∫<0,π/2>[c(2mx)*ln(16(cosx)^4)]dx= =ln16∫<0,π/2>cos(2mx)dx+4∫<0,π/2>[cos(2mx)*ln(cosx)]dx 只需要计算： I(m)=∫<0,π/2>[cos(2mx)*ln(cosx)]dx m=0,1,2,... 1. m=0, I(0)=∫<0,π/2>ln(cosx)dx=(u=π/2-x) =∫<0,π/2>ln(sinu)du= =(1/2)[∫<0,π/2>[ln(cosx)+ln(sinx)]dx= =(1/4)∫<0,π/2>ln(sin2x)d(2x)-(ln2/2)∫<0,π/2>dx= =(1/4)∫<0,π>ln(sinx)dx-πln2/4= =(1/4)[∫<π/2,π>ln(sinx)dx+I(0)]-πln2/4=(u=π-x)= =(1/4)[I(0)+I(0)]-πln2/4 ==> I(0)=-πln2/2. 2. m=1,2,... I(m)=(1/(2m)∫<0,π/2>ln(cosx)]d(sin2mx)= =-(1/(2m)∫<0,π/2>[sinx*sin2mx/cosx]dx J(m)=∫<0,π/2>[sinx*sin2mx/cosx]dx I(m)=-(1/2m)J(m). 3. J(m)+J(m+1)= =∫<0,π/2>{sinx*[sin2mx+sin2(m+1)x]/cosx}dx= =2∫<0,π/2>sinx*sin(2m+1)xdx= =∫<0,π/2>cos(2m)xdx-∫<0,π/2>cos(2m+2)xdx= =0 ==> J(2s)=J(2)=-J(2s-1)=-J(1) J(1)=∫<0,π/2>[sinx*sin2x/cosx]dx= =2∫<0,π/2>[sinx]^2dx=π/2 ==> I(2s)=∫<0,π/2>[cos(4sx)*ln(cosx)]dx= =-(1/4s)J(2s)=π/(8s) I(2s-1)=∫<0,π/2>[cos((4s-2)x)*ln(cosx)]dx= =-(1/(4s-2）)J(2s-1)=-π/(8s-4)