sint2dt+0到的积分

俞子惠4726设L是曲线x=cost,y=sint上由t1=0到t2=∏/2的一段弧,计算∫L ydx - xdy. -
裘终素13255538887 ______[答案] ∫L ydx-xdy=∫L sintd(cost)-costd(sint)=∫L -sin^2t dt-cos^2t dt=-∫L (sin^2t +cos^2t)dt=-∫L dt=-t|L=-∏/2

俞子惠4726求极限limx - >0 [∫(2,0)sintdt/x^4]
裘终素13255538887 ______ ∫(2,0)sint dt > 0, 是个常数 limx->0 [∫(2,0)sintdt / x^4] = 0

俞子惠4726设连续函数f(x)满足f(x)=sinax - (tf(x - t)0到x的积分)(a>0),求f(x) -
裘终素13255538887 ______[答案] f(x)=sinax-(tf(x-t)0到x的积分)(a>0) 所以f(0)= sin(0*a) = 0 对f(x)=sinax-(tf(x-t)0到x的积分)求导知道: f(x)′ = a*cosax - x*f(x-x) = a*cosax 所以f(x) = sinax + c (c为常数) 又有f(0)= 0 f(x) = sinax

俞子惠4726设F(x)=∫(x到x+2π) sinte^sintdt,则F(x)为正数.为什么? -
裘终素13255538887 ______[答案] 显然F(x)=∫(x到x+2π) sint*e^sintdt =∫(0到2π) sint*e^sintdt不明白的话对F(x)求导,得到F'(x)= sin(x+2π)e^sin(x+2π) - sinx *e^sinx=0,一阶导数为0,即F(x)是一个常数,与变量x无关那么F(x)=∫(0到2π) sint*e^...

俞子惠4726(2sintcost)dt等于多少?答案说是sint2t,中间过程是怎么算的? -
裘终素13255538887 ______[答案] (2sintcost)dt =2sint (cost dt) =2sint dsint = d(sin²t)

俞子惠4726如果f2(x)=∫x0f(t)sint2+costdt,求f(x). -
裘终素13255538887 ______[答案] 由题意,f2(x)=∫x0f(t)sint2+costdt两边对x求导,得2f(x)f′(x)=f(x)sinx2+cosx若f(x)≠0,则f′(x)=sinx2+cosx∴上式两边积分,得f(x)=∫sinx2+cosxdx=c−ln(2+cosx)又f(x)≡0,显然也满足题设条件∴f...

俞子惠4726要有具体过程求曲线x=a(cost+tsint),y=a(sint - tcost),(0≤t≤)的长度L 这题我知道是用弧微分来做 但是做出来的是∫a√(1+t^2)dt 从0积分到2π.答案上是∫atdt 从0积... -
裘终素13255538887 ______[答案] x=a(cost+tsint),y=a(sint-tcost)L = ∫√(dx² + dy²) dx = a t cost dt dy = a t sint dt= ∫a t√( (cos²t + sin²t)) dt= ∫a t dt (t = 0 →2π)= 2π²a你检查一下,是不是 dx dy 求错了...

俞子惠4726∫(0到2)X^2√(4 - X^2)dx -
裘终素13255538887 ______ 设x=2sint x²√(4-x²)dx=4sin²t*2cost*2costdt=16sin²tcos²t=4sin²2tdt=2(1-cos4t)dt 所以原积分=∫(0到π/2)2(1-cos4t)dt=π-2∫(0到π/2) cos4tdt=π

俞子惠4726ddx∫sinx0ln(t2+1)dt=______. -
裘终素13255538887 ______[答案] 利用积分上限函数的求导公式以及复合函数的求导法则可得, d dx ∫sinx0ln(t2+1)dt =ln(sin2x+1)•(sinx)′ =cosx•ln(sin2x+1). 故答案为:cosx•ln(sin2x+1).

俞子惠4726∫cost/(sint+cost)dt在0到π取积分 -
裘终素13255538887 ______[答案] ∫ cost/(sint + cost) dt= (1/2)∫ [(cost + sint) + (cost - sint)]/(sint + cost) dt= (1/2)∫ [1 + (cost - sint)/(sint + cost)] dt= t/2 + (1/2)ln|sint + cost| + C设ƒ(t) = cost/(sint + cost) ∫(0→...