$\sin^2x\:+\frac{1-tan^2x}{sec^2x}\:=\:cos^2x$

$\int\left(x^3+5x^2+2\right)e^{2x}dx$

$\frac{2}{3}ab\left(a^3b^2+2a^2b^3-3ab^4\right)$

$\left(\frac{1}{6}x^3-\frac{2}{3}xy^2\right)^2$

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

$6\left(-2\:-\:x\right)\:=\:-5\left(2x\:+\:4\right)$

$\left(1^n\right)\frac{n+4}{n^2}$