$y=\frac{\left(sin^2x\:tan^6x\right)}{\left(x^2+3\right)^2}$

$\int x\left(4x^2+3\right)^6dx$

$\frac{11a}{x^2-16a^2}+2$

$sin^5x\:=\:\left(\frac{1}{8}\:sin\:x\right)\left(3\:-\:4\:cos\:2x\:+\:cos\:4x\right)\:$

$-10x^2-3x+1$

$\left(15ab^5+12c\right)^2$

$3-\left(+4-8\right)$