求(x sin(x)) (x sin(5x))的极限

I=∫(0->π)(xsinx)^2dx=(1/2)∫(0->π)x^2(1-cos2x)dx=(1/2)[x^3/3](0->π)-(1/4)∫(0->π)x^2.dsin2x=π^3/6-(1/4

令y=1/x则原式=sin(y)/y,当y趋向于0和无穷的极限趋向于0的时候siny=y既为1趋向于无穷=0既x趋向于0时候为0趋向于无穷时候为1

因为lim(x->0)x=0而|sin1/x|≤1即sin1/x是有界函数所以由无穷小与有界函数的乘积是无穷小这个性质,得原式=0

由于被积函数是奇函数被积区间[-1,1]关于原点对称所以积分=0

∫cos2x/(sinx*cosx)dx=∫cos2x/(1/2*sin2x)dx=4∫cos2x/(sin2x)dx=4∫csc2x*cot2xdx=-2∫csc2x*cot2xd(2x)=-2cs

如果题目是x（sinЛ/x）+（Л/x）*sinx用重要极限lim(x->0)(sinx)/x=1lim(x->∞)(sinx)/x=0则原式极限=pi*[sin(pi/x)]/(pi/x)+pi*(

=limxsin1/x-limsinx/xx趋近于0=0-1=-1

lim(x→0){(2x-sin2x)/(x*sin^2x)}=lim(x→0){(2x-sin2x)/(x*x^2*(sin^2x/x^2))}=lim(x→0){(2x-sin2x)/(x*x^2

答:y=xsin(1/x)水平渐近线求x趋于无穷时极限y=lim(x→∞)xsin(1/x)=lim(x→∞)sin(1/x)/(1/x)=lim(t→0)sint/t=1所:y=xsin(1/x)水

解法一的（cos2x+1)/2dx应该是（1-cos2x)/2dx高手犯了个低级错误哦!sin^2x=（1-cos2x)/2

1、∫(cot)^2•xdx=∫[(csc)^2•x-1]dx=-cotx-x+c2、∫cos2x/(cos^2xsin^2x)dx=∫(cos^2x-sin^2x)/(cos

求lim{[(sinx)/x]+xsin(1/2x)}(x→∞)用极限的可加性拆成lim(sinx/x)和lim[xsinx(1/2x)]sinx/x,因为x→∞,所以1/x趋向0,sinx在1和-1

lim(sinx/x+xsin(1/x))=lim(sinx/x+sin(1/x)/(1/x))sin(1/x)和1/x是等价无穷小量|sinx|

解y=2xsin(2x+5)y'=2(x)'sin(2x+5)+2x[sin(2x+5)]'(2x+5)'=2sin(2x+5)+2xcos(2x+5)×2=2sin(2x+5)+4xcos(2x+5

答：你的解法当然不对了你自己把结果求导一下就知道是错误的你的结果求导是：2*(1/8)sin²2xcos2x=(1/4)cos2xsin²2x,不是积分函数

y=xsin(1/x)=sin(1/x)/1/x当x无穷大时,1/x无穷接近于0所以y=sin(1/x)/1/x=1/x/1/x=1所以x>0,求y=xsin(1/x)的渐近线是y=1

由y=f*g（f,g是两个函数）的导数公式可知：y=f'*g+f*g'又由f(g)'=f'*g'所以y'=(2x)'*sin(2x+5)+2x*[sin(2x+5)]'=2sin(2x+5)+2xco

integralsin^4(x)cos^5(x)dx=(3sin(x))/128-1/192sin(3x)-1/320sin(5x)+(sin(7x))/1792+(sin(9x))/2304+C再问

lim(xsin*2/x+2/x*sinx)=lim(xsin*2/x)+lim(2/x*sinx)=2lim[(x/2)sin*2/x]+2lim(sinx/x)=0+2=2再问：为什么lim（x/

∫（xsinx）²dx=Sx^2*(sinx)^2dx=Sx^2*(1-cos2x)/2dx=1/2*Sx^2dx-1/2*Sx^2cos2xdx=1/6*x^3-1/4*Sx^2dsin2