$-2\:-35\:+\:3\:-7\:-\:2$

$\tan^2\left(\infty\right)\cdot\left(1-\sin^2\left(\infty\right)\right)=\sin^2\left(\infty\right)$

$\lim_{x\to0}\frac{6x^3}{sinx-x}$

$48=y-3$

$y'\left(x+2\right)+4x\left(y+1\right)^2$

$-27-\left(-17\right)-\left(-34\right)-16$

$\int y\sqrt[3]{y+4}dy$