x + +(cot4x) 3 1+cot 2 4x (cot4x) 1+cot 2 4x dx+cot4x = 1 4 Z (cot4x) 7 (cot4x) 5 +(cot4x) 3 (cot4x) d(cot4x)+ Z cot4xdx = 1 4 (cot4x) 8 8 (cot4x) 6 6 + (cot4x) 4 4 (cot4x) 2 2 ! + 1 4 lnjsin4xj+C 5. Bi¸n êi nguy¶n h m ta câ I = Z (tanx+cotx) 5 dx = Z (tanx) 5 +5(tanx) 4 cotx+10(tanx) 3 (cotx) 2 +10(tanx) 2 (cotx) 3 +5tgx(cotx) 4 +(cotx) 5 = Z (tanx) 5 +(cotx) 5 +5(tanx) 3 +5(cotx) 3 +10tanx+10cotx dx = Z (tanx) 5 +5(tanx) 3 +10tanx dx+ Z (cotx) 5 +5(cotx) 3 +10cotx dx = Z (tanx) 3 1+tan 2 x +4tanx 1+tan 2 x +6tanx dx + Z (cotx) 3 1+cot 2 x +4cotx 1+cot 2 x +6cotx dx = Z (tanx) 3 +4tanx d(tanx)+6 Z tanxdx Z (cotx) 3 +4cotx d(cotx)+6 Z cotxdx = (tanx) 4 4 +2tan 2 x 6lnjcosxj (cotx) 4 4 2cot 2 x+6lnjsinxj+C Tâm l¤i. Qua 5 v½ dö tr¶n ta ¢ ph¦n n o hiºu ÷ñc ph÷ìng ph¡p l m c¡c b i tªp cõa d¤ng to¡n n y, m§u chèt l ÷a v· nguy¶n h m t½ch ph¥n h m a thùc qua c¡c ph²p bi¸n êi v th¶m bît, v çng thíi công c¦n ¡p döng linh ho¤t cæng thùc khai triºn h» thùc Newton º gi£i quy¸t b i to¡n d¹ d ng. V· ph¦n b i tªp luy»n tªp câ l³ khæng c¦n th¶m v¼ c¡c b¤n câ thº bàa b§t k¼ mët b i to¡n t÷ìng tü vîi c¡c b i m¨u! Chinh phöc Olympic To¡n 9 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC D¤ng 5 T½nh t½ch ph¥n têng qu¡t I = Z (tanx) m (cosx) n dx;I = Z (cotx) m (sinx) n dx Ph÷ìng ph¡p Ta s³ x²t d¤ng I = Z (tanx) m (cosx) n dx v¼ ¥y l 2 d¤ng t÷ìng tü nhau. Tr÷íng hñp 1. N¸u m;n ch®n ta bi¸n êi nh÷ sau I = Z (tanx) m (cosx) n dx = Z (tanx) m 1 cos 2 x k 1 dx cos 2 x = Z (tanx) m 1+tan 2 x k 1 d(tanx) = Z (tanx) m C 0 k 1 +C 1 k 1 tan 2 x 1 +:::+C p k 1 tan 2 x p +:::+C k 1 k 1 tan 2 x k 1 d(tanx) =C 0 k 1 (tanx) m+1 m+1 +C 1 k 1 (tanx) m+3 m+3 +:::+C p k 1 (tanx) m+2p+1 m+2p+1 +:::+C k 1 k 1 (tanx) m+2k 1 m+2k 1 +C Tr÷íng hñp 2. N¸u m v n ·u l´ th¼ ta bi¸n êi nh÷ sau I = Z (tanx) 2k+1 (cosx) 2h+1 dx = Z (tanx) 2k 1 cosx 2h tanx cosx dx = Z tan 2 x k 1 cosx 2h sinx cos 2 x dx = Z 1 cos 2 x 1 k 1 cosx 2h d 1 cosx = Z u 2 1 k u 2h du u = 1 cosx = Z u 2h C 0 k u 2 k C 1 k u 2 k 1 +:::+( 1) p C p k u 2 k p +:::+( 1) k C k k du =C 0 k u 2k+2h+1 2k+2h+1 C 1 k u 2k+2h 1 2k+2h 1 +:::+( 1) p C p k u 2k+2h 2p+1 2k+2h 2p+1 +:::+( 1) k C k k u 2h+1 2h+1 +C Tr÷íng hñp 3. N¸u m ch®n v n l´ th¼ ta bi¸n êi nh÷ sau I = Z (tanx) 2k (cosx) 2h+1 dx = Z (sinx) 2k cosx (cosx) 2(k+h+1) dx = Z (sinx) 2k 1 sin 2 x k+h+1 d(sinx) °t u = sinx ta câ I = Z u 2k du (1 u 2 ) k+h+1 = Z u 2k 2 [1 (1 u 2 )] (1 u 2 ) k+h+1 du = Z u 2k 2 du (1 u 2 ) k+h+1 Z u 2k 2 du (1 u 2 ) k+h H» thùc tr¶n l h» thùc truy hçi c¡c b¤n câ thº tham kh£o ð ph¦n sau, do â t½nh ÷ñc I. Nh¼n chung c¡c b i to¡n tr¶n mang t½nh têng qu¡t v câ l³ nh¼n v o c¡c líi gi£i têng qu¡t â ta s³ th§y nâ thªt l¬ng nh¬ng v phùc t¤p, nh÷ng khi v o c¡c v½ dö cö thº ta s³ th§y c¡ch l m c¡c d¤ng to¡n n y kh¡ d¹. Sau ¥y ta s³ i v o c¡c b i minh håa. Chinh phöc Olympic To¡n 10 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh T¼m c¡c nguy¶n h m sau I = Z (cot5x) 10 (sin5x) 8 dx I = Z (tan4x) 7 (cos4x) 95 dx I = Z (cot3x) 9 (sin3x) 41 dx I = Z (tan3x) 7 (cos3x) 6 dx B i 5 Líi gi£i 1. Bi¸n êi nguy¶n h m ta câ I = Z (cot5x) 10 (sin5x) 8 dx = Z (cot5x) 10 1 (sin5x) 2 3 dx (sin5x) 2 = 1 5 Z (cot5x) 10 1+cot 2 5x 3 d(cot5x) = 1 5 Z (cot5x) 10 1+3(cot5x) 2 +3(cot5x) 4 +(cot5x) 6 d(cot5x) = 1 5 " (cot5x) 11 11 +3 (cot5x) 13 13 +3 (cot5x) 15 15 + (cot5x) 17 17 # +C 2. Bi¸n êi nguy¶n h m ta câ I = Z (tan4x) 7 (cos4x) 95 dx = Z (tan4x) 6 1 cos4x 94 tan4x cos4x dx = 1 4 Z 1 (cos4x) 2 1 3 1 cos4x 94 d 1 cos4x = 1 4 Z u 94 u 2 1 3 du = 1 4 Z u 94 u 6 3u 4 +3u 2 1 du = 1 4 u 101 101 3 u 99 99 +3 u 97 97 u 95 95 +C 3. Bi¸n êi nguy¶n h m ta câ I = Z (cot3x) 9 (sin3x) 41 dx = Z (cot3x) 8 1 sin3x 40 cot3x sin3x dx = 1 3 Z 1 sin 2 x 1 4 1 sin3x 40 d 1 sin3x = 1 3 Z u 40 u 2 1 4 du Chinh phöc Olympic To¡n 11 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC = 1 3 Z u 40 u 8 4u 6 +6u 4 4u 2 +1 4 du = 1 3 u 49 49 4 u 47 47 +6 u 45 45 4 u 43 43 + u 41 41 +C 4. Bi¸n êi nguy¶n h m ta câ I = Z (tan3x) 7 1 (cos3x) 2 2 dx (cos3x) 2 = 1 3 Z (tan3x) 7 1+tan 2 3x 2 d(tan3x) = 1 3 Z (tan3x) 7 1+2(tan3x) 2 +(tan3x) 4 d(tan3x) = 1 3 " (tan3x) 8 8 +2 (tan3x) 10 10 + (tan3x) 12 10 # +C Tâm l¤i. Qua 4 v½ dö tr¶n ta th§y â, m§u chèt ch¿ l cæng thùc l÷ñng gi¡c v ph¥n t½ch hñp lþ, c¡i n y ð ph¦n h÷îng d¨n ¢ câ ¦y õ rçi. T÷ìng tü m§y ph¦n tr÷îc b i tªp tü luy»n câ l³ khæng c¦n v¼ c¡c b¤n câ thº tü ngh¾ ra mët c¥u º m¼nh l m. Ta còng chuyºn ti¸p sang ph¦n sau! 2 C¡c d¤ng to¡n bi¸n êi n¥ng cao C¡c b i to¡n nguy¶n h m t½ch ph¥n l÷ñng gi¡c r§t phong phó v do â s³ khæng døng l¤i c¡c d¤ng to¡n b¶n tr¶n. Ð ph¦n n y ta s³ còng t¼m hiºu c¡c d¤ng to¡n n¥ng cao hìn, vîi nhúng ph²p bi¸n êi phùc t¤p hìn. Sau ¥y chóng ta s³ còng i v o tøng d¤ng to¡n cö thº! D¤ng 1 T½nh t½ch ph¥n têng qu¡t I = Z dx sin(x+a)sin(x+b) Ph÷ìng ph¡p Dòng çng nh§t thùc 1 = sin(a b) sin(a b) = sin[(x+a) (x+b)] sin(a b) = sin(x+a)cos(x+b) cos(x+a)sin(x+b) sin(a b) Tø â suy ra I = 1 sin(a b) Z sin(x+a)cos(x+b) cos(x+a)sin(x+b) sin(x+a)sin(x+b) dx = 1 sin(a b) Z cos(x+b) sin(x+b) cos(x+a) sin(x+a) dx = 1 sin(a b) [lnjsin(x+b)j lnjsin(x+a)j]+C Chó þ. Vîi c¡ch n y, ta câ thº t¼m ÷ñc c¡c nguy¶n h m Chinh phöc Olympic To¡n 12 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh J = Z dx cos(x+a)cos(x+b) b¬ng c¡ch dòng çng nh§t thùc 1 = sin(a b) sin(a b) K = Z dx sin(x+a)cos(x+b) b¬ng c¡ch dòng çng nh§t thùc 1 = cos(a b) cos(a b) Sau ¥y l c¡c v½ dö minh håa cho c¡c b i to¡n n y. T¼m c¡c nguy¶n h m sau I = Z dx sinxsin x+ 6 I = Z dx cos3xcos 3x+ 6 I = Z dx sin x+ 3 cos x+ 12 B i 1 Líi gi£i 1. Ta câ 1 = sin 6 sin 6 = sin h x+ 6 x i 1 2 = 2 h sin x+ 6 cosx cos x+ 6 sinx i )I = 2 Z h sin x+ 6 cosx cos x+ 6 sinx i sinxsin x+ 6 dx = 2 Z 2 4 cosx sinx cos x+ 6 sin x+ 6 3 5 dx = 2 Z d(sinx) sinx 2 Z d sin x+ 6 sin x+ 6 = 2ln sinx sin x+ 6 +C 2. Ta câ 1 = sin 6 sin 6 = sin h 3x+ 6 3x i 1 2 = 2 h sin 3x+ 6 cos3x cos 3x+ 6 sin3x i )I = 2 Z h sin 3x+ 6 cos3x cos 3x+ 6 sin3x i cos3xcos 3x+ 6 dx = 2 Z sin 3x+ 6 cos 3x+ 6 dx 2 Z sin3x cos3x dx Chinh phöc Olympic To¡n 13 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC = 2 3 Z d cos 3x+ 6 cos 3x+ 6 + 2 3 Z d(cos3x) cos3x = 2 3 ln cos3x cos 3x+ 6 +C 3. Ta câ 1 = cos 4 cos 4 = cos h x+ 3 x+ 12 i p 2 2 = p 2 h cos x+ 3 cos x+ 12 +sin x+ 3 sin x+ 12 i )I = p 2 Z cos x+ 3 cos x+ 12 +sin x+ 3 sin x+ 12 sin x+ 3 cos x+ 12 dx = p 2 Z cos x+ 3 sin x+ 3 dx+ p 2 Z sin x+ 12 cos x+ 12 dx = p 2 Z d sin x+ 3 sin x+ 3 p 2 Z d cos x+ 12 cos x+ 12 = p 2ln sin x+ 3 cos x+ 12 +C D¤ng 2 T½nh t½ch ph¥n têng qu¡t I = Z tan(x+a)tan(x+b)dx Ph÷ìng ph¡p Ta câ tan(x+a)tan(x+b) = sin(x+a)sin(x+b) cos(x+a)cos(x+b) = sin(x+a)sin(x+b)+cos(x+a)cos(x+b) cos(x+a)cos(x+b) 1 = cos(a b) cos(x+a)cos(x+b) 1 Tø â suy ra I = cos(a b) Z dx cos(x+a)cos(x+b) 1. ¸n ¥y ta g°p b i to¡n t¼m nguy¶n h m ð D¤ng 1. Chó þ. Vîi c¡ch n y, ta câ thº t½nh ÷ñc c¡c nguy¶n h m J = Z cot(x+a)cot(x+b)dx K = Z tan(x+a)tan(x+b)dx Chinh phöc Olympic To¡n 14 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh Sau ¥y l c¡c v½ dö minh håa cho c¡c b i to¡n n y. T¼m c¡c nguy¶n h m sau I = Z cot x+ 3 cot x+ 6 dx K = Z tan x+ 3 cot x+ 6 dx B i 2 Líi gi£i 1. Ta câ cot x+ 3 cot x+ 6 = cos x+ 3 cos x+ 6 sin x+ 3 sin x+ 6 = cos x+ 3 cos x+ 6 +sin x+ 3 sin x+ 6 sin x+ 3 sin x+ 6 1 = cos h x+ 3 x+ 6 i sin x+ 3 sin x+ 6 1 = p 3 2 : 1 sin x+ 3 sin x+ 6 1 Tø â ta t½nh ÷ñc I = p 3 2 Z 1 sin x+ 3 sin x+ 6 dx Z dx = p 3 2 I 1 x+C B¥y gií ta s³ i t½nh I 1 = Z dx sin x+ 3 sin x+ 6 . Ta câ 1 = sin 6 sin 6 = sin h x+ 3 x+ 6 i 1 2 = 2 h sin x+ 3 cos x+ 6 cos x+ 3 sin x+ 6 i Tø â suy ra I 1 = 2 Z sin x+ 3 cos x+ 6 cos x+ 3 sin x+ 6 sin x+ 3 sin x+ 6 dx = 2 Z cos x+ 6 sin x+ 6 dx 2 Z cos x+ 3 sin x+ 3 dx = 2ln sin x+ 6 sin x+ 3 +C Chinh phöc Olympic To¡n 15 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC Nh÷ vªy th¼ I = p 3 2 :2ln sin x+ 6 sin x+ 3 x+C = p 3ln sin x+ 6 sin x+ 3 x+C 2. Ta câ tan x+ 3 cot x+ 6 = sin x+ 3 cos x+ 6 cos x+ 3 sin x+ 6 = sin x+ 3 cos x+ 6 cos x+ 3 sin x+ 6 cos x+ 3 sin x+ 6 +1 = sin h x+ 3 x+ 6 i cos x+ 3 sin x+ 6 +1 = 1 2 : 1 cos x+ 3 sin x+ 6 +1 Nh÷ vªy ta ÷ñc K = 1 2 Z 1 cos x+ 3 sin x+ 6 dx+ Z dx = 1 2 K 1 +x+C Ta t½nh ÷ñc K 1 = Z dx cos x+ 3 sin x+ 6 = 2 p 3 ln sin x+ 6 cos x+ 3 +C )K = p 3 3 ln sin x+ 6 cos x+ 3 +x+C D¤ng 3 T½nh t½ch ph¥n têng qu¡t I = Z dx asinx+bcosx Ph÷ìng ph¡p Ta bi¸n êi asinx+bcosx = p a 2 +b 2 a p a 2 +b 2 sinx+ b p a 2 +b 2 cosx )asinx+bcosx = p a 2 +b 2 sin(x+ ) )I = 1 p a 2 +b 2 Z dx sin(x+ ) = 1 p a 2 +b 2 ln tan x+ 2 +C Chinh phöc Olympic To¡n 16 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh Sau ¥y l c¡c v½ dö minh håa cho c¡c b i to¡n n y. T¼m c¡c nguy¶n h m sau I = Z 2dx p 3sinx+cosx J = Z dx cos2x p 3sin2x B i 3 Líi gi£i 1. Ta câ I = Z 2dx p 3sinx+cosx = Z dx p 3 2 sinx+ 1 2 cosx = Z dx sinxcos 6 +cosxsin 6 = Z dx sin x+ 6 = Z d x+ 6 sin x+ 6 = ln tan x+ 6 2 +C = ln tan x 2 + 12 +C 2. Ta câ J = Z dx cos2x p 3sin2x = 1 2 Z dx 1 2 cos2x p 3 2 sin2x = 1 2 Z dx sin 6 cos2x cos 6 sin2x = 1 2 Z dx sin 6 2x = 1 4 Z d 6 2x sin 6 2x = 1 4 ln tan 6 2x 2 +C = 1 4 ln tan 12 x +C D¤ng 4 T½nh t½ch ph¥n têng qu¡t I = Z dx asinx+bcosx+c Chinh phöc Olympic To¡n 17 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC Ph÷ìng ph¡p °t tan x 2 =t) 8 > > > > > > > > < > > > > > > > > : dx = 2dt 1+t 2 sinx = 2t 1+t 2 cosx = 1 t 2 1+t 2 tanx = 2t 1 t 2 Sau ¥y l c¡c v½ dö minh håa cho c¡c b i to¡n n y. T¼m c¡c nguy¶n h m sau I = Z dx 3cosx+5sinx+3 J = Z 2dx 2sinx cosx+1 K = Z dx sinx+tanx I = Z 2 0 ln 1+sinx 1+cosx dx B i 4 Líi gi£i 1. Ta °t tan x 2 =t) 8 > > > > > < > > > > > : dx = 2dt 1+t 2 sinx = 2t 1+t 2 cosx = 1 t 2 1+t 2 Tø â ta câ I = Z 2dt 1+t 2 3: 1 t 2 1+t 2 +5 2t 1+t 2 +3 = Z 2dt 3 3t 2 +10t+3+3t 2 = Z 2dt 10t+6 = 1 5 Z d(5t+3) 5t+3 = 1 5 lnj5t+3j+C = 1 5 ln 5tan x 2 +3 +C 2. °t tan x 2 =t) 8 > < > : dx = 2dt 1+t 2 sinx = 2t 1+t 2 ;cosx = 1 t 2 1+t 2 )J = Z 2: 2dt 1+t 2 2: 2t 1+t 2 1 t 2 1+t 2 +1 = Z 4dt 4t 1+t 2 +1+t 2 = Z 4dt 2t 2 +4t = 2 Z dt t(t+2) Chinh phöc Olympic To¡n 18 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh = Z 1 t 1 t+2 dt = lnjtj lnjt+2j+C = ln tan x 2 ln tan x 2 +2 +C 3. °t tan x 2 =t) 8 > < > : dx = 2dt 1+t 2 sinx = 2t 1+t 2 ;tanx = 2t 1 t 2 )K = Z 2dt 1+t 2 2t 1+t 2 + 2t 1 t 2 = 1 2 Z 1 t 2 t dt = 1 2 Z dt t 1 2 Z tdt = 1 2 lnjtj 1 4 t 2 +C = 1 2 ln tan x 2 1 4 tan 2 x 2 +C 4. Bi¸n êi gi£ thi¸t ta ÷ñc Z 2 0 ln 1+sinx 1+cosx dx = Z 2 0 ln 0 @ sin 2 x 2 +cos 2 x 2 +2sin x 2 cos x 2 2cos 2 x 2 1 A dx = 1 2 Z 2 0 ln tan 2 x 2 +2tan x 2 +1 dx °t tan x 2 =t)I = 1 2 Z 1 0 t 2 +1 ln t 2 +t+1 dt ¸n ¥y sû döng t½ch ch§t Z b a f (x)dx = Z b a f (a+b x)dx ta s³ t½nh ÷ñc t½ch ph¥n c¦n t½nh. C¡ch 2. Ta câ I = Z 2 0 ln(1+sinx)dx Z 2 0 ln(1+cosx)dx Sû döng t½ch ph¥n tøng ph¦n ta câ Z 2 0 ln(1+sinx)dx = 2 ln2 Z 2 0 xcosx 1+sinx dx Z 2 0 ln(1+cosx)dx = Z 2 0 xsinx 1+cosx dx )I = 2 ln2 Z 2 0 xcosx 1+sinx dx+ Z 2 0 xsinx 1+cosx dx ! Tø ¥y ta s³ i t½nh Z 2 0 xcosx 1+sinx dx. °t t = 2 x ta ÷ñc Z 2 0 xcosx 1+sinx dx = 2 Z 2 0 sinx 1+cosx dx Z 2 0 xsinx 1+cosx dx)I = 0 D¤ng 5 T½nh t½ch ph¥n têng qu¡t I = Z dx a:sin 2 x+b:sinxcosx+c:cos 2 x Chinh phöc Olympic To¡n 19 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC Ph÷ìng ph¡p Ta bi¸n êi v· d¤ng I = Z dx (atan 2 x+btanx+c):cos 2 x . Ta °t tanx =t) dx cos 2 x =dt)I = Z dt at 2 +bt+c Sau ¥y l c¡c v½ dö minh håa cho c¡c b i to¡n n y. T¼m c¡c nguy¶n h m sau I = Z dx 3sin 2 x 2sinxcosx cos 2 x J = Z dx sin 2 x 2sinxcosx 2cos 2 x B i 5 Líi gi£i 1. Ta câ I = Z dx 3sin 2 x 2sinxcosx cos 2 x = Z dx (3tan 2 x 2tanx 1)cos 2 x °t tanx =t) dx cos 2 x =dt)I = Z dt 3t 2 2t 1 = Z dt (t 1)(3t+1) = 1 4 Z 1 t 1 3 3t+1 dt = 1 4 Z dt t 1 1 4 Z d(3t+1) 3t+1 = 1 4 ln t 1 3t+1 +C = 1 4 ln tanx 1 3tanx+1 +C 2. Ta câ J = Z dx sin 2 x 2sinxcosx 2cos 2 x = Z dx (tan 2 x 2tanx 2)cos 2 x °ttanx =t) dx cos 2 x =dt )J = Z dt t 2 2t 2 = Z d(t 1) (t 1) 2 p 3 2 = 1 2 p 3 ln t 1 p 3 t 1+ p 3 +C = 1 2 p 3 ln tanx 1 p 3 tanx 1+ p 3 +C Chinh phöc Olympic To¡n 20 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh D¤ng 6 X²t t½ch ph¥n têng qu¡t I = Z a 1 sinx+b 1 cosx a 2 sinx+b 2 cosx dx Ph÷ìng ph¡p Ta t¼m A; B sao cho a 1 sinx+b 1 cosx =A(a 2 sinx+b 2 cosx)+B(a 2 cosx b 2 sinx) T¼m c¡c nguy¶n h m sau I = Z 4sinx+3cosx sinx+2cosx dx J = Z dx sin 2 x 2sinxcosx 2cos 2 x B i 6 Líi gi£i 1. Ta t¼m A; B sao cho 4sinx+3cosx =A(sinx+2cosx)+B(cosx 2sinx) ) 4sinx+3cosx = (A 2B)sinx+(2A+B)cosx) 8 < : A 2B = 4 2A+B = 3 , 8 < : A = 2 B = 1 Tø â I = Z 2(sinx+2cosx) (cosx 2sinx) sinx+2cosx dx = 2 Z dx Z d(sinx+2cosx) sinx+2cosx = 2x lnjsinx+2cosxj+C 2. Ta t¼m A; B sao cho 3cosx 2sinx =A(cosx 4sinx)+B( sinx 4cosx) ) 3cosx 2sinx = (A 4B)cosx+( 4A B)sinx) 8 < : A 4B = 3 4A+B = 2 , 8 < : A = 11 17 B = 10 17 Tø â ta câ J = Z 11 17 (cosx 4sinx) 10 17 ( sinx 4cosx) cosx 4sinx dx = 11 17 Z dx 10 17 Z d(cosx 4sinx) cosx 4sinx = 11 17 x 10 17 lnjcosx 4sinxj+C D¤ng 7 X²t t½ch ph¥n têng qu¡t I = Z a(sinx) 2 +bsinxcosx+c(cosx) 2 msinx+ncosx dx Chinh phöc Olympic To¡n 21 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC Ph÷ìng ph¡p °t S =a(sinx) 2 +bsinxcosx+c(cosx) 2 Gi£ sû S = (psinx+qcosx)(msinx+ncosx)+r sin 2 x+cos 2 x ,S = (mp+r)(sinx) 2 +(np+mq)sinxcosx+(nq+r)(cosx) 2 , 8 > > > < > > > : mp+r =a np+mq =b nq+r =c , 8 > > > < > > > : mp+r =a np+mq =b mp nq =a c , 8 > > > > > < > > > > > : p = (a c)m+bn m 2 +n 2 q = (a c)n bm m 2 +n 2 r = an 2 +cm 2 bmn m 2 +n 2 Khi â ta câ I = Z (a c)m+bn m 2 +n 2 sinx+ (a c)n bm m 2 +n 2 cosx dx+ an 2 +cm 2 bmn m 2 +n 2 Z dx msinx+ncosx = (a c)n bm m 2 +n 2 sinx (a c)m+bn m 2 +n 2 cosx+ an 2 +cm 2 bmn m 2 +n 2 Z dx msinx+ncosx T½ch ph¥n cuèi còng ta ¢ ÷ñc t¼m hiºu ð d¤ng tr÷îc! Sau ¥y l c¡c v½ dö minh håa cho c¡c b i to¡n n y. T½nh c¡c t½ch ph¥n sau I = Z 3 0 (cosx) 2 dx sinx+ p 3cosx I = Z 3 p 3 2 (sinx) 2 + 4 p 3+3 sinxcosx+2(cosx) 2 3sinx+4cosx dx B i 7 Líi gi£i 1. Gi£ sû (cosx) 2 = (asinx+bcosx) sinx+ p 3cosx +c sin 2 x+cos 2 x , (cosx) 2 = (a+c)(sinx) 2 + a p 3+b sinxcosx+ b p 3+c (cosx) 2 ,a = 1 4 ;b = p 3 4 ;c = 1 4 ) I = 1 2 Z 3 0 p 3 2 cosx 1 2 sinx ! dx+ 1 4 Z 3 0 dx sinx+ p 3cosx Chinh phöc Olympic To¡n 22 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh = 1 2 Z 3 0 cos 6 cosx sin 6 sinx dx+ 1 8 Z 3 0 dx cos 3 sinx+sin 3 cosx = 1 2 Z 3 0 cos x+ 6 dx+ 1 8 Z 3 0 dx sin x+ 3 = 1 2 sin x+ 6 + 1 8 ln tan x 2 + 6 3 0 = 1 2 + 1 8 ln p 3 1 4 1 8 ln p 3 = 1 4 + 1 4 ln p 3 = 1 4 1+ln p 3 2. Gi£ sû 3 p 3 2 (sinx) 2 + 4 p 3+3 sinxcosx+2(cosx) 2 = (asinx+bcosx)(3sinx+4cosx)+c sin 2 x+cos 2 x , 8 > > > < > > > : 3a+c = 3 p 3 2 4a+3b = 4 p 3+3 4b+c = 2 , 8 > > > < > > > : a = p 3 b = 1 c = 2 )I = 1 2 Z 3 0 p 3 2 sinx+ 1 2 cosx ! dx 2 Z 3 0 dx 3sinx+4cosx = 1 2 Z 3 0 sin 3 sinx+cos 3 cosx dx 2 5 Z 3 0 dx sin arcsin 3 5 sinx+cos arcsin 3 5 cosx = 1 2 Z 3 0 cos x 3 dx 2 5 Z 3 0 dx cos(x u) = 1 2 Z 3 0 cos x 3 dx 2 5 Z 3 0 d[sin(x u)] 1 sin 2 (x u) = 1 2 sin x 3 1 5 ln 1+sin(x u) 1 sin(x u) 3 0 = p 3 4 1 5 ln 1+sinxcosu sinucosx 1 sinxcosu+sinucosx 3 0 = p 3 4 + 1 5 ln 5 4sin 3 +3cos 3 5+4sin 3 3cos 3 1 5 ln4 = 1 5 ln 13 4 p 3 4 7+4 p 3 p 3 4 D¤ng 8 X²t t½ch ph¥n têng qu¡t I = Z msinx+ncosx a(sinx) 2 +2bsinxcosx+c(cosx) 2 dx Ph÷ìng ph¡p Gåi 1 ; 2 l nghi»m cõa ph÷ìng tr¼nh a b b c = 0 , 2 (a+c)+ac b 2 = 0, 1;2 = a+c q (a c) 2 +4b 2 2 Chinh phöc Olympic To¡n 23 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC Bi¸n êi mët x½u: a(sinx) 2 +2bsinxcosx+c(cosx) 2 = 1 A 2 1 + 2 A 2 2 = 1 1+ b 2 (a 1 ) 2 cosx b a 1 sinx 2 + 2 1+ b 2 (a 2 ) 2 cosx b a 2 sinx 2 °t u 1 = cosx b a 1 sinx;u 2 = cosx b a 2 sinx;k 1 = 1 a 1 ;k 2 = 1 a 2 A 1 = 1 p 1+b 2 k 2 1 (cosx bk 1 sinx);A 2 = 1 p 1+b 2 k 2 2 (cosx bk 2 sinx) º þ A 2 1 +A 2 2 = 1) 1 A 2 1 + 2 A 2 2 = ( 1 2 )A 2 1 + 2 = ( 2 1 )A 2 2 + 1 Gi£ sû msinx+ncosx =p sinx+ b a 1 cosx +q sinx+ b a 2 cosx , 8 < : p+q =m p a 1 + q a 2 = n b ,p = bm n(a 2 ) b( 2 1 ) (a 1 );q = bm n(a 1 ) b( 1 2 ) (a 2 ) )I = Z msinx+ncosx a(sinx) 2 +2bsinxcosx+c(cosx) 2 dx = Z pdu 1 ( 1 2 )A 2 1 + 2 + Z qdu 2 ( 2 1 )A 2 2 + 1 = p q 1+b 2 k 2 1 Z dA 1 ( 1 2 )A 2 1 + 2 q q 1+b 2 k 2 2 Z dA 2 ( 2 1 )A 2 2 + 1 Sau ¥y l c¡c v½ dö minh håa cho c¡c b i to¡n n y. T½nh t½ch ph¥n sau I = Z (sinx+cosx)dx 2sin 2 x 4sinxcosx+5cos 2 x B i 8 Líi gi£i Gåi 1 ; 2 l 2 nghi»m cõa ph÷ìng tr¼nh 2 2 2 5 = 0, 1 = 1; 2 = 6 Ta câ: 2sin 2 x 4sinxcosx+5cos 2 x = 1 5 (cosx+2sinx) 2 + 24 5 cosx 1 2 sinx 2 A 1 = 1 p 5 (cosx+2sinx);A 2 = 2 p 5 cosx 1 2 sinx ;A 2 1 +A 2 2 = 1 Chinh phöc Olympic To¡n 24 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh )I = Z (sinx+cosx)dx 2sin 2 x 4sinxcosx+5cos 2 x = 3 5 Z (2sinx+cosx)dx (2cosx sinx) 2 +1 1 5 Z (sinx 2cosx)dx 6 (cosx+2sinx) 2 = 3 5 Z d(sinx 2cosx) (sinx 2cosx) 2 +1 + 1 5 Z d(cosx+2sinx) 6 (cosx+2sinx) 2 = 3 5 arctan(sinx 2cosx)+ 1 10 p 6 ln p 6+cosx+2sinx p 6 cosx 2sinx +C D¤ng 9 Bi¸n êi n¥ng cao vîi 2 d¤ng t½ch ph¥n Z dx (sinx) n v R dx (cosx) n Thüc ch§t m¼nh chia d¤ng to¡n n y th nh 1 d¤ng to¡n nhä v¼ trong khi t½nh nguy¶n h m ho°c t½ch ph¥n ta s³ câ thº g°p c¡c b i to¡n kiºu th¸ n y, do â m¼nh muèn giîi thi»u cho c¡c b¤n c¡c c¡ch º xû lþ nâ. X²t b i to¡n Z dx (sinx) n I 1 = Z dx sinx = Z dx 2sin x 2 cos x 2 = Z dx 2tan x 2 cos 2 x 2 = Z d tan x 2 tan x 2 = ln tan x 2 +C I 2 = Z dx sin 2 x = Z d(cotx) = cotx+C I 3 = Z dx sin 3 x = Z dx 2sin x 2 cos x 2 3 = Z dx 8 tan x 2 3 cos x 2 6 = 1 4 Z 1+tan 2x 2 2 d tan x 2 tan x 2 3 = 1 4 Z 1+2tan 2x 2 +tan 4x 2 tan x 2 3 = 1 4 " 1 2 tan x 2 2 +2ln tan x 2 + 1 2 tan x 2 2 # +C I 4 = Z dx sin 4 x = Z 1+cot 2 x d(cotx) = cotx+ 1 3 cot 3 x +C I 5 = Z dx sin 5 x = Z dx 2sin x 2 cos x 2 5 = Z dx 32 tan x 2 5 cos x 2 10 = 1 16 Z 1+tan 2x 2 4 d tan x 2 tan x 2 5 = 1 16 Z 1+4tan 2x 2 +6tan 4x 2 +4tan 6x 2 +tan 8x 2 tan x 2 5 d tan x 2 = 1 16 " 1 4 tan x 2 4 2 tan x 2 2 +6ln tan x 2 +2 tan x 2 2 + 1 4 tan x 2 4 # +C I 6 = Z dx sin 6 x = Z 1+cot 2 x 2 d(cotx) = cotx+ 2 3 cot 3 x+ 1 5 cot 5 x +C Chinh phöc Olympic To¡n 25 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC I 7 = Z dx sin 7 x = Z dx 2sin x 2 cos x 2 7 = Z dx 2 7 tan x 2 7 cos x 2 14 = 1 2 6 Z 1+tan 2x 2 6 d tan x 2 tan x 2 7 = 1 64 " 1 6 tan x 2 6 3 2 tan x 2 4 15 2 tan x 2 2 +20ln tan x 2 + 15 2 tan x 2 2 + 3 2 tan x 2 4 + 1 6 tan x 2 6 +C I 8 = Z dx sin 8 x = Z 1+cot 2 x 3 d(cotx) = Z 1+3cot 2 x+3cot 4 x+cot 6 x d(cotx) = cotx+cot 3 x+ 3 5 cot 5 x+ 1 7 cot 7 x +C I 9 = Z dx (sinx) 2n+1 = Z dx 2sin x 2 cos x 2 2n+1 = Z dx 2 2n+1 tan x 2 2n+1 cos x 2 4n+2 = 1 2 2n Z 1+tan 2x 2 2n d tan x 2 tan x 2 2n+1 = 1 2 2n " C 0 2n 2n tan x 2 2n ::: C n 1 2n 2 tan x 2 2 +C n 2n ln tan x 2 + C n+1 2n 2 tan x 2 2 +:::+ C 2n 2n 2n tan x 2 2n +C I 10 = Z dx sin 2n+2 x = Z 1+cot 2 x n d(cotx) = Z h C 0 n +C 1 n cot 2 x+:::+C k n cot 2 x k +:::+C n n cot 2 x n i d(cotx) = C 0 11 (cotx)+ C 1 n 3 cot 3 x+:::+ C k n 2k+1 (cotx) 2k+1 +:::+ C n n 2n+1 (cotx) 2n+1 +C X²t b i to¡n I = Z dx (cosx) n I 1 = Z dx cosx = Z d x+ 2 sin x+ 2 = Z du sinu = Z du 2sin u 2 cos u 2 = Z du 2tan u 2 cos 2 u 2 = Z d tan u 2 tan u 2 = ln tan u 2 +C = ln tan x 2 + 4 +C I 2 = Z dx cos 2 x = Z d(tanx) = tanx+C Chinh phöc Olympic To¡n 26 h T¤p ch½ v t÷ li»u to¡n håcTP CH V T× LIU TON HÅC Ph¤m Vi»t Anh I 3 = Z dx cos 3 x = Z d x+ 2 sin 3 x+ 2 = Z du sin 3 u = Z du 2sin u 2 cos u 2 3 = Z du 8 tan u 2 3 cos u 2 6 = 1 4 Z 1+tan 2u 2 2 d tan u 2 tan u 2 3 = 1 4 " 1 2 tan u 2 2 +2ln tan u 2 + 1 2 tan u 2 2 # +C = 1 4 " 1 2 tan x 2 + 4 2 +2ln tan x 2 + 4 + 1 2 h tan x 2 + 4 i 2 # +C I 4 = Z dx cos 4 x = Z 1+tan 2 x d(tanx) = tanx+ 1 3 tan 3 x+C I 5 = Z dx cos 5 x = Z d x+ 2 sin 5 x+ 2 = Z du sin 5 u = Z du 2sin u 2 cos u 2 5 = Z du 32 tan u 2 5 cos u 2 10 = 1 16 Z 1+tan 2u 2 4 d tan u 2 tan u 2 5 = 1 16 Z 1+4tan 2u 2 +6tan 4u 2 +4tan 6u 2 +tan 8u 2 tan u 2 5 d tan u 2 = 1 16 2 6 4 1 4 tan u 2 4 2 tan u 2 2 +6ln tan u 2 +2 tan u 2 2 + 1 4 tan u 2 4 3 7 5 +C I 6 = Z dx cos 6 x = Z 1+tan 2 x 2 d(tanx) = tanx+tan 3 x+ 1 5 tan 5 x+C I 7 = Z dx cos 7 x = Z d x+ 2 sin 7 x+ 2 = Z du sin 7 u = Z du 2 7 tan u 2 7 cos u 2 14 = 1 2 6 Z 1+tan 2 u 2 6 d tan u 2 tan u 2 7 = 1 2 6 Z 1+6tan 2 u 2 +15tan 4 u 2 +20tan u 2 +15tan 8 u 2 +6tan 10 u 2 +tan 12 u 2 tan u 2 7 d tan u 2 = 1 64 " 1 6 tan u 2 6 3 2 tan u 2 4 15 2 tan u 2 2 +20ln tan u 2 + 15 2 tan u 2 2 + 3 2 tan u 2 4 + 1 6 tan u 2 6 +C Chinh phöc Olympic To¡n 27 h T¤p ch½ v t÷ li»u to¡n håcNguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC I 8 = Z dx cos 8 x = Z 1+tan 2 x 3 d(tanx)= R (1+3tanx 2 +3tanx 4 +tanx 6 )d(tanx) = tanx+tan 3 x+ 3 5 tan 5 x+ 1 7 tan 7 x+C I 9 = Z dx cos 2n+1 x = Z d x+ 2 sin 2n+1 x+ 2 = Z du (sinu) 2n+1 = Z du 2sin u 2 cos u 2 2n+1 = Z du 2 2n+1 tan u 2 2n+1 cos u 2 4n+2 = 1 2 2n Z 1+tan 2 u 2 2n d tan u 2 tan u 2 2n+1 = 1 2 2n Z C 0 2n +C 1 2n +C 1 2n tan 2u 2 +:::+C n 2n tan 2u 2 n +:::+C 2n 2n tan 2u 2 2n tan u 2 2n+1 d tan u 2 = 1 2 2n " C 0 2n 2n tan u 2 2n ::: C n 1 2n 2 tan u 2 2 +C n 2n ln tan u 2 + C n+1 2n 2 tan u 2 2 +:::+ C 2n 2n 2n tan u 2 2n +C I 10 = Z dx cos 2n+2 x = Z 1+tan 2 x n d(tanx) = Z C 0 11 +C 1 n tan 2 x+:::+C k n tan 2 x k +:::+C n n tan 2 x n d(tanx) = C 0 n (tanx)+ C 1 n 3 tanx 3 +:::+ C k n 2k+1 (tanx) 2k+1 +:::+ C n n 2n+1 (tanx) 2n+1 +C Tâm l¤i. Qua c¡c b i to¡n vîi nhúng líi gi£i kinh khõng ð tr¶n chc ¢ l m b¤n åc cho¡ng rçi, tuy nhi¶n h¢y º þ nâ câ m§u chèt c£ nh². ¦u ti¶n l 2 d¤ng n y t÷ìng tü nhau n¶n m¼nh s³ ch¿ nâi mët d¤ng. C¡c b¤n h¢y chó þ tîi c¡c b i sè mô ch®n, m§u chèt ch¿ l sû döng cæng thùc theo tan v sin, cán nhúng b i sè mô l´ ta ·u sû döng c¡ch t¡ch sinx = 2sind x 2 cos x 2 â ch½nh l ch¼a kho¡ cõa c¡c b i to¡n tr¶n, líi gi£i khõng ch¯ng qua l bi¸n êi d i thæi chù khæng câ g¼ khâ kh«n c£! Chinh phöc Olympic To¡n 28 h T¤p ch½ v t÷ li»u to¡n håcCh÷ìng 2 B i To¡n · Xu§t T½nh c¡c nguy¶n h m ho°c t½ch ph¥n sau B i 1. 3 Z 4 tan 4 xdx B i 2. 2 Z 4 cos 6 x sin 4 x dx B i 3. 4 Z 0 sin 2 x cos 6 x dx B i 4. 2 Z 0 sin2x 4 cos 2 x dx B i 5. 4 Z 0 1 2sin 2 x 1+sin2x dx B i 6. I = 2 Z 0 sin 10 x+cos 10 x sin 4 xcos 4 x dx B i 7. I = 3 Z 6 1 sinxsin x+ 6 dx B i 8. I = 2 Z 0 sin2x+sinx p 1+3cosx dx B i 9. I = 2 Z 0 sin2xcosx 1+cosx dx 29Nguy¹n Minh Tu§n TCH PH N HM L×ÑNG GIC B i 10. I = 2 Z 0 sin2x p cos 2 x+4sin 2 x dx B i 11. I = 2 Z 0 cos3x sinx+1 dx B i 12. I = 2 Z 0 cos2x (sinx cosx+3) 3 dx B i 13. I = 4 Z 0 cos2x 1+2sin2x dx B i 14. I = 6 Z 0 sin3x sin 3 3x 1+cos3x dx Chinh phöc Olympic To¡n 30 h T¤p ch½ v t÷ li»u to¡n håcT i li»u tham kh£o [1] Tuyºn tªp c¡c chuy¶n · v kÿ thuªt t½nh t½ch ph¥n - Tr¦n Ph÷ìng [2] T i li»u Internet 31