$\int\left(\frac{\left(8-8tan^2x\right)}{-4sec^2x}\right)dx$
$-\frac{7}{12}y=-28$
$3z-z$
$x-2\left(6x^3-2x^2+2x-4\right)$
$\int-1\:do$
$\left(x+8\right)^{-\frac{1}{5}}-\left(x+8\right)^{-\frac{6}{5}}$
$49-x^2$
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