$\left(x^4-1\right)\left(x+1\right)$
$4cos^2x-3=0$
$4-7-8+1$
$\lim_{m\to6}\left(\frac{m^3-6m+180}{m^2-m-42}\right)$
$\frac{1}{4}-\frac{1}{4}\cos\left(2x\right)=\frac{1}{8}\sin\left(x\right)^2$
$m-n+y;\:m=-2;\:n=3y=10$
$\int_2^6xe^{\left(-2x\right)}\:\:dx$
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