sionsony
幼苗
共回答了18个问题采纳率:77.8% 举报
∫[0,a] [f(x)+f(2a-x)]dx
=∫[0,a] f(x)dx+∫[0,a] f(2a-x)dx
令t=2a-x,x=2a-t,dx=-dt,
x=0时,t=2a,x-a时,t=a
因此上式变为
=∫[0,a] f(x)dx+∫[0,a] f(2a-x)dx
=∫[0,a] f(x)dx-∫[2a,a] f(t)dt
=∫[0,a] f(x)dx+∫[a,2a] f(t)dt
=∫[0,2a] f(x)dx
∫ [0,π[(xsinx) / (1+cos^2 x)]dx
=∫[0,π/2] [(xsinx) / (1+cos^2 x)]dx+∫[π/2,π] [(π-x)sin(π-x) / (1+cos^2 (π-x)]dx
∫[(xsinx) / (1+cos^2 x)]dx
=-∫[(x) / (1+cos^2 x)]dcosx
=-xarctan
1年前
追问
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yongyou110
举报
http://zhidao.baidu.com/question/283303455.html 我的提问和加分在这里面