Question

求解一道定积分

Answer 1

[x/(1+cos2x)]dx
=∫[x/(1+2cos^2 x-1)]dx
=∫[x/(2cos^2 x)]dx
=(1/2)∫(x/cos^2 x)dx
=(1/2)∫x*sec^2 xdx
=(1/2)∫xd(tanx)
=(1/2)[x*tanx-∫tanxdx]
=(1/2)[x*tanx-∫(sinx/cosx)dx]
=(1/2)[x*tanx+∫(1/cosx)d(cosx)]
=(1/2)[x*tanx+ln|cosx|]
因为x∈[0,π/4],则cosx＞0
所以：原定积分=(1/2)[x*tanx+ln(cosx)]|
=(1/2){[(π/4)*1+ln(√2/2)]-[0*0+0]}
=(1/2)*[(π/4)-(1/2)ln2]
=(π/8)-(1/4)ln2