TTBDKTPT VQ GV: PHAN VĂN VINH ‹ CÁC BÀI TOÁN TÍCH PHÂN LUYỆN THI ĐẠI HỌC‹ 1/ Cho hàm số : f(x)= x.sinx+x2 Tìm nguyên hàm hàm số g(x)= x.cosx biết nguyên hàm triệt tiêu x=k π 2/Định m để hàm số: F(x) = mx +(3m+2)x2 -4x+3 nguyên hàm hàm số: f(x) = 3x2 +10x-4 3/Tìm họ nguyên hàm hàm số: f(x)= cos x.sin8x 3.TÍNH : π 4/I = ∫ 3tg x dx ∫ sin x dx 12 / I = π 13/I = ∫ 5/I = ∫ (2cotg x + 5) dx π π − cos x π ∫ + cos x dx 7/ I = ∫ sin2 x.cos2xdx ∫π x x dx sin cos 2 15/I = π π ∫ 2 (2cos x-3sin x)dx 16/I = − π π − x ) dx π s in ( + x) cotg2x dx 17/I = ∫ esin − x sin 2x dx π ∫π e tgx+ 2 I= ∫ cos x (tgx-cotgx)2 dx 18/ π π 11/ I = ∫ cos x dx ∫ 4 21/I = cos2x(sin x + cos x)dx π 22/ I = ∫ cos3 xdx π 4sin x 23/ I = ∫ dx + cosx ∫ x − x dx 25/I = ∫ x + x dx x dx 2x + 1 27/I = ∫ x dx 0e +4 dx 28/I = ∫ −x 11− e e2x 29/I = ∫ x dx e + 1 e− x dx 30/I = ∫ − x e + e ln x dx 31/I = ∫ x(ln x + 1) 26/I = ∫ π π dx π s in ( 10 / I = ∫ π x π 24/ I = 9/ I= ∫ π ∫ sin x dx 14/I = π π sin x − sin x cot gx dx sin x ∫ cos 20/ I = π 8/I = π π 6/I = π π 19/ I = ∫ sin x dx π ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 TTBDKTPT VQ 3 34/I = ∫ x 4−x 35/I = ∫ 36*/I = x 16 − x ∫ 2 dx dx x x −9 dx −1 38/I = ∫ x (x + 4)3 dx x2 − dx x ∫ 39/I = 3 − ∫ 40*/I = −2 ln 41/I = ∫ x2 +1 dx x x +1 e x − 1dx 1 42/I = ∫ dx − 2x ln x + ln x dx 48/I = ∫ x e sin(ln x) dx 49/I = ∫ x e 5 43/I = ∫ sin xdx π 44*/I = ∫ dx cos x e−2x dx 45/I = ∫ − x e + ln dx 46/I = ∫ ex + dx ln e − e x +1 62/I = ∫ ln xdx x 61/I = 51/I = ∫ (1 + 2x)(1 + 3x + 3x )3 dx 52/I = ∫ 1x π 1+ x π 64/I = ∫ sin x.sin 2x.sin 3xdx π 65/I = ∫ cos 2x(sin x + cos x)dx π dx 66*/I = ∫ ( cos x − sin x )dx 2 53/I = ∫ tg x + cot g x − 2dx π x7 67/I = ∫ dx + x − 2x π 68*/I = ∫ 4cos x − 3sin x + dx 54/I = ∫ (1 − x )3 dx 0 55*/I = ∫ 2x dx e + ln ex ∫ x (e + 1) 69/I = ∫ x − xdx dx 57/I = ∫ x(e2x + x + 1)dx π 58/I = ∫ − cos3 x sin x.cos5 xdx 59*/I = ∫ x x2 + π x dx + cos 2x 60/I = ∫ ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ 4sin x + 3cos x + 56/I = x x2 dx 63/I = ∫ (x + 1) x + 0 ∫ −1 π dx 50/I = ∫ x (x − 1) dx 2 ∫ x − x dx sin x cot gx π 37/I = 47/I = ∫ 33/I = ∫ (x − 3) x − 6x + dx e2x ln π x +1 dx 3x + 32/I = ∫ GV: PHAN VĂN VINH x +1 dx 3x + π x 71*/I = ∫ sin dx 2 x 72*/I = ∫ dx + x + − x 70/I = ∫ 73/I = dx ∫ x + x dx ln(1 + x) dx x + 74**/I = ∫ π sin x dx sin x + cos x 75/I = ∫ DĐ: 0905.838.969 TTBDKTPT VQ eπ 76/I = ∫ cos(ln x)dx 77*/I = ∫ + x dx x dx + x − 1 78/I = ∫ e 79/I = ∫ GV: PHAN VĂN VINH + 3ln x ln x dx x 90*/I = ∫ ln( + x − x)dx x −1 x +1 dx 92/I = ∫ x x3 93/I = ∫ dx x − 16 82/I = ∫ e e2 83/I = ∫ cos x dx − 5sin x + sin x 94/I = ∫ 85/I = ln x dx x ln x dx ln x 86/I = ∫ 87/I = ∫ x + dx 1 − x2 dx π2 ∫ sin xdx π ln(sin x) 88/I = ∫ π cos x e2 dx 89/I = ∫ cos(ln x)dx 96/I = 1 − )dx ln x ln x ∫ x 2ex dx 110*/I = ∫ (x + 2) π 111/I = ∫ e2x sin xdx cos 2x + 1dx 99/I = ∫ cos x e ln x dx (x + 1) sin xdx 113/I = ∫ 2π e ∫ + sin xdx 3π 114/I = ∫ x.ln ∫ sin 2x dx t 102/I = ∫ − sin xdx π 3 103/I = ∫ ln(x + x + 1) dx −1 1+ x dx 1− x ln x 115/I = ∫ dx ⇒ I < x π π 1 x 112/I = ∫ x ln(1 + )dx π π 101/I = ∫ x cos xdx 97/I = ∫ x − 2x − x + dx −1 3π 108/I = π2 109/I = ∫ x.sin x cos xdx ∫ x − dx 100/I = ∫ x sin xdx π −4 98/I = π2 e 84/I = ∫ x ln(x + 1)dx 107/I = e2 dx 104*/I = ∫ π 95*/I = ∫ ( 81/I = ∫ (ln x) dx 2 80/I = ∫ ln(x − x)dx e ∫ 91*/I = x sin x dx + cos x 1 105*/I = ∫ dx x (x + 1)(4 + 1) −1 x4 106*/I = ∫ dx x + −1 π 116/I = ∫ sin x.ln(cos x)dx π e2 117/I = ∫ cos (ln x)dx ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 TTBDKTPT VQ π dx cos x 118/I = ∫ GV: PHAN VĂN VINH dx cos x 120/I = ∫ x 3e x dx π 2 121/I = ∫ esin x sin x cos3 xdx π sin 2x 122/I = ∫ dx + cos x 123/I = ∫ dx x − 4x − 5 dx 124/I = ∫ x − 6x + 1 dx 125/I = ∫ 2x + 8x + 26 −5 2x + 126/I = ∫ dx x + dx 127/I = ∫ x (x + 1) sin 2x 128*/I = ∫ dx −π (2 + sin x) x −3 129/I = ∫ dx (x + 1)(x + 3x + 2) 4x dx (x + 1) 1 131/I = ∫ dx (x + 4x + 3) 130/I = ∫ π 3 sin x dx (sin x + 3) 132/I = ∫ 4sin x dx π − cos x 145/I = ∫ x − xdx π 146/I = ∫ 133/I = ∫ π 119*/I = ∫ π 134/I = ∫ dx cos x.sin x π −1 π 148/I = ∫ 135/I = ∫ sin x.tgxdx π −1 139/I = π 140/I = 152/I = dx 2 π sin x + 9cos x ∫ 153/I = cos x − 1 + sin x cos x dx sin x + cos x + dx 142/I = ∫ x (x + 1) 141/I = ∫ −3 x + + (x + 4)3 π sin x dx 144/I = ∫ cos x ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ ∫ + e2x x 9+x dx dx π π ∫ 3e4x + e2x 154/I = ∫ e x sin xdx ∫ + 3cos x dx 143/I = ∫ dx π ∫ cos x + dx π − π 2 x + 4x + 13 151/I = ∫ dx x 3+ e −2 sin3 x dx 137/I = ∫ 2 (tg x + 1) cos x π 2x − ∫ 150/I = π − 4x − x + dx 149/I = ∫ dx 136/I = ∫ sin 2x π 138/I = ∫ 147/I = x−4 dx x+2 x+2 dx x + 2x + dx 4x − x dx cos x 155/I = ∫ dx 4 cos x + sin x dx 156/I = ∫ x+9 − x π 157/I = ∫ x sin xdx π 158/I = ∫ x cos xdx 159/I = ∫ cos x dx DĐ: 0905.838.969 TTBDKTPT VQ 160/I = ∫ sin x dx GV: PHAN VĂN VINH 161/I = π2 ∫ x sin x dx 162/I = π ln x 176/I = ∫ dx x e ln x dx 177/I = ∫ (x + 1) ∫ x cos x dx π 163/I = ∫ x cos x sin x dx 164/I = π ∫ x cos x sin x dx x 165/I = ∫ e e π π 180/ ∫ esin π 181/I= 2x sin x dx x x e 168/I = ∫ dx (x + 2) e 169/I = ∫ (1 + x) ln x dx e 170/I = ∫ x ln x dx 1 e 171/I = ∫ ln x dx e 172/I = ∫ x(2 − ln x) dx 173/I = ∫ ( e 189/I = ∫ 1 − )dx ln x ln x 174/I = ∫ (x + x) ln x dx x e +e e 190/I= dx ∫ ln x dx π 191/I = ∫ (esin x + cos x) cos x dx x sin x cos3 x dx π sin 2x.cos x dx + cos x 192/I = ∫ π sin 2x + sin x dx + 3cos x 193/I = ∫ π − 2sin x dx + sin 2x 194/I = ∫ π −x e 166/I = ∫ e3x sin 4x dx e2 1+ x dx 1− x 179/I = ∫ cos x.ln(1 − cos x) dx π dx ex 178/I = ∫ x ln π 167/I = ∫ e x 175/I = ∫ x ln(1 + ) dx sin 2x ∫ + sin x dx π sin 2x dx 182/I = ∫ + cos x 183/I = ∫ dx x − 6x + 1 x + 3x + dx 184/I = ∫ x + 185/I = ∫ dx x (x + 1) 1 ln(1 + x) dx 186/I = ∫ x + 1 1+ x4 187/I ∫ dx + x 188/I = ∫ x15 + x dx ∫ 195/I = x +1 π 196/I = ∫ π x + 2x dx tgx cos x + cos x 197/I = ∫ ( −1 π dx x −1 ) dx x+2 198/I = ∫ x.tg x dx 199/I = ∫ ( x + − x − ) dx −3 200/I = ∫ −1 201/I = ∫ dx x +5 +4 x dx x+2 + 2−x ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 TTBDKTPT VQ GV: PHAN VĂN VINH ln(1 + x) 202/I = ∫ dx x2 π sin 2x 203/I = ∫ dx + cos x π sin 2008 x dx 2008 x + cos 2008 x sin 204/I = ∫ 216/I = 2 x ∫ π 2 dx 1− x 1− x2 dx 217/I = ∫ 11+ x 218/I = ∫ x π ln sin x dx cos x 207/I = ∫ π 2 208/I = ∫ cos x.cos 4x dx 1 dx 2x x e + e e ln x 210/I = ∫ dx (x + 1) 209/I = ∫ e 1 211/I = ∫ dx x + + x x2 dx 212/I = ∫ − x x 213/I = ∫ dx 4−x x4 dx 214/I = ∫ x −1 π sin 3x 215/I = ∫ dx cos x + 1 x − 3x + 232*/I = ∫ x sin x.cos xdx dx π cos x dx cos 2x + 234/I = ∫ dx x (x + 1) 233/I = ∫ π 221/I = ∫ x + 1dx 235/I = ∫ sin 2x(1 + sin x) dx π 222/I = ∫ (cos3 x + sin x) dx x2 +1 223/I = ∫ dx + x ∫ x x +9 dx π 238/I = ∫ x sin x cos xdx cos x cos x + dx x +1 226/I = ∫ dx 3x + π + sin 2x + cos 2x dx cos x + sin x 227/I = ∫ π x (1 + e ) dx 2x e + 228/I = ∫ x +1 dx 3x + 236/I = ∫ 224/I = ∫ (1 + x) e2x dx 237/I = 225/I = ∫ dx π π 4x − 231/I = ∫ 1+ x − ex 205/I = ∫ sin x.ln(1 + cos x) dx 219/I = ∫ dx x + e x +1 206/I = ∫ dx 220/I = ∫ x − x dx x π sin x.cos3 x dx 230/I = ∫ cos x + 229/I = ∫ x (1 − x) dx ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ π 239/I = ∫ cosx cosx − cos3 xdx − π 240*/I = ∫ ln( x + a + x)dx −1 π − sin x dx x (1 cos x)e + 241/I = ∫ π sin 2x + sin x dx + cos3x 242/I = ∫ π sin 2x dx 2 sin x + 2cos x 243/I = ∫ DĐ: 0905.838.969 TTBDKTPT VQ 244/I = 2 ∫ 245/I = 2 ∫ 246/I = ∫ 2 247/I = ∫ 248/I = ∫ x π 3 1− x2 x3 1− x GV: PHAN VĂN VINH dx 256/I = ∫ tg xdx dx π + sin x 257*/I = ∫ + cos x 1− x dx x2 − x2 dx x x −1 dx π sin x dx + sin x 250/I = ∫ π cos x dx + cos 2x 252/I = ∫ dx (1 + x)x x +1 dx 253/I = ∫ 3x + 251/I = ∫ π e x dx 258/I = ∫ (1 − x ) dx π 259/I = ∫ x.tg xdx cos x + sin x dx + sin 2x π dx 2 (4 + x ) 3x 261/I = ∫ dx x +2 − x5 262*/I = ∫ dx x(1 + x ) 260/I= ∫ π cos x dx − sin x 263/I = ∫ π sin x 264/I = ∫ dx cos x π sin x + sin x 265/I = ∫ dx cos 2x π dx π sin x + cos x 254*/I = ∫ 266/I = ∫ π 255/I = ∫ cosx cosx − cos3 xdx − π sin x − cos x 270/I = ∫ dx sin x + cos x + π sin x − cos x 271/I = ∫ dx sin x + cos x + π sin x cos x + cos x dx sin x + 272/I = ∫ 249/I = ∫ x (1 − x )6 dx π 269/I = ∫ sinxcosx(1+cosx)2dx π x π sin x dx cos x + 267/I = ∫ π π2 268/I = ∫ sin x dx x ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ 273/I = ∫ ex dx x x + 2x + 10x + 274/I = ∫ dx x + 2x + x3 dx 275/I = ∫ (x + 1) 276/I = ∫ dx x + 1 x +1 dx 277*/I = ∫ + x 1 x 278/I = ∫ dx (2x + 1) dx 279/I = ∫ 2 + x +1 a 280/I = ∫ 281*/I = ∫ x 1− x dx x ln(x + + x ) 1+ x dx 282/I = ∫ (x − 1)2 ln x dx DĐ: 0905.838.969 TTBDKTPT VQ GV: PHAN VĂN VINH 283/I = ∫ x ln(x + 1)dx 296/I = 3x 284/I = ∫ dx x + 2x + 1 4x − 285/I = ∫ dx + + + x 2x x 2 286/I = ∫ −1 (3+ 2x) 2 5+12x +4x 287/I = ∫ x + 1+ x π dx dx cos x dx + cos 2x 288/I = ∫ π cos x + sin x dx + sin 2x π 289/I = ∫ π 290/I = ∫ (cos3 x + sin x)dx π 291/I = ∫ cos5 x sin xdx π 292/I = ∫ cos2x(sin4 x+cos4 x)dx π dx + sin x 293/I = ∫ π dx − cos x 294/I = ∫ 295/I = ∫ x x −1 dx ∫ 297*/I = ∫ 1 298/I = ∫ 1+ x2 π sin x dx cos x + sin x dx + 1+ x2 −1 + π dx 2x −1 + e 309*/I = ∫ dx x 1+ x x3 0x 299/I = ∫ x3 310*/I = ∫ π dx x + 1+ x2 dx 300/I = ∫ dx π sin x cos x π cos x dx cos x + π cos x dx 302/I = ∫ − cos x π sin x 303/I = ∫ dx sin x + π cos3 x 304/I = ∫ dx cos x + 306/I = π tgx 312*/I = ∫ − ln (cos x) dx π sin x dx cos x + sin x 1 dx 314*/I = ∫ x −1 (e + 1)(x + 1) 313*/I = ∫ 301/I = ∫ 305/I = sin x 311/I = ∫ dx 4 cos x + sin x π ∫ 2cos x + sin x + dx π cos x dx ∫ (1 − cos x) π π 307/I = ∫ tg x dx sin x dx 308*/I= ∫ x + −π π ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ 315*/I = ∫ e 3x +1 dx x2 316*/I = ∫ π 2 x +4 dx cos3 x dx 317*/I = ∫ cos − 3cos x + 318*/Tìm x> để π 319*/I = ∫ π t 2et dt = ∫ (t + 2) x tan x cos x cos x + dx 320*/I = ∫ −3x + 6x + 1dx π 321*/I = ∫ tg x dx DĐ: 0905.838.969 TTBDKTPT VQ π 336,Cmr 322/I = ∫ cotg x dx π π 50 dt ∫ ( 3+ 2cost ) 339, Cmr ∫ + tgx dx ∫ 2x −1 x −1 dx < ∫ dx x x +1 341, I = ∫ π x x2 + cos 2x dx π − cos 2x 342, I = ∫ x3 − x2 dx sin 2x + sin x dx 343, I = ∫ + 3cos x 0 π π 327*/I = ∫ ( t g x − ) d x tg x + 1 328*/I = 2 (e x + 1) e x − π −1 e4 e 332/I = ∫ dx 346, I = ∫ x + x +1 dx x cos (ln x + 1) ln 348, I = + 3ln x ln x dx x dx ∫ln3 ex + 2e− x − sin 2x cos x dx + cos x 350, I = ∫ ln ( x − x ) dx π π 333*/I = ∫ ln(1 + tgx)dx π π dx π ≤∫ ≤ 16 + 3cos x 10 ( −7 ) 358, I = dx ∫ x+x π 359, I = − cos 2xdx ∫ π − 360, I = ∫ ( x − 1) dx x2 − x − x ln x + + x 361, I = ∫ ( + x2 e2 ln x + ln ( ln x ) x π )dx dx 363, J = ∫ cos2 x dx π sin x x2 dx x6 − 364, I = ∫ π dx 365, I = ∫ π sin x cos x π sin 2x dx sin x + sin x + 366, I = ∫ π 367 I = ∫ sin xex dx π 368, I = ∫ cos x co s 7x d x 352, I = ∫ ( x − ) e dx 2x 54 ≤ ∫ x+7+ 11−x dx ≤108 − 2sin x dx + s in2x 357, I = ∫ 351, I = ∫ ( esin x + cos x ) cos xdx 0 π π 349, I = ∫ dx x (1 + x ) 11 e 347, I = ∫ 356, I = ∫ x sin xdx e dx π2 362, I = ∫ dx 345, I = ∫ + x2 ex ln 331/I = ∫ x − x3 dx x4 329*/I = ∫ 344, I = ∫ − x dx x dx x +1 ∫ 330/I = ∫ 355, I = ∫ x3ex dx dx 326/I = ∫ ) 2 340, I = ∫ 325/I = ∫ sin x dx cos x + ( 3+ x dx 1+ x −1 1 ≤ 2 ≤ ∫ e x − x dx ≤ e e π 335,Cmr 354, I = ∫ 338,Cmr π 324*/I = x dx + cos 2x ≤ ∫ dx ≤ 337,Cmr ≤ π π x3 −1 323/I = ∫ tg x dx 334,Cmr GV: PHAN VĂN VINH 353, I = ∫ x − x dx ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ 369, I = π sin x + cos4 x ∫π 3x + dx − DĐ: 0905.838.969 TTBDKTPT VQ GV: PHAN VĂN VINH AN ACT OF KINDNESS IS NEVER WASTED ! ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 10 ... x dx 334,Cmr GV: PHAN VĂN VINH 353, I = ∫ x − x dx ĐC: KIỆT I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ 369, I = π sin x + cos4 x ∫π 3x + dx − DĐ: 0905.838.969 TTBDKTPT VQ GV: PHAN VĂN VINH AN ACT... 0905.838.969 TTBDKTPT VQ eπ 76/I = ∫ cos(ln x)dx 77*/I = ∫ + x dx x dx + x − 1 78/I = ∫ e 79/I = ∫ GV: PHAN VĂN VINH + 3ln x ln x dx x 90*/I = ∫ ln( + x − x)dx x −1 x +1 dx 92/I = ∫ x x3 93/I = ∫ dx... I, TRUNG ĐÔNG, PHÚ THƯỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 TTBDKTPT VQ π dx cos x 118/I = ∫ GV: PHAN VĂN VINH dx cos x 120/I = ∫ x 3e x dx π 2 121/I = ∫ esin x sin x cos3 xdx π sin 2x 122/I =