Bài 6 trang 127 SGK Giải tích 12. Tính

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a) 02cos2xsin2xdx ; b) -112x-2-xdx ;
c) 12x+1x+2x+3x2dx ; d) 021x2-2x-3dx ;
e) 0π2sinx+cosx2dx ; g) 0πx+sinx2dx.

 


a) 0π2cos2xsin2xdx=0π2cos2x1-cos2x2dx=120π2cos2x-cos22xdx=120π2cos2x-1+cos4x2dx=120π2cos2xdx-0π214dx-140π2cos4xdx=12sin2x0π2-14x0π2-116sin4x0π2=0-π8-0=-π8.

b) -112x-2-xdx

+) Ta có :

  • 2x-2-x02x2-xx-xx0.
  • 2x-2-x<02x<2-xx<-x2x<0x<0.

+) Do đó : -112x-2-xdx=-102x-2-xdx+012x-2-xdx=-10-2x-2-xdx+012x-2-xdx=-2xln2+2-xln2-10+2xln2+2-xln201=-2ln2+12ln2+2ln2+2ln2+12ln2-2ln2=1ln2.

c) 12x+1x+2x+3x2dx=12x2+3x+2x+3x2dx=12x3+3x2+3x2+9x+2x+6x2dx=12x3+6x2+11x+6x2dx=12x+6+11x+6x2dx=x22+6x+11lnx-6x12=11+11ln2-12=212+11ln2.

d) 021x2-2x-3dx=021x+1x-3dx=0214x-3-14x+1dx=14lnx-3x+102=14ln13-ln3=14ln3-1-ln3=-12ln3.

e) 0π2sinx+cosx2dx=0π2sin2x+2sinxcosx+cos2xdx=0π21+sin2xdx=x-12cos2x0π2=π2+12--12=π2+1.

g) I=0πx+sinx2dx=0πx2+2xsinx+sin2xdx=0πx2dx+20πxsinxdx+0πsin2xdx=A+2B+C.

  • A=0πx2dx=x330π=π33.
  • B=0πxsinxdx : Đặt u=xdv=sinxdxdu=dxv=-cosxB=-xcosx0π-0π-cosxdx=π+sinx0π=π.
  • C=0πsin2xdx=0π1-cos2x2dx=12x-14sin2x0π=π2.
  • Vậy I=π33+2π+π2=π33+5π2.