Tính: 
  1)I=120x4x21dx2)I(t)=t0tan4xdxcos2x(0<t<π4)
Và chứng minh bất đẳng thức :
     tan(t+π4)>e23(tan3t+3tant)(0<t<π4)
 1)I=120x4x21dx=120(x2+1+1x21)dx=[x33+x+12ln(x1x+1)]120=132412ln3
    2)I=t0tg4x(1+tg2x)1tg2xdx=t0tg4xd(tgx)1tg2x
   Đặt     u=tgxdu=d(tgx),{x=0u=0x=tu=tgt
   Suy ra:   I=tgt0u4u21du
=[u33+u+12ln|u1u+1|]tgt0=12ln|tg(t+π4)|tg3t3tgt
   t(0,π4):tg4xcos2x>0I(t)>012ln|tg(t+π4)|>tg3t3tgtln[tg(t+π4)]>23(tg3t+3tgt)tg(t+π4)>e23(tg3t+3tgt)

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