To solve the integral \(\int_{\pi/6}^{\pi/3} \frac{\tan x}{\tan x + \cot x} \, dx\), we first simplify the expression inside the integral.

The integrand is \(\frac{\tan x}{\tan x + \cot x}\). We know:

• \(\tan x = \frac{\sin x}{\cos x}\)
• \(\cot x = \frac{\cos x}{\sin x}\)

Therefore, the expression \(\tan x + \cot x\) becomes:

\(\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x}\)

Using the trigonometric identity \(\sin^2 x + \cos^2 x = 1\), we can simplify it further:

\(\tan x + \cot x = \frac{1}{\sin x \cos x}\)

Substitute back into the integrand:

\(\frac{\tan x}{\tan x + \cot x} = \frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x \cos x}}\)

Simplifying this gives:

\(\frac{\tan x}{\tan x + \cot x} = \sin^2 x\)

Now, the integral becomes:

\(\int_{\pi/6}^{\pi/3} \sin^2 x \, dx\)

Using the identity \(\sin^2 x = \frac{1 - \cos 2x}{2}\), we rewrite the integral:

\(\int_{\pi/6}^{\pi/3} \frac{1 - \cos 2x}{2} \, dx\)

This splits into two separate integrals:

\(\frac{1}{2} \int_{\pi/6}^{\pi/3} 1 \, dx - \frac{1}{2} \int_{\pi/6}^{\pi/3} \cos 2x \, dx\)

Calculate each of these:

• The first integral:
• The second integral involves a substitution:
  • When \(x = \pi/6\), \(u = \pi/3\)
  • When \(x = \pi/3\), \(u = 2\pi/3\)
• \(-\frac{1}{4} [\sin u]_{\pi/3}^{2\pi/3}\)
• \(-\frac{1}{4} (\sin(2\pi/3) - \sin(\pi/3)) = -\frac{1}{4} \left(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right) = 0\)

Thus, the original integral evaluates to:

\(\frac{\pi}{12} + 0 = \frac{\pi}{12}\)

Therefore, the value of the integral is \(\frac{\pi}{12}\).