证明f(cosx)dx在0到π 2的积分等于f(sinx)dx
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补充楼上的回答∫[0,π/2]f(sinx,cosx)dxx=π/2-ux=0,u=π/2,x=π/2,u=0=∫[π/2,0]f(sin(π/2-u),cos(π/2-u))d(π/2-u)=-∫[
算嘛再答:再问:额,这样额再问:再问:那如果是这样的也是算?再答:你那是大几的题目啊再问:大一额再答:问你们数学老师去
证明:由题意可得∫f(sinx)dx求导可得f(sinx)∫f(cosx)dx求导可得f(cosx)因为f(x)一定,当x在(0,π/2)时f(sinx)在f(0~1)之间取值同理f(cosx)也在f
设t=x-π/2左边=∫(-π/2,π/2)f(丨cos(t+π/2)丨)dt=∫(-π/2,π/2)f(丨sint丨)dt因为f(丨sint丨)是偶函数所以=2∫(0,π/2)f(丨sint丨)dt
I=∫[0,π/2]f(cosx)dx换元,令u=π/2-x,dx=(﹣1)du=∫[π/2,0]f(sinu)(-1)du=∫[0,π/2]f(sinu)du=∫[0,π/2]f(sinx)dx
∫(0,x)(x-t)f(t)dt=1-cosx即为x∫(0,x)f(t)dt--∫(0,x)tf(t)dt=1-cosx求导有∫(0,x)f(t)dt+xf(x)--xf(x)=sinx令x=π/2
∫sinx/(sinx+cosx)dx=x/2-1/2*(log(sinx+cosx))将[0,π/2]代入得=π/4∫cosx/(sinx+cosx)dx=1/2*(x+log(sinx+cosx)
证明:由于sinx,cosx是连续函数,而由已知f(u,v)在区域D=上连续,所以复合函数f(sinx,cosx)和f(cosx,sinx)是在0≤x≤π/2是连续的,因此在0≤x≤π/2上f(sin
换元法,令t=π/2-x即可将sinx化为cost,将cosx化为sint,自己动手试一下吧(数学题还是要自己多练习,学习效果才好)
(f(cosx)sinx)'=-f(cosx)*sin^2(x)+f(cosx)cosx所以I=f(cosπ)sinπ-f(cos0)sin0=0
∫1/根号2cos(x-π/4)dx=根号2/2∫1/cos(x-π/4)d(x-π/4)=根号2/2ln|sec(x-π/4)+tan(x-π)/4|+C用牛顿莱布尼兹公式代入x=π/2和x=0计算
令y=π/2-x,则x=π/2-y∫(π/2~0)f(cosx)dx=∫(0~π/2)f(cos(π/2-y))d(π/2-y)=∫(0~π/2)-f(siny)dy=-∫(0~π/2)f(siny)
∫因为:∫f(t)dt【t=0→x】=1-cosx所以:∫f(t)dt=C-cost因此:∫f(x)dx【x=0→π】=C-cosx【x=0→π】=(C-cosπ)-(C-cos0)=(C+1)-(C
令y=π/2-x,则x=π/2-y∫(π/2~0)f(cosx)dx=∫(0~π/2)f(cos(π/2-y))d(π/2-y)=∫(0~π/2)-f(siny)dy=-∫(0~π/2)f(siny)