Answer
$$\int^0_{-\pi/3}2\sec^3xdx=2\sqrt3-\ln(2-\sqrt3)$$
Work Step by Step
$$A=\int^0_{-\pi/3}2\sec^3xdx$$ $$A=2\int^0_{-\pi/3}\sec x\sec^2xdx$$
Take $u=\sec x$ and $dv=\sec^2xdx$
That makes $du=\sec x\tan xdx$ and $v=\tan x$
According to Integration by Parts rule, we have $$A=2\Big(\sec x\tan x\Big]^0_{-\pi/3}-\int^0_{-\pi/3}\sec x\tan^2xdx\Big)$$ $$A=2\Big(0-\sec(-\frac{\pi}{3})\tan(-\frac{\pi}{3})-\int^0_{-\pi/3}\sec x(\sec^2x-1)dx\Big)$$ $$A=2\Big(-2(-\sqrt3)-\int^0_{-\pi/3}\sec^3xdx+\int^0_{-\pi/3}\sec xdx\Big)$$ $$A=2\Big(2\sqrt3-\int^0_{-\pi/3}\sec^3xdx+\ln|\sec x+\tan x|\Big]^0_{-\pi/3}\Big)$$
The integral $\int^0_{-\pi/3}\sec^3xdx$ is actually half the given problem, so
$$A=2\Big(2\sqrt3-\frac{1}{2}A+\ln|1+0|-\ln|2-\sqrt3|\Big)$$ $$A=2(2\sqrt3-\frac{1}{2}A-\ln(2-\sqrt3))$$ $$A=4\sqrt3-A-2\ln(2-\sqrt3)$$ $$2A=4\sqrt3-2\ln(2-\sqrt3)$$ $$A=2\sqrt3-\ln(2-\sqrt3)$$
