$\frac{\sin\:^2\left(x\right)-1}{\sin\:\left(x\right)+1}$
$8-8\cdot8+8$
$\lim_{x\to\infty}\left(\frac{\sin^3\left(2x\right)}{2\tan^3\left(3x\right)}\right)$
$x\frac{dy}{dx}=x-y$
$3r^4\left(-12r^3\right)$
$6>\sqrt{40}$
$\int2x^2+3x^2-4xdx$
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