Đặt $u=\frac{2x}{1+x^2},t=sinu $ thì $-1\leq u\leq 1,[-1,1]\subset [\frac{-\pi}{2},\frac{\pi}{2} ]$
$sin(-1)\leq sinu\leq sin1\Rightarrow -sin1\leq t\leq sin1$ và $y=-2t^2+t+2=f(t)$
Ta có $f^/(t)=-4t+1>0,\forall t\in[-\sin 1;\sin 1]$
$\Rightarrow
y_{Max}=f(\sin 1)= -2\sin^21+\sin 1+2$ khi $t=\sin 1\Leftrightarrow
u=1\Leftrightarrow x=1$
$y_{Min}=f(-\sin 1)= -2sin^21-sin 1+2$ khi $t=-\sin 1\Leftrightarrow u=-1\Leftrightarrow x=-1$