$\int\sin^2\left(\frac{x}{4}\right)\:dx$
$u=x+y^3-3$
$\frac{dy}{dx\:}\cdot\frac{1}{y}+\frac{3y}{4}=7$
$6x+2y.4x+3y$
$1+216z^9$
$\frac{dy}{dx}-2y=6x+6$
$\frac{1}{1-cosx}=csc^2x\left(1+cosx\right)$
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