Solutions - 0040-02

Notice ${\sec }^{2}xdx$. Therefore integrate with $u$-sub setting $u=\tan x$ and $du={\sec }^{2}xdx$:

$$\begin{array}{cc}\int {\tan }^{2}x\cdot {\sec }^{2}xdx& \gg \gg \int u^{2}du\\ & \\ & \gg \gg \frac{u^3}{3}+C\\ & \\ & \gg \gg \frac{{\tan }^{3}x}{3}+C\end{array}$$