$-8x\:-\:10\:>\:18\:-\:16x$

$\int tan^5\left(\frac{x}{4}\right)\cdot sec^2\left(\frac{x}{4}\right)$

$dy=e^{-3x}dx$

$\int\frac{\sec^2\left(a\right)}{tan\left(a\right)+1}dx$

$\frac{\theta}{2}=-0.88$

$\int-\left(16x+10\right)\left(1-\left(4x^2+5x\right)\right)^5dx$

$\left(m^2+4\right)\left(m^2-4\right)$