分类：教育/科学-理工学科-数学
问题：计算积分∫&lt;0,π/2&gt;[cos2mx*ln(16(? 
计算积分∫<0,π/2>[c2mx*ln(16(cosx)^4)]dx
回答：∫<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)