设t=tanx/2,x=2arctant,dx=2dt/(1+t^2)怎么求出sinx=2t/(1+t^2),cosx=

解法一：∵∫2arctantdt=2xarctanx-2∫tdt/(1+t²) (应用分部积分法)
=2xarctanx-ln(1+x²)
lim(x->+∞)[ln(1+x²)/x]=lim(x->+∞)[2x/(1+x²)] (∞/∞型极限,应用罗比达法则)
=lim(x->+∞)[(2/x)/(1+1/x²)]
=0
∴原式=lim(x->+∞)[(2xarctanx-ln(1+x²))/√(1+x²)]
=lim(x->+∞)[(2arctanx-ln(1+x²)/x)/√(1+1/x²)] (分子分母同除x)
=[2(π/2)-0]/√(1+0)
=π；
解法二：原式=lim(x->+∞)[2arctanx/(x/√(1+x²))] (∞/∞型极限,应用罗比达法则)
=2[lim(x->+∞)(arctanx)]*{lim(x->+∞)[√(1+1/x²]}
=2(π/2)*√(1+0)
=π.

这是参数方程求导
dx/dt=2t/(1+t²)
dy/dt=2/(1+t²)-1=(1-t²)/(1+t²)
故dy/dx=(dy/dt)/(dx/dt)=(1-t²)/2t

令u=tan(x/2),则du=(1/2)sec²(x/2)dx,得dx=2/(u²+1)du
∫1/(1+sinx)dx
=∫1/[1+2u/(u²+1)]·2u/(u²+1)du
=∫2/(u²+2u+1)du
=2∫1/(u+1)² du
=2∫1/(u+1)² d(u+1)
=-2/(u+1)+C
=-2/[tan(x/2)+1]+C
如果令x=2arctant话,因为arctanx∈（-π/2,π/2）,所以首先不能保证x∈R

x=2arctant,sinx=?
tan x/2 = t
tanx=(2tanx/2)/(1-tan²x/2)
= 2t / (1-t²)
解三角形
sinx = ±2t/ (t²+1)