Answer
$\displaystyle \frac{3}{2}\ln 2$
Work Step by Step
Area between curves $f(x)$ and $g(x)$, where $g(x)\geq f(x)$ on $[a,b], $ is
$A=\displaystyle \int_{a}^{b}[g(x)-f(x)]dx$
On $[-\pi/4,0]$ the graph of tan is below the x-axis
On $[0,\pi/3]$, it is above the x-axis.
$A=\displaystyle \int_{-\pi/4}^{0}-\tan xdx+\int_{0}^{\pi/3}\tan xdx$
Because $\displaystyle \tan x=\frac{\sin x}{\cos x}$,
$\displaystyle \int\tan xdx=\quad \displaystyle \left[\begin{array}{l}
u=\cos x\\
du=-\sin x
\end{array}\right]=\int u^{-1}du=\ln|u|+C=\ln|\cos x|+C$
So,
$A=[\ln|\cos x|]_{-\pi/4}^{0}-[\ln|\cos x|]_{0}^{\pi/3}$
$=(\displaystyle \ln 1-\ln\frac{1}{\sqrt{2}})-(\ln\frac{1}{2}-\ln 1)$
$=0-(-\displaystyle \frac{1}{2}\ln 2)-(-1)\ln 2+0$
$=\displaystyle \frac{3}{2}\ln 2$
