∫f(sinx)dx=2∫f(fcosx)dx
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![∫f(sinx)dx=2∫f(fcosx)dx](/uploads/image/f/933528-48-8.jpg?t=%E2%88%ABf%28sinx%29dx%EF%BC%9D2%E2%88%ABf%28fcosx%29dx)
分部积分.先把sinx积出来.变成-∫f(cosX)d(cosX)然后再把cosX看成变量,再积一次,变成-F(cosX)
cos的几次方呀?
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
∫f(x)dx=1/2x^2+Cf(x)=[∫f(x)dx]'=(1/2x^2+C)'=xf(sinx)=sinx∫f(sinx)dx=∫sinxdx=-cosx+C再问:f(sinx)=sinx是不
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
∫(上π,下π/2)xf(sinx)dx=(令t=x-π/2)=∫(上π/2,下0)(t+π/2)f(sint)dt=∫(上π/2,下0)tf(sint)dt+π/2∫(上π/2,下0)f(sint)
题目写错了吧如果是不定积分的话是不成立的
左边=-cosπ+cos0=2右边=2(-cosπ/2+cos0)=2原式成立再问:是f(sinx),不是sinx再答:抱歉,没仔细看题呵。令x=(π/2)-t则∫(0,π/2)f(sinx)dx=∫
原式=∫f(arcsinx)darcsinx=sin(arcsinx)+c=x+c
f(cosx)=sin²x=1-cos²x===>f(x)=1-x²令x-1=t====>x=t+1dx=dt原式=∫f(x-1)dx=∫f(t)dt=∫(1-t&sup
记sinx=t∫cosxf(sinx)dx=∫f(sinx)dsinx=∫f(t)dt=F(t)+C=F(sinx)+C
证明:令x=π-t,则x由0到π,t由π到0,dx=-dt原式记为I则I=-(积分区间π到0)∫(π-t)f(sin(π-t)dt=-(积分区间π到0)∫(π-t)f(sin(t)dt=(积分区间0到
令u=π-x,du=-dx,u:π--->0,则∫[0--->π]xf(sinx)dx=-∫[π--->0](π-u)f(sin(π-u))du=∫[0--->π](π-u)f(sinu)du=π∫[
证明:因为∫(0→π)f(sinx)dx=∫(0→π/2)f(sinx)dx+∫(π/2→π)f(sinx)dx令x=π-t则当x=π/2时t=π/2当x=π时t=0所以∫(π/2→π)f(sinx)
两端求导得f(x)=cos(x/2)
letx=siny∫f(x)dx=∫f(siny)d(siny)=∫[y/(siny)^2]d(siny)=-∫yd[1/(siny)]=-y/siny+∫(1/siny)dy=-y/siny+ln|
设x=π-y,dx=-dy当x=0,y=π当x=π,y=0∫(0→π)xf(sinx)dx=-∫(π→0)(π-y)f(sin(π-y))dy=π∫(0→π)f(siny)dy-∫(0→π)yf(si