To find the area of the region bounded by the x-axis and the curves defined by \( y = \tan x \) (for \( -\frac{\pi}{3} \leq x \leq \frac{\pi}{3} \)) and \( y = \cot x \) (for \( \frac{\pi}{6} \leq x \leq \frac{2\pi}{3} \)), we will follow these steps:
### Step 1: Identify the curves and their intersections
We have two curves:
1. \( y = \tan x \) for \( -\frac{\pi}{3} \leq x \leq \frac{\pi}{3} \)
2. \( y = \cot x \) for \( \frac{\pi}{6} \leq x \leq \frac{2\pi}{3} \)
To find the area between these curves, we first need to find their points of intersection. Setting \( \tan x = \cot x \) gives us:
\[
\tan^2 x = 1 \implies \tan x = \pm 1
\]
The solutions for \( x \) are:
\[
x = \frac{\pi}{4} \quad \text{and} \quad x = -\frac{\pi}{4}
\]
### Step 2: Determine the relevant intervals
From the intervals given:
- \( y = \tan x \) is valid from \( -\frac{\pi}{3} \) to \( \frac{\pi}{3} \).
- \( y = \cot x \) is valid from \( \frac{\pi}{6} \) to \( \frac{2\pi}{3} \).
The intersection point \( x = \frac{\pi}{4} \) lies within both intervals. Therefore, we will calculate the area from \( \frac{\pi}{6} \) to \( \frac{\pi}{4} \) for \( y = \tan x \) and from \( \frac{\pi}{4} \) to \( \frac{2\pi}{3} \) for \( y = \cot x \).
### Step 3: Set up the integrals for the area
The area \( A \) can be expressed as:
\[
A = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \tan x \, dx + \int_{\frac{\pi}{4}}^{\frac{2\pi}{3}} \cot x \, dx
\]
### Step 4: Calculate the integrals
1. **Integral of \( \tan x \)**:
\[
\int \tan x \, dx = -\log(\cos x) + C
\]
Thus,
\[
\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \tan x \, dx = \left[-\log(\cos x)\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -\log\left(\cos\frac{\pi}{4}\right) + \log\left(\cos\frac{\pi}{6}\right)
\]
Calculating the values:
\[
\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}, \quad \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}
\]
So,
\[
\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \tan x \, dx = -\log\left(\frac{1}{\sqrt{2}}\right) + \log\left(\frac{\sqrt{3}}{2}\right) = \log\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{2}\log(2)
\]
2. **Integral of \( \cot x \)**:
\[
\int \cot x \, dx = \log(\sin x) + C
\]
Thus,
\[
\int_{\frac{\pi}{4}}^{\frac{2\pi}{3}} \cot x \, dx = \left[\log(\sin x)\right]_{\frac{\pi}{4}}^{\frac{2\pi}{3}} = \log\left(\sin\frac{2\pi}{3}\right) - \log\left(\sin\frac{\pi}{4}\right)
\]
Calculating the values:
\[
\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}, \quad \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}
\]
So,
\[
\int_{\frac{\pi}{4}}^{\frac{2\pi}{3}} \cot x \, dx = \log\left(\frac{\sqrt{3}}{2}\right) - \log\left(\frac{1}{\sqrt{2}}\right) = \log\left(\frac{\sqrt{3}}{2}\cdot\sqrt{2}\right) = \log\left(\frac{\sqrt{6}}{2}\right)
\]
### Step 5: Combine the results
Now, we combine both integrals:
\[
A = \log\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{2}\log(2) + \log\left(\frac{\sqrt{6}}{2}\right)
\]
This simplifies to:
\[
A = \log\left(\frac{\sqrt{3} \cdot \sqrt{6}}{4}\right) = \log\left(\frac{\sqrt{18}}{4}\right) = \log\left(\frac{3\sqrt{2}}{4}\right)
\]
### Final Result
Thus, the area of the region bounded by the x-axis and the curves is:
\[
\boxed{\log\left(\frac{3\sqrt{2}}{4}\right)}
\]
