cost除以sint等于

蒲环真1355∫cost/(sint^2) dt =∫dsint/sint^2 = - 1/sint + C -
幸是蚂18432243261 ______ 中间那步不用那样的.因为d(sint)=costdt,先把cost换到d里面就是:原式=∫【1/(sint^2)】dsint 设sint=x 化为∫(1/x^2)dx=-1/x+C 再把x换回sint

蒲环真1355x=sint+cost y=sint乘cost t为参数 化为一般方程 -
幸是蚂18432243261 ______ x+4=cost y-3=sint ∵cos²t+sin²t=1 ∴直角坐标方程为 (x+4)²+(y-3)²=1

蒲环真1355(2sintcost)dt等于多少?答案说是sint2t,中间过程是怎么算的? -
幸是蚂18432243261 ______[答案] (2sintcost)dt =2sint (cost dt) =2sint dsint = d(sin²t)

蒲环真1355设x=cost,y=sint则(dy)/(dx)= -
幸是蚂18432243261 ______[答案] (cost dt)/(-sint dt)=-cot t

蒲环真13551/(cost+sint) 最小值 -
幸是蚂18432243261 ______ 求解: t = 225度时,取得最小值, 1/(-√2/2-√2/2) =-1/√2 =-√2/2

蒲环真1355x=sintcost y=sinx/cosx求导 -
幸是蚂18432243261 ______[答案] y = sinx/cosx dy/dx = [(sinx)' cosx - sinx (cosx)'] / cos²x = (cos²x + sin²x) / cos²x = 1/cos²x x = sint cost dx/dt = (sint)' cost + sint (cost)' = cos²t - sin²t dy/dt = (dy/dx)(dx/dt) = (1/cos²x)(cos²t - sin²t) = (cos²t - sin²t) / cos²(sint cost) 如果题目是...

蒲环真1355x=sint+5 y= - 2cost - 1 -
幸是蚂18432243261 ______ x=sint+5 y=-2cost-1 sint=x-5 ,cost=-(y+1)/2(sint)^2+(cost)^2=1(x-5)^2+(y+1)^2/4=1

蒲环真1355为什么x=sint ,dx=costdt -
幸是蚂18432243261 ______[答案] 公式啊 x对t微分 推导公式按定义 sin'(x)=lim [sin(x+dx)-sin(x)]/dx (dx-->0) 令dx--->0,则cos(dx)-->1,sin(dx)-->dx ∴sin'(x)=lim [sin(x+dx)-sin(x)]/dx = lim [sin(x)cos(dx)+sin(dx)cos(x)-sin(x)]/dx =lim [sin(x)+dx cos(x)-sin(x)]/dx =cos(x) 所以 dx/dt=d(sint )dt=cost...

蒲环真1355求解一道傅里叶变换的题:求f(t)=sintcost的傅里叶变换求f(t)=sintcost的傅里叶变换请问这答题怎么做 谢谢 -
幸是蚂18432243261 ______[答案] G(ω)=F[sintcost]=∫[-∞,+∞]sintcoste^(-iωt)dt =1/2∫[-∞,+∞]sin2te^(-iωt)dt

蒲环真1355参数方程x=cost+sint,y=sint*cost*(t为参数)的普通方程是多少 -
幸是蚂18432243261 ______[答案] 需要注意的是有个隐藏条件:(sint)^2+(cost)^2=1 即 (sint+cost)^2-2sint*cost=1 将x=cost+sint,y=sint*cost 代入得 x^2-2y=1,即 y=(x^2)/2-1/2