TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 1 ‹ CÁC BÀI TOÁN TÍCH PHÂN LUYỆN THI ĐẠI HỌC‹ 1/ Cho hàm số : f(x)= x.sinx+x 2 . Tìm nguyên hàm của hàm số g(x)= x.cosx biết rằng nguyên hàm này triệt tiêu khi x=k π 2/Định m để hàm số: F(x) = mx 3 +(3m+2)x 2 -4x+3 là một nguyên hàm của hàm số: f(x) = 3x 2 +10x-4. 3/Tìm họ nguyên hàm của hàm số: f(x)= cos 3 x.sin8x. 3.TÍNH : 4/I = 3 2 4 3tg x dx π π ∫ 5/I = 4 2 6 (2cotg x 5)dx π π + ∫ 6/I = 2 0 1 cosx dx 1 cos x π − + ∫ 7/ I = ∫ 2 0 π sin 2 x.cos 2 xdx 8/I = ∫ 3 0 π (2cos 2 x-3sin 2 x)dx 9 / I = 2 2 s i n ( x ) 4 d x s i n ( x ) 4 π − π π − π + ∫ 10 / I = ∫ − 3 6 π π (tgx-cotgx) 2 dx 11/ I = 4 4 0 cos x dx π ∫ 12 / I = 2 3 0 sin x dx π ∫ 13/I = 3 3 2 3 sin x sin x cotgxdx sin x π π − ∫ 14/I = 2 4 0 sin xdx π ∫ 15/I = ∫ 3 4 22 2 cos 2 sin 1 π π xx dx 16/I = ∫ 4 6 π π cotg2x dx 17/I = 2 2 sin x 4 e sin 2x dx π π ∫ 18/ I = ∫ + 4 0 2 2 cos π x e tgx 19/ I = ∫ 2 4 4 sin 1 π π x dx 20/ I = ∫ 4 0 6 cos 1 π x dx 21/I = dxxxnsix )cos(2cos 44 2 0 + ∫ π 22/ I = 2 3 0 cos xdx π ∫ 23/ I = 3 2 0 4sin x dx 1 cosx π + ∫ 24/ I = 1 3 2 0 x 1 x dx − ∫ 25/I = 1 5 2 0 x 1 x dx + ∫ 26/I = 1 0 x dx 2x 1 + ∫ 27/I = 1 x 0 1 dx e 4 + ∫ 28/I = 2 x 1 1 dx 1 e − − ∫ 29/I = 2x 2 x 0 e dx e 1 + ∫ 30/I = x 1 x 0 e dx e 1 − − + ∫ 31/I = e 2 1 ln x dx x(ln x 1) + ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 2 32/I = 7 3 3 0 x 1 dx 3x 1 + + ∫ 33/I = 2 3 2 0 (x 3) x 6x 8 dx − − + ∫ 34/I = 1 2 2 3 1 dx x 4 x − ∫ 35/I = 4 2 2 1 dx x 16 x − ∫ 36*/I = 6 2 2 3 1 dx x x 9 − ∫ 37/I = 2 2 2 1 x 4 x dx − − ∫ 38/I = 2 2 3 0 x (x 4) dx + ∫ 39/I = 2 4 4 3 3 x 4 dx x − ∫ 40*/I = 2 2 2 2 x 1 dx x x 1 − − + + ∫ 41/I = ln 2 x 0 e 1dx − ∫ 42/I = 1 0 1 dx 3 2x − ∫ 43/I = 2 5 0 sin xdx π ∫ 44*/I = 3 0 1 dx cosx π ∫ 45/I = 2x 1 x 0 e dx e 1 − − + ∫ 46/I = ln3 x 0 1 dx e 1 + ∫ 47/I = 4 2 6 1 dx sin x cotgx π π ∫ 48/I = 3 2 e 1 ln x 2 ln x dx x + ∫ 49/I = e 1 sin(ln x) dx x ∫ 50/I = 1 3 4 5 0 x (x 1) dx − ∫ 51/I = 1 2 3 0 (1 2x)(1 3x 3x ) dx + + + ∫ 52/I = 2 3 1 1 dx x 1 x + ∫ 53/I = 3 2 2 6 tg x cotg x 2dx π π + − ∫ 54/I = 1 2 3 0 (1 x ) dx − ∫ 55*/I = 1 2x 0 1 dx e 3 + ∫ 56/I = x ln3 x 3 0 e dx (e 1)+ ∫ 57/I = 0 2x 3 1 x(e x 1)dx − + + ∫ 58/I = 2 6 3 5 0 1 cos x sin x.cos xdx π − ∫ 59*/I = 2 3 2 5 1 dx x x 4 + ∫ 60/I = 4 0 x dx 1 cos2x π + ∫ 61/I = 2x ln5 x ln 2 e dx e 1 − ∫ 62/I = 2 e 1 x 1 .ln xdx x + ∫ 63/I = 2 1 0 x dx (x 1) x 1 + + ∫ 64/I = 2 0 sin x.sin 2x.sin3xdx π ∫ 65/I = 2 4 4 0 cos2x(sin x cos x)dx π + ∫ 66*/I = 2 3 3 0 ( cosx sin x)dx π − ∫ 67/I = 7 3 8 4 2 x dx 1 x 2x+ − ∫ 68*/I = 2 0 4cosx 3sin x 1 dx 4sin x 3cos x 5 π − + + + ∫ 69/I = 9 3 1 x. 1 xdx − ∫ 70/I = 2 3 0 x 1 dx 3x 2 + + ∫ 71*/I = 6 0 x sin dx 2 π ∫ 72*/I = 2 0 x dx 2 x 2 x + + − ∫ 73/I = 3 3 2 0 x . 1 x dx + ∫ 74**/I = 1 2 0 ln(1 x) dx x 1 + + ∫ 75/I = 2 0 sin x dx sin x cos x π + ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 3 76/I = e 1 cos(ln x)dx π ∫ 77*/I = 2 2 0 4 x dx + ∫ 78/I = 2 1 x dx 1 x 1 + − ∫ 79/I = e 1 1 3ln x ln x dx x + ∫ 80/I = 3 2 2 ln(x x)dx − ∫ 81/I = e 2 1 (ln x) dx ∫ 82/I = 2 e e ln x dx x ∫ 83/I = 2 e 1 ln x dx ln x ∫ 84/I = 2 2 1 x ln(x 1)dx + ∫ 85/I = 3 2 3 1 dx x 3 + ∫ 86/I = 1 2 0 1 dx 4 x − ∫ 87/I = 2 4 0 sin xdx π ∫ 88/I = 3 2 6 ln(sin x) dx cos x π π ∫ 89/I = 2 1 cos(ln x)dx ∫ 90*/I = 2 2 0 ln( 1 x x)dx + − ∫ 91*/I = 3 2 2 1 dx x 1 − ∫ 92/I = 3 8 1 x 1 dx x + ∫ 93/I = 3 3 2 1 x dx x 16 − ∫ 94/I = 6 2 0 cosx dx 6 5sin x sin x π − + ∫ 95*/I = 2 e 2 e 1 1 ( )dx ln x ln x − ∫ 96/I = 3 2 4 x 4 dx − − ∫ 97/I = 2 3 2 1 x 2x x 2 dx − − − + ∫ 98/I = 3 4 4 cos2x 1dx π π + ∫ 99/I = 0 cosx sin xdx π ∫ 100/I = 2 0 1 sin xdx π + ∫ 101/I = 3 4 4 sin 2x dx π π ∫ 102/I = 0 1 sin xdx π − ∫ 103/I = 1 3 2 1 ln(x x 1) dx −   + +     ∫ 104*/I = 2 0 xsin x dx 1 cos x π + ∫ 105*/I = 1 2 x 1 1 dx (x 1)(4 1) − + + ∫ 106*/I = 4 1 x 1 x dx 1 2 − + ∫ 107/I = 2 4 0 xsin xdx π ∫ 108/I = 2 4 0 xcos xdx π ∫ 109/I = 6 2 0 x.sin xcos xdx π ∫ 110*/I = 2 x 1 2 0 x e dx (x 2) + ∫ 111/I = 2x 2 0 e sin xdx π ∫ 112/I = 2 2 1 1 x ln(1 )dx x + ∫ 113/I = e 2 1 e ln x dx (x 1) + ∫ 114/I = 1 2 0 1 x x.ln dx 1 x + − ∫ 115/I = 2 t 1 ln x dx I 2 x   ⇒ <     ∫ 116/I = 3 0 sin x.ln(cos x)dx π ∫ 117/I = 2 e 2 1 cos (ln x)dx π ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 4 118/I = 4 0 1 dx cosx π ∫ 119*/I = 4 3 0 1 dx cos x π ∫ 120/I = 2 1 3 x 0 x e dx ∫ 121/I = 2 2 sin x 3 0 e .sin x cos xdx π ∫ 122/I = 2 4 0 sin 2x dx 1 cos x π + ∫ 123/I = 1 2 0 3 dx x 4x 5 − − ∫ 124/I = 2 2 1 5 dx x 6x 9 − + ∫ 125/I = 1 2 5 1 dx 2x 8x 26 − + + ∫ 126/I = 1 0 2x 9 dx x 3 + + ∫ 127/I = 4 2 1 1 dx x (x 1) + ∫ 128*/I = 0 2 2 sin 2x dx (2 sin x) −π + ∫ 129/I = 1 2 0 x 3 dx (x 1)(x 3x 2) − + + + ∫ 130/I = 1 3 0 4x dx (x 1) + ∫ 131/I = 1 4 2 0 1 dx (x 4x 3) + + ∫ 132/I = 3 3 2 0 sin x dx (sin x 3) π + ∫ 133/I = 3 3 6 4sin x dx 1 cosx π π − ∫ 134/I = 3 2 6 1 dx cosx.sin x π π ∫ 135/I = 3 0 sin x.tgxdx π ∫ 136/I = 3 4 1 dx sin 2x π π ∫ 137/I = 3 4 2 2 5 0 sin x dx (tg x 1) .cos x π + ∫ 138/I = 3 2 2 3 1 dx sin x 9cos x π π − + ∫ 139/I = 2 2 cosx 1 dx cosx 2 π π − − + ∫ 140/I = 2 0 1 sin x dx 1 3cos x π + + ∫ 141/I = 2 0 cosx dx sin x cos x 1 π + + ∫ 142/I = 4 2 1 1 dx x (x 1) + ∫ 143/I = 1 3 3 1 dx x 4 (x 4) − + + + ∫ 144/I = 3 3 0 sin x dx cosx π ∫ 145/I = 1 0 x 1 xdx − ∫ 146/I = 6 4 x 4 1 . dx x 2 x 2 − + + ∫ 147/I = 0 2 1 1 dx x 2x 9 − + + ∫ 148/I = 3 2 1 1 dx 4x x − ∫ 149/I = 2 2 1 4x x 5dx − − + ∫ 150/I = 2 2 2 2x 5 dx x 4x 13 − − + + ∫ 151/I = 1 x 0 1 dx 3 e + ∫ 152/I = 1 4x 2x 2 2x 0 3e e dx 1 e + + ∫ 153/I = 4 2 7 1 dx x 9 x + ∫ 154/I = 2 x 2 0 e sin xdx π ∫ 155/I = 4 2 4 4 0 cos x dx cos x sin x π + ∫ 156/I = 1 0 3 dx x 9 x + − ∫ 157/I = 0 xsin xdx π ∫ 158/I = 2 2 0 x cos xdx π ∫ 159/I = 1 0 cos x dx ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 5 160/I = 1 0 sin x dx ∫ 161/I = 2 4 0 xsin x dx π ∫ 162/I = 2 4 0 xcos x dx π ∫ 163/I = 2 0 x cos xsin x dx π ∫ 164/I = 6 2 0 xcos xsin x dx π ∫ 165/I = 4 x 1 e dx ∫ 166/I = 4 3x 0 e sin 4xdx π ∫ 167/I = 2x 2 0 e sin xdx π ∫ 168/I = 2 x 1 2 0 x e dx (x 2)+ ∫ 169/I = e 1 (1 x)ln xdx + ∫ 170/I = e 2 1 x ln xdx ∫ 171/I = 1 e 2 1 ln xdx ∫ 172/I = e 1 x(2 ln x)dx − ∫ 173/I = 2 e 2 e 1 1 ( )dx ln x ln x − ∫ 174/I = 2 2 1 (x x)ln x dx + ∫ 175/I = 2 2 1 1 x ln(1 )dx x + ∫ 176/I = 2 5 1 ln x dx x ∫ 177/I = e 2 1 e ln x dx (x 1)+ ∫ 178/I = 1 2 0 1 x xln dx 1 x + − ∫ 179/I = 2 3 cosx.ln(1 cosx)dx π π − ∫ 180/ 2 2 sin x 3 0 e sin xcos x dx π ∫ 181/I= 2 4 0 sin 2x dx 1 sin x π + ∫ 182/I = 2 4 0 sin 2x dx 1 cos x π + ∫ 183/I = 2 2 1 5 dx x 6x 9 − + ∫ 184/I = 2 1 0 x 3x 2 dx x 3 + + + ∫ 185/I = 4 2 1 1 dx x (x 1) + ∫ 186/I = 1 2 0 ln(1 x) dx x 1 + + ∫ 187/I 4 1 6 0 1 x dx 1 x + + ∫ 188/I = 1 15 8 0 x 1 x dx + ∫ 189/I = x 1 x x 0 e dx e e − + ∫ 190/I= e 1 e ln x dx ∫ 191/I = 2 sin x 0 (e cosx)cosx dx π + ∫ 192/I = 2 0 sin 2x.cos x dx 1 cos x π + ∫ 193/I = 2 0 sin 2x sin x dx 1 3cos x π + + ∫ 194/I = 2 4 0 1 2sin x dx 1 sin 2x π − + ∫ 195/I = 5 3 3 2 0 x 2x dx x 1 + + ∫ 196/I = 3 2 4 tgx dx cosx 1 cos x π π + ∫ 197/I = 2 2 1 x 1 ( ) dx x 2 − − + ∫ 198/I = 4 2 0 x.tg xdx π ∫ 199/I = 5 3 ( x 2 x 2)dx − + − − ∫ 200/I = 4 1 2 dx x 5 4 − + + ∫ 201/I = 2 1 x dx x 2 2 x + + − ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 6 202/I = 2 2 1 ln(1 x) dx x + ∫ 203/I = 2 0 sin 2x dx 1 cos x π + ∫ 204/I = 2008 2 2008 2008 0 sin x dx sin x cos x π + ∫ 205/I = 2 0 sin x.ln(1 cos x)dx π + ∫ 206/I = 2 3 2 1 x 1 dx x + ∫ 207/I = 3 4 2 0 sin x dx cos x π ∫ 208/I = 2 2 0 cos x.cos4x dx π ∫ 209/I = 1 2x x 0 1 dx e e+ ∫ 210/I = e 2 1 e ln x dx (x 1)+ ∫ 211/I = 1 0 1 dx x 1 x + + ∫ 212/I = 2 1 2 0 x dx 4 x− ∫ 213/I = 1 2 0 x dx 4 x− ∫ 214/I = 1 4 2 2 0 x dx x 1 − ∫ 215/I = 2 0 sin3x dx cosx 1 π + ∫ 216/I = 2 2 2 2 0 x dx 1 x − ∫ 217/I = 2 2 4 1 1 x dx 1 x − + ∫ 218/I = 3 7 3 2 0 x dx 1 x + ∫ 219/I = x ln 2 x 0 1 e dx 1 e − + ∫ 220/I = 1 0 x 1 x dx − ∫ 221/I = 1 2 0 x 1dx + ∫ 222/I = 2 3 3 0 (cos x sin x)dx π + ∫ 223/I = 2 3 0 x 1 dx x 1 + + ∫ 224/I = 1 2 2x 0 (1 x) .e dx + ∫ 225/I = 2 2 0 cosx dx cos x 1 π + ∫ 226/I = 7 3 3 0 x 1 dx 3x 1 + + ∫ 227/I = 2 6 1 sin 2x cos2x dx cosx sin x π π + + + ∫ 228/I = x 2 1 2x 0 (1 e ) dx 1 e + + ∫ 229/I = 3 2 3 0 x (1 x) dx − ∫ 230/I = 3 2 2 0 sin x.cos x dx cos x 1 π + ∫ 231/I = 1 2 2 0 4x 1 dx x 3x 2 − − + ∫ 232*/I = 2 0 xsin x.cos xdx π ∫ 233/I = 2 0 cosx dx cos2x 7 π + ∫ 234/I = 4 2 1 1 dx x (x 1) + ∫ 235/I = 2 2 3 0 sin2x(1 sin x) dx π + ∫ 236/I = 2 3 0 x 1 dx 3x 2 + + ∫ 237/I = 4 2 7 1 dx x x 9 + ∫ 238/I = 3 4 0 xsin x cos xdx π ∫ 239/I = 2 3 2 cosx cosx cos xdx π π − − ∫ 240*/I = 1 2 1 ln( x a x)dx − + + ∫ 241/I = 2 x 0 1 sin x dx (1 cos x)e π − + ∫ 242/I = 2 0 sin 2x sin x dx cos3x 1 π + + ∫ 243/I = 4 2 2 0 sin 2x dx sin x 2cos x π + ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 7 244/I = 2 3 2 2 0 x dx 1 x − ∫ 245/I = 2 3 2 2 0 x dx 1 x − ∫ 246/I = 2 1 2 2 2 1 x dx x − ∫ 247/I = 2 1 2 0 x dx 4 x − ∫ 248/I = 2 2 2 3 1 dx x x 1 − ∫ 249/I = 1 5 3 6 0 x (1 x ) dx − ∫ 250/I = 2 0 sin x dx 1 sin x π + ∫ 251/I = 2 0 cosx dx 7 cos2x π + ∫ 252/I = 4 2 1 1 dx (1 x)x+ ∫ 253/I = 2 3 0 x 1 dx 3x 2 + + ∫ 254*/I = 3 4 cosx sin x dx 3 sin 2x π π + + ∫ 255/I = 2 3 2 cosx cosx cos xdx π π − − ∫ 256/I = 3 4 4 tg xdx π π ∫ 257*/I = 2 x 0 1 sin x e dx 1 cos x π + + ∫ 258/I = 1 2 3 0 (1 x ) dx − ∫ 259/I = 4 2 0 x.tg xdx π ∫ 260/I= 2 2 2 0 1 dx (4 x )+ ∫ 261/I = 2 1 3 0 3x dx x 2 + ∫ 262*/I = 5 2 5 1 1 x dx x(1 x ) − + ∫ 263/I = 3 2 0 cosx dx 1 sin x π − ∫ 264/I = 2 3 6 0 sin x dx cos x π ∫ 265/I = 3 6 0 sin x sin x dx cos2x π + ∫ 266/I = 2 3 1 dx sin x 1 cos x π π + ∫ 267/I = 2 2 0 sin x dx cos x 3 π + ∫ 268/I = 2 0 sin x dx x π ∫ 269/I = 2 2 0 sinxcosx(1 cosx) dx π + ∫ 270/I = 4 4 4 0 sin x cos x dx sin x cos x 1 π − + + ∫ 271/I = 4 4 4 0 sin x cos x dx sin x cos x 1 π − + + ∫ 272/I = 2 0 sin x cosx cosx dx sin x 2 π + + ∫ 273/I = 1 1 x 3 a e dx x ∫ 274/I = 3 2 1 2 0 x 2x 10x 1 dx x 2x 9 + + + + + ∫ 275/I = 3 1 2 3 0 x dx (x 1) + ∫ 276/I = 1 3 0 3 dx x 1 + ∫ 277*/I = 4 1 6 0 x 1 dx x 1 + + ∫ 280/I = 3 2 2 1 2 1 dx x 1 x− ∫ 281*/I = 2 1 2 0 x ln(x 1 x ) dx 1 x + + + ∫ 282/I = 4 2 1 (x 1) ln x dx − ∫ 278/I = 1 3 0 x dx (2x 1)+ ∫ 279/I = 7 2 1 dx 2 x 1 + + ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 8 283/I = 3 2 0 x ln(x 1)dx + ∫ 284/I = 3 2 2 1 3x dx x 2x 1 + + ∫ 285/I = 1 3 2 0 4x 1 dx x 2x x 2 − + + + ∫ 286/I = 1 2 2 1 2 1 dx (3 2x) 5 12x 4x − + + + ∫ 287/I = 1 0 1 dx x 1 x + + ∫ 288/I = 2 0 cosx dx 2 cos2x π + ∫ 289/I = 2 4 cosx sin x dx 3 sin 2x π π + + ∫ 290/I = 2 3 3 0 (cos x sin x)dx π + ∫ 291/I = 2 5 4 0 cos xsin xdx π ∫ 292/I = 2 4 4 0 cos2x(sin x cos x)dx π + ∫ 293/I = 2 0 1 dx 2 sin x π + ∫ 294/I = 2 0 1 dx 2 cosx π − ∫ 295/I = 2 2 2 3 1 dx x x 1 − ∫ 296/I = 3 7 3 2 0 x dx 1 x + ∫ 297*/I = 2 3 1 1 dx x 1 x + ∫ 298/I = 3 1 2 0 x dx x 1 x + + ∫ 299/I = 1 2 1 1 dx 1 x 1 x − + + + ∫ 300/I = 3 4 6 1 dx sin x cosx π π ∫ 301/I = 2 0 cosx dx cosx 1 π + ∫ 302/I = 2 0 cosx dx 2 cosx π − ∫ 303/I = 2 0 sin x dx sin x 2 π + ∫ 304/I = 3 2 0 cos x dx cosx 1 π + ∫ 305/I = 2 0 1 dx 2cos x sin x 3 π + + ∫ 306/I = 2 2 3 cosx dx (1 cosx) π π − ∫ 307/I = 4 3 0 tg x dx π ∫ 308*/I= 2 x sin x dx 3 1 π −π + ∫ 309*/I = 1 2x 1 1 dx 3 e − + ∫ 310*/I = 2 0 sin x dx cosx sin x π + ∫ 311/I = 4 2 4 4 0 sin x dx cos x sin x π + ∫ 312*/I = 2 2 0 tgx dx 1 ln (cosx) π − ∫ 313*/I = 2 0 sin x dx cosx sin x π + ∫ 314*/I = 1 x 2 1 1 dx (e 1)(x 1) − + + ∫ 315*/I = 1 3x 1 0 e dx + ∫ 316*/I = 2 1 2 0 x dx x 4 + ∫ 317*/I = 3 2 4 2 0 cos x dx cos 3cos x 3 π − + ∫ 318*/ Tìm x> 0 để 2 t x 2 0 t e dt 1 (t 2) = + ∫ 319*/I = 3 2 4 tan x dx cosx cos x 1 π π + ∫ 320*/I = 1 2 0 3x 6x 1dx − + + ∫ 321*/I = 4 5 0 tg x dx π ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 9 322/I = 4 3 6 cotg x dx π π ∫ 323/I = 3 4 4 tg x dx π π ∫ 324*/I = 4 0 1 dx 2 tgx π + ∫ 325/I = 5 2 0 sin x dx cos x 1 π + ∫ 326/I = 3 2 6 cos 2x dx 1 cos 2x π π − ∫ 327*/I = 4 2 0 t gx 1 ( ) dx tgx 1 π − + ∫ 328*/I = 1 3 1 2 x dx x 1 + ∫ 329*/I = 3 3 2 4 1 x x dx x − ∫ 330/I = x ln 3 x x 0 e dx (e 1) e 1 + − ∫ 331/I = 1 4 e 2 1 e 1 dx x cos (ln x 1) π − + ∫ 332/I = 3 6 2 1 1 dx x (1 x ) + ∫ 333*/I = 4 0 ln(1 tgx)dx π + ∫ 334,Cmr 2 2 0 dx 16 5 3cos x 10 π π π ≤ ≤ + ∫ 335,Cmr ( ) 11 7 54 2 7 11 108 x x dx − ≤ + + − ≤ ∫ 336,Cmr 3 1 1 1 2 4 x dx − ≤ ≤ ∫ 337,Cmr ( ) ( ) 1 2 2 2 0 1 1 50 3 2cos 2 3 3 dt t ≤ ≤ + + ∫ 338,Cmr 2 2 4 2 0 2 2 x x e dx e e − ≤ ≤ ∫ 339, Cmr 2 2 1 1 1 2 1 1 x x dx dx x x − − < + ∫ ∫ 340, 2 1 x I dx 1 x 1 = + − ∫ 341, 2 3 2 5 dx I x x 4 = + ∫ 342, 1 3 2 0 I x 1 x dx = − ∫ 343, 2 0 sin2x sin x I dx 1 3cosx π + = + ∫ 344, 1 2 0 I 1 x dx = − ∫ 345, 1 2 0 dx I 1 x = + ∫ 346, 1 2 0 dx I x x 1 = + + ∫ 347, e 1 1 3ln x lnx I dx x + = ∫ 348, ln5 x x ln3 dx I e 2e 3 − = + − ∫ 349, 2 0 sin2x cos x I dx 1 cosx π = + ∫ 350, ( ) 3 2 2 I ln x x dx = − ∫ 351, ( ) 2 sinx 0 I e cosx cos xdx π = + ∫ 352, ( ) 1 2x 0 I x 2 e dx = − ∫ 353, 2 2 0 I x x dx = − ∫ 354, 4 0 x I dx 1 cos2x π = + ∫ 355, 2 1 3 x 0 I x e dx = ∫ 356, 2 0 I x sin xdx π = ∫ 357, 2 4 0 1 2sin x I dx 1 sin2x π − = + ∫ 358, 3 3 1 dx I x x = + ∫ 359, 4 3 I 1 cos2xdx π π − = − ∫ 360, ( ) 2 2 1 5 x 1 I dx x x 6 − = − − ∫ 361, ( ) 2 1 2 0 x ln x 1 x I dx 1 x + + = + ∫ 362, ( ) 2 e e ln x ln ln x I dx x + = ∫ 363, 3 2 2 6 cos x J dx sin x π π = ∫ 364, 1 2 6 0 x dx I x 9 = − ∫ 365, 3 3 5 4 4 dx I sin x cos x π π = ∫ 366, 2 4 2 0 sin 2x I dx sin x 6sin x 5 π = + + ∫ 367 x 0 I sin xe dx π = ∫ 368, 2 5 0 I cos x cos 7xdx π = ∫ 369, 4 4 4 x 4 sin x cos x I dx 3 1 π π − + = + ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 10 AN AN AN AN ACT ACT ACT ACT OF OF OF OF KINDNESS KINDNESS KINDNESS KINDNESS IS IS IS IS NEVER NEVER NEVER NEVER WASTED WASTEDWASTED WASTED ! !! ! . 367 x 0 I sin xe dx π = ∫ 368, 2 5 0 I cos x cos 7xdx π = ∫ 369, 4 4 4 x 4 sin x cos x I dx 3 1 π π − + = + ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG,. 1 + ∫ 30/I = x 1 x 0 e dx e 1 − − + ∫ 31/I = e 2 1 ln x dx x(ln x 1) + ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 2 . 74**/I = 1 2 0 ln(1 x) dx x 1 + + ∫ 75/I = 2 0 sin x dx sin x cos x π + ∫ TTBDKTPT VQ GV: PHAN VĂN VINH ĐC: KI ỆT I, TRUNG ĐÔNG, PHÚ TH Ư ỢNG, PHÚ VANG, HUẾ DĐ: 0905.838.969 3