巴洛克
令π/4-x=t ,则dx=-dt积分限: x=0,t=π/4; x=π/4,t=0原式= ∫(0, π/4) (π/4-t)dt/[costcos(π/4-t)]=∫(0, π/4) π/4dt/[costcos(π/4-t)]-∫(0, π/4) tdt/[costcos(π/4-t)]因为不同的变量并不会影响最终积分值,所以:∫(0, π/4) tdt/[costcos(π/4-t)]=∫(0, π/4) xdx/[cosxcos(π/4-x)]则:∫(0, π/4) xdx/[cosxcos(π/4-x)]=1/2∫(0, π/4) π/4dt/[costcos(π/4-t)]而∫(0, π/4) π/4dt/[costcos(π/4-t)]=∫(0, π/4) π/4dt/[costcosπ/4cost+sinπ/4sint]=√2π/4∫(0, π/4) dt/[cost(cost+sint)]=√2π/4∫(0, π/4) d(tant)/(1+tant)=√2π/4ln(1+tant)|(0, π/4)=√2πln2/4则:∫(0, π/4) xdx/[cosxcos(π/4-x)]=√2πln2/8