Question

求解定积分 [x/(1+cos2x)]dx 积分上限是π/4 ，积分下限是0

Answer 1

简单计算一下即可，答案如图所示

Answer 2

求解定积分
[x/(1+cos2x)]dx
积分上限是π/4
，积分下限是0
先看不定积分，后代值
∫[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)]|<0,π/4>
=(1/2){[(π/4)*1+ln(√2/2)]-[0*0+0]}
=(1/2)*[(π/4)-(1/2)ln2]
=(π/8)-(1/4)ln2