sint2+dt

嵇实奋3530∫ t 平方*cosωtdt 原函数怎么求? -
滑玲房13529574453 ______[答案] ∫t^2coswtdt=∫t^2/w dsinwt =t^2sinwt/w-∫sinwt/wdt^2 =t^2sinwt/w+∫2t/w^2dcoswt=t^2sinwt/w+2tcoswt/w^2-∫2coswt/w^2 dt =t^2sinwt/w+2tcoswt/w^2-2sinwt/w^3 +c

嵇实奋3530设函数f(x)= ∫1−cosx0sint2dt,g(x)= x5 5+ x6 6,则当x→0时,f(x)是g(x)的( ) -
滑玲房13529574453 ______[选项] A. 低阶无穷小 B. 高阶无穷小 C. 等价无穷小 D. 同阶但不等价的无穷小

嵇实奋3530sin(2t)+cos(πt)是周期信号 - 上学吧普法考试
滑玲房13529574453 ______ ∫dt/sint=∫sintdt/(sint)^2 =∫-dcost/(1-cost)(1+cost) =(-1/2)∫(1-cost+1+cost)dt/(1-cost)(1+cost) =(-1/2)[∫dt/(1+cost)+∫dt/(1-cost)] =(-1/2)ln|(1+cost)/(1-cost)| =(1/2)ln|(1-cost)/(1+cost)| =(1/2)ln|(1-cost)^2/(1-cost^2)| =ln|(1-cost)/sint|=ln|csct-cott|

嵇实奋3530∫√x *sin√x dx -
滑玲房13529574453 ______ t=根号x,则:x=t^2,dx=2tdt ∫√x *sin√x dx=∫tsint*2tdt=-∫2t^2dcost=-2tcost+∫costd(2t^2)=-2tcost+∫4tcostdt=-2tcost+∫4tdsint=-2tcost+4tsint-∫4sintdt==-2tcost+4tsint+4cost+c=-2√x*cos√x+4√x*sin√x+4cos√x+c

嵇实奋3530t*sin^3(t) dt.不定积分.求救! -
滑玲房13529574453 ______[答案] ∫ t*sin³t dt = ∫ t*(1-cos²t)*sint dt = ∫ t*sint dt - ∫ t*cos²t*sint dt = ∫ t*sint dt - (1/2)∫ t*(1+cos2t)*sint dt = (1/2)∫ t*sint dt - (1/2)∫ t*cos2t*sint dt = (1/2)∫ t*sint dt - (1/4)∫ t*(sint3t-sint) dt = (1/2)∫ t*sint dt - (1/4)∫ t*sin3t dt + (1/4)∫ t*sint dt = (3/4)∫ t*sint dt - (1/4...

嵇实奋3530sqrt(x - 1/x)=x - sqrt(1 - 1/x)求解 -
滑玲房13529574453 ______ 过程是这样的: 两边同平方:x-1/x=x2+1-1/x-2*x*sqrt(1-1/x) 2*x*sqrt(1-1/x) =x2-x+1 x*sqrt(x2-x)=x2-x+1 设sqrt(x2-x)=t 2*t=t2+1 t2-2*t+1=0 t=1 所以sqrt(x2-x)=1 x2-x=1 x2-x-1=0 所以X1=(1+sqrt5)/2 或X2=(1-sqrt5)/2 补充:x2是x的平方,X1,X2是X的最终两个解

嵇实奋3530如果f2(x)=∫x0f(t)sint2+costdt,求f(x). -
滑玲房13529574453 ______[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...

嵇实奋3530∫ t^2 * sin(t) dt -
滑玲房13529574453 ______[答案] ∫ t^2 * sin(t) dt =- ∫ t²dcost =-∫ t²cost+ ∫costd t² =- t²cost+ 2∫tcostdt =- t²cost+ 2∫tdsint =- t²cost+ 2tsint-2∫sintdt =- t²cost+ 2tsint+2cost+C