THẦY ĐÀO PHÚ HÙNG TRƯỜNG THPT TRƯNG VƯƠNG –QUY NHƠN ================================================================= Giaûi: Caâu 21 ( ) ( ) 2 2 2 3 2 2 4 3 3 3 0 0 0 sin xcosx 1 3sin xcos x sin x dx= sin xcosxdx sin xcosx.d sin xcosx dx π π π + − + ∫ ∫ ∫ =I+J I= 2 3 0 sin xcosxdx π ∫ = 2 2 4 3 / 0 0 sin x 1 sin x(sinx) dx 4 4 π π = = ∫ J= ( ) ( ) 2 2 3 3 3 0 sin xcosx sin xcosx.d sin xcosx dx= =0 2 2 0 π π ∫ ĐS: 1 4 Giaûi: Caâu 22: 2 2 2 2 2 e e e 2 1 1 1 (lnx 1) 1 ln x (lnx 1) (lnx 1) (lnx 1) 1 dx= dx .d x x x x x + − + + + + + ÷ ÷ ∫ ∫ ∫ =I+J I= 2 e 1 (lnx 1) dx x + ∫ = 3 e e 2 / 1 1 (lnx 1) 7 (lnx 1) (lnx 1) dx 3 3 + + + = = ∫ J= 2 2 4 e 2 2 1 (lnx 1) (lnx 1) (lnx 1) 16 1 e .d = = 1 x x 2 2x 2e + + + − ∫ ĐS: 2 16 11 6 2e + Giaûi: Caâu 23: ( ) ( ) 3 5 2 2 7 3 0 3 3 2 x 5x 3 x dx dx 1 x x 1 + ÷ + ÷ + ÷ + ∫ ( ) ( ) 3 3 2 2 4 2 3 7 7 2 2 0 0 3 3 2 3x x 1 2x x dx x dx x 1 1 x x 1 + + = + + + + ∫ ∫ =I+J THẦY ĐÀO PHÚ HÙNG TRƯỜNG THPT TRƯNG VƯƠNG –QUY NHƠN ================================================================= I= 3 2 7 0 3 3 x dx 1 x+ ∫ = 3 3 1 1 3 3 3 / 7 7 2 3 3 0 0 (1 x ) x (1 x ) dx (1 x ) dx 3 − − + + = + ∫ ∫ = ( ) 1 3 3 3 1 3 2 7 7 3 3 0 0 1 (1 x ) 1 3 1 x 1 3 2 2 1 3 − + + = + = − + J= ( ) ( ) ( ) ( ) 3 3 2 2 4 3 3 3 6 3 7 7 2 2 2 2 2 2 0 0 2 2 3 3x x 1 2x x x x x 49 7 dx= .d dx= = 0 x 1 x 1 x 1 2 49 1 x 1 2 x 1 + + = ÷ ÷ + + + + + + ∫ ∫ ĐS: ( ) 2 3 49 3 2 2 49 1 + + Giaûi: Caâu 24: ( ) ( ) 2 2 1 x x 0 e x 1 1 2x e dx − − + − ∫ = 2 2 2 1 1 x x 0 0 x 1 2x e xdx e x. dx e − − − + ∫ ∫ =I+J I= 2 1 x 0 e xdx − ∫ Ñaët t = 2 x e − ⇒ dt = –2x . 2 x e − dx , ° x = 1 ⇒ t = e–1 = 1 e x = 0 ⇒ t = 1 ⇒ I = 1 1 e e 2 1 1 x dt dt t t 2t 2e − − = − ÷ ÷ ∫ ∫ = 1 1 e e 1 1 1 1 1 1 dt t 1 2 2 2 e − = − = − − ÷ ∫ J= ( ) 2 2 2 2 1 x x 2 0 2x x 1 1 e x.d e x = = 0 2e 2e − − ∫ ĐS: 2 1 1 1 1 2 e 2e − − ÷ Giaûi: Caâu 25: x 4 4 2 1 1 e lnx 1 lnx dx dx x x x − ÷ + + ÷ ÷ ∫ ∫ =I+J THẦY ĐÀO PHÚ HÙNG TRƯỜNG THPT TRƯNG VƯƠNG –QUY NHƠN ================================================================= I = x 4 1 e dx x ∫ , Ñaët t = x e ⇒ dt = x e dx 2 x ° x = 4 ⇒ t = e 2 , x = 1 ⇒ t = e ⇒ I = 2 2 e e 2 e e 2tdt 2t 2(e e)= = − ∫ J= 2 2 4 4 2 2 1 1 lnx 1 lnx lnx lnx ln x ln 2 4 dx= d = = 1 x x x 8 x 2x − ÷ ÷ ÷ ∫ ∫ ĐS: 2 2 ln 2 2(e e) 8 − + Giaûi: Caâu 26: I = 1 5 3 6 0 x (1 x ) dx− ∫ Ñaët t = 1 – x 3 ⇒ dt = –3x 2 dx ° x = 1, t = 0 , x = 0, t = 1 ⇒ I = 7 8 1 0 1 6 6 7 1 0 0 1 1 1 t t 1 (1 t).t dt (t t )dt 3 3 3 7 8 168 − − = − = − = ÷ ÷ ÷ ∫ ∫ Giaûi: Caâu 27: 1 6 0 (2x x 1) x dx− ∫ . Ñaët t = 2x x – 1 = 3 2 2x – 1 ⇒ dt = 2. 1 2 3 x 2 dx = 3 x dx ° x = 1, t = 1 , x = 0, t = –1 ⇒ I = 7 1 1 1 6 6 1 1 1 1 1 1 t 2 t dt t dt 3 3 3 7 21 − − − = = = ÷ ∫ ∫ Giaûi: THẦY ĐÀO PHÚ HÙNG TRƯỜNG THPT TRƯNG VƯƠNG –QUY NHƠN ================================================================= Câu 28: ( ) ( ) 2 2 2 2 0 1 sin x sin2x 3 2sin x dx π + + ∫ = ( ) ( ) ( ) 2 2 2 2 2 2 2 0 0 1 sin x sin2xdx 1 sin x .2 1 sin x sin2xdx π π + + + + ∫ ∫ =I+J I = ( ) 2 2 2 0 1 sin x sin2xdx π + ∫ Đặt t = 1 + sin 2 x ⇒ dt = 2sinxcosx dx ° x = 2 π , t = 2 , x = 0, t = 1 ⇒ I = 3 2 2 2 1 1 t 7 t dt 3 3 = = ∫ ( ) ( ) ( ) 4 2 2 2 2 2 2 0 1 sin x 16 J 1 sin x .d 1 sin x = = =8 2 2 2 0 π π + = + + ∫ ĐS: 31 3 Giải: Câu 29: ( ) ( ) 3 2 2 4 3 1 4 x 1 3x x 1 dx x 1 x 1 − − − + + + ∫ = ( ) 2 4 3 2 2 4 4 2 1 1 4 x 1 x x 1 4x dx . dx x 1 x 1 x 1 − − − + − + + + + ∫ ∫ =I+J I= 2 2 4 1 x 1 dx x 1 − − + ∫ chia tử và mẫu cho x 2 . I= / 2 2 2 2 1 1 2 2 1 1 x 1 x x dx dx 1 1 x x 2 x x + − ÷ = + + − ÷ ∫ ∫ . t = 1 x x + ⇒ dt = / 1 x x + ÷ dx ° x = 1 ⇒ t = 2 , x = 2 ⇒ t = 5 2 . THẦY ĐÀO PHÚ HÙNG TRƯỜNG THPT TRƯNG VƯƠNG –QUY NHƠN ================================================================= Vaäy: I = ( ) ( ) ( ) ( ) 5 2 2 2 5 5 2 2 2 2 dt 1 t 2 1 = ln = ln 2 2 2 t 2 2 2 t 2 5 2 2 2 2 2 − + − + − + − ∫ ( ) ( ) 4 3 2 2 2 4 2 4 4 2 1 1 4 4 x x 1 4x x x x 1 4 1 2 J . dx= d = = -1 2 289 4 x 1 x 1 x 1 x 1 2 x 1 − − + − = − ÷ ÷ + + + + + ∫ ∫ ĐS: ( ) ( ) ( ) ( ) 5 2 2 2 2 1 1 4 1 ln 2 289 4 2 2 5 2 2 2 2 − + + − ÷ + − Giaûi: Caâu 30: 4 1 1 3 3 3 27 27 3 3 3 3 2 1 1 1 3 3 1 x 4 I 9 dx 9 d 1 x 1 x 9 2 27 3 1 9 1 x − − − − − ÷ ÷ = − = − + = + = − − ÷ ÷ ÷ ÷ ÷ ÷ ÷ + ÷ ÷ ∫ ∫ Giaûi: Caâu 31: ( ) ( ) ( ) ( ) ( ) 2 2 6 6 2 2 2 ln sinx ln 2 6 I cotx.ln sinx dx ln sinx .d ln sinx 2 2 π π π π π π = = = = ∫ ∫ Giaûi: Caâu 32: ( ) ( ) ( ) ( ) ( ) 2 2 3 3 0 0 0 ln cosx ln 2 I t anx.ln cosx dx ln cosx .d ln cosx 3 2 2 π π π = = − = − = − ∫ ∫ Giaûi: THẦY ĐÀO PHÚ HÙNG TRƯỜNG THPT TRƯNG VƯƠNG –QUY NHƠN ================================================================= Caâu 33: ( ) ( ) ( ) 2 3 3 4 4 4 ln tanx 1 1 ln 3 3 I dx .d ln tanx sin2x 2 2 8 π π π π π π = = = = ∫ ∫ Giaûi: 4 1 1 1 3 3 3 3 27 27 3 3 3 2 1 1 1 3 3 2 1 3 3 3 3 x Caâu 34: I 9 1 x dx 9 1 x d 1 x 9 1 x 2 1 x 9 2. 4 9 27 9. 1 2 2 − − − − − − − ÷ ÷ ÷ = − + = − + + = ÷ ÷ ÷ ÷ ÷ ÷ ÷ + ÷ ÷ ÷ + − ÷ ÷ ÷ ÷ − = − ∫ ∫ Giaûi: ( ) ( ) ( ) ( ) ( ) ( ) 6 6 2 2 2 2 2 2 Caâu 35: I cot x. 1 ln sinx dx dx ln sinx .d ln sinx ln sinx ln 2 6 6 ln sinx ln2 2 2 π π π π π π = + = + = π π + == − + ∫ ∫ Giaûi: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 3 0 0 0 3 3 0 0 d cosx Caâu 36: I tanx. 1 ln cosx dx dx ln cosx .d ln cosx cosx ln cosx ln 2 ln cosx ln2 3 3 3 3 π π π = + = − = π π − = − + ∫ ∫ ∫ Giaûi: THẦY ĐÀO PHÚ HÙNG TRƯỜNG THPT TRƯNG VƯƠNG –QUY NHƠN ================================================================= ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 4 4 4 2 2 4 4 ln tanx 2cos2x d sin2x 1 Caâu 37: I dx ln tanx .d ln tanx sin2x 2 sin2x ln tanx ln 3 3 3 3 ln sin2x ln 4 16 2 π π π π π π π π + = = + = π π + = + ∫ ∫ ∫ Giaûi: ( ) 5 3 3 1 1 1 4 3 3 3 81 81 4 4 4 2 1 1 1 3 4 4 1 4 3 4 4 3 3 x Caâu 38: I 9 1 x dx 9 1 x d 1 x 12 1 x 1 x 4 2 3 81 9 9. 1 4 4 − − − − − − − ÷ ÷ ÷ = − + = − + + = ÷ ÷ ÷ ÷ ÷ ÷ ÷ + ÷ ÷ ÷ + ÷ − ÷ ÷ − = − ∫ ∫ Giaûi: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 4 6 6 6 2 2 2 5 5 2 2 d sinx Caâu 39: I cot x. 1 ln sinx dx ln sinx .d ln sinx sinx ln sinx ln 2 6 6 ln sinx ln2 5 5 π π π π π π π π = + = + = π π + = − − ∫ ∫ ∫ Giaûi: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 3 3 3 3 0 0 0 5 5 3 3 0 0 d cosx Caâu 40: I tanx. 1 ln cosx dx dx ln cosx .d ln cosx cosx 3ln cosx 3ln 2 ln cosx ln2 3 3 5 5 π π π = + = − = π π − = − + ∫ ∫ ∫