y=xsin 2x,求dy

∫x√(x²+1)dx=(1/2)∫√(x²+1)d(x²+1)
=(1/2)(2/3)√(x²+1)³+C (C是积分常数)
=(1/3)√(x²+1)³+C ；
∫e^xsin(e^x-2)dx=∫sin(e^x-2)d(e^x-2)
=C-cos(e^x-2) (C是积分常数)；
∫(lnx/x)dx=∫lnxd(lnx)
=(1/2)ln²x+C (C是积分常数)；
∫(1/x^2)乘tan(1/x)dx=-∫[sin(1/x)/cos(1/x)]d(1/x)
=∫[1/cos(1/x)]d[cos(1/x)]
=ln│cos(1/x)│+C (C是积分常数)；
∫cos²x乘sinxdx=-∫cos²xd(cosx)
=C-(1/3)cos³x (C是积分常数)；
∫(tanx)^5*sec²xdx=∫(tanx)^5d(tanx)
=(1/6)(tanx)^6+C (C是积分常数).

x趋向于无穷,sinx/x最大也就是1/x,即0
x趋向去无穷的时候,sin（1/2x)的极限,相当于1/2x趋向于0时sin（1/2x)的极限,即1/2x
（因为有公式,x趋向于0时,sinx趋向于x）
,于是xsin（1/2x)极限就是x乘以1/2x,也就是1/2了
理解了吧?

dy=sin(lnx)dx+xcos(lnx)d(lnx)
=xsin(lnx)d(lnx)+xcos(lnx)d(lnx)
=x[xsin(lnx)+cos(lnx)]d(lnx)

lim(x→∞) (xsin(1/x)+1/x(sinx))
设1/x=t
原式
=lim(t->0)(sint/t+t*sin(1/t))
=1+0
=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|

sin2x/x=2sin2x/(2x)
2x趋于0
所以2sin2x/2x极限=2×1=2
2/x趋于无穷
所以sin(2/x)有界
x是无穷小
所以x*sin(2/x)趋于0
即x趋于0,极限是0
所以原极限=2-0=2

lim｛sinx/x - xsin(1/x)｝=lim sinx/x - lim xsin(1/x)
当x趋于无穷时sinx有界,所以第一个极限为0;
令y=1/x,则第二个极限转化为 y趋于0时 1/y*siny 的极限.因为 lim siny/y=1
当y趋于0时,所以原极限的值为 0-1=-1

lim (x→0)[∫上x下0 cos(t²)dt]/x
=lim(x→0)cos(x²) 0/0型,用洛比达法则
=1
lim( x→0)[∫上x下0 ln(1+t)dt]/(xsinx)
=lim(x→0)ln(1+x)/(sinx+xcosx) 0/0型,用洛比达法则
=lim(x→0) [1/(1+x)]/(cosx+cosx-xsinx) 0/0型,用洛比达法则
=1/2

原式=(1/3)∫sin(x^3+5)d(x^3+5)
=(-1/3)cos(x^3+5)+C (C是积分常数).

∫xsin(x²+3)dx
=1/2 ∫sin(x²+3)d(x²+3)
=-1/2 cos(x²+3)+C

∫xsin(x²+3)dx
=1/2∫sin(x²+3)d(x²+3)
=1/2∫sinudu
=1/2(-cosu)+c
=-1/2cos(x²+3)+c

原式=∫sin(x/2)de^x
=e^x*sin(x/2)-∫e^xdsin(x/2)
=e^x*sin(x/2)-1/2*∫e^xcos(x/2)dx
=e^x*sin(x/2)-1/2*∫cos(x/2)de^x
=e^x*sin(x/2)-1/2*e^xcos(x/2)+1/2*∫e^xdcos(x/2)
=e^x*sin(x/2)-1/2*e^xcos(x/2)-1/4*∫e^xsin(x/2)dx
所以(5/4)∫e^xsin(x/2)dx=e^x*sin(x/2)-1/2*e^xcos(x/2)
所以∫e^xsin(x/2)dx=5/4*e^x*sin(x/2)-5/8*e^xcos(x/2)+C

∫xsin(1-x) dx
=-∫xsin(1-x) d(1-x)
=∫x dcos(1-x)
=x*cos(1-x)-∫cos(1-x) dx
=x*cos(1-x)+∫cos(1-x) d(1-x)
=x*cos(1-x)+sin(1-x)+C (C为常量）

把x=1代入z=f(xy)-xsin(xy)
得f(y)=y+siny
因此,f(xy)=xy+sinxy
z=f(xy)-xsin(xy)
=xy+sinxy-xsin(xy)
=xy+(1-x)sin(xy)