$\lim_{x\to\infty}\frac{3}{7}\cdot\left(\frac{1}{6}\right)^{n+1}$

$\int\sec^5\frac{1}{2}x\:\tan\frac{1}{2}xdx$

$16cos^2x-4=0$

$\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+6$

$\frac{2\sec\left(x\right)-3\tan\left(x\right)+1}{2\csc\left(x\right)+\cot\left(x\right)-3}$

$\int x\ln\left|2x\right|dx$

$a^3+12a^2+48a+64$