What is the value of `int_(pi//6)^(pi//4) (dx)/(sinx cos x)` ?

Options

A. `2 ln sqrt(3)`
B. `ln sqrt(3)`
C. `2 ln 3`
D. `4 ln 3`

Explanation

To solve the integral \( I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{dx}{\sin x \cos x} \), we can follow these steps:

### Step 1: Rewrite the integrand
We know that \( \sin x \cos x = \frac{1}{2} \sin 2x \). Therefore, we can rewrite the integral as:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{dx}{\sin x \cos x} = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{2 \, dx}{\sin 2x} \]

### Step 2: Factor out the constant
Now, we can factor out the constant 2 from the integral:
\[ I = 2 \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{dx}{\sin 2x} \]

### Step 3: Use the integral formula
The integral of \( \frac{1}{\sin u} \) is \( \log \left| \tan \frac{u}{2} \right| + C \). Thus, we can apply this to our integral:
\[ I = 2 \left[ -\log \left| \tan \frac{2x}{2} \right| \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = 2 \left[ -\log \left| \tan x \right| \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \]

### Step 4: Evaluate the limits
Now we evaluate the limits:
\[ I = 2 \left( -\log \left| \tan \frac{\pi}{4} \right| + \log \left| \tan \frac{\pi}{6} \right| \right) \]
Calculating the tangent values:
- \( \tan \frac{\pi}{4} = 1 \)
- \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \)

Thus:
\[ I = 2 \left( -\log 1 + \log \frac{1}{\sqrt{3}} \right) = 2 \left( 0 + \log \frac{1}{\sqrt{3}} \right) = 2 \left( -\frac{1}{2} \log 3 \right) = -\log 3 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{-\log 3} \]