考研高等数学

Question

求大佬看看怎么解。。。

Answer 1

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

Answer 2

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