Answer
0
Work Step by Step
$\int$4sec(x)tan(x)dx = 4$\int$sec(x)tan(x)dx
remember that $\frac{d}{dx}$[sec(x)] = sec(x)tan(x)
therefore 4$\int$sec(x)tan(x)dx = 4sec(x)+C
However this is a definite integral so we must evaluate the antiderivative from -$\frac{\pi}{3}$ to $\frac{\pi}{3}$.
Both sec(-$\frac{\pi}{3}$) and sec($\frac{\pi}{3}$) are equal to 2.
We can then write 4sec($\frac{\pi}{3}$) - 4sec(-$\frac{\pi}{3}$), which simplifies to 4(2)-4(2). Which clearly equals zero.
Another way to solve this is graphically. The function has point symmetry about the origin. Therefore if the integral is evaluated from -x to x, it will always equal zero.
