$\sin^2b=1-cos^2b$

$\int\frac{t^3}{\left(4-2t^4\right)^8}dt$

$14+-1+-30$

$3\left(11\right)^4-3\left(11\right)^3-\left(11\right)^2+3\left(11\right)-1\:$

$\left|-7\right|\cdot\left|-4\right|\cdot\left|-2\right|$

$5\:\cos^4\:\theta\:-\:5\:sin^4\:\theta\:=\:5\:cos\:\theta$

$dy\left(x+4y\right)=\left(2x-y\right)dx$