`int_(pi//6)^(pi//3)(1)/(1+tan x) dx=`

Options

A. `pi`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(12)`

Answer

To solve the integral \( \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \tan x} \, dx \), we can use a property of definite integrals. Let's go through the solution step by step.

### Step 1: Define the Integral
Let
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \tan x} \, dx \]

### Step 2: Use the Property of Definite Integrals
We can use the property that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
In our case, \( a = \frac{\pi}{6} \) and \( b = \frac{\pi}{3} \). Thus,
\[ a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \]
So, we can rewrite the integral as:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \tan\left(\frac{\pi}{2} - x\right)} \, dx \]

### Step 3: Simplify the Integral
Using the identity \( \tan\left(\frac{\pi}{2} - x\right) = \cot x \), we have:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \cot x} \, dx \]
Now, we can rewrite \( \frac{1}{1 + \cot x} \):
\[ \cot x = \frac{1}{\tan x} \implies 1 + \cot x = 1 + \frac{1}{\tan x} = \frac{\tan x + 1}{\tan x} \]
Thus,
\[ \frac{1}{1 + \cot x} = \frac{\tan x}{\tan x + 1} \]
So we can express our integral as:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\tan x}{\tan x + 1} \, dx \]

### Step 4: Add the Two Integrals
Now we have two expressions for \( I \):
1. \( I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \tan x} \, dx \) (Equation 1)
2. \( I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\tan x}{\tan x + 1} \, dx \) (Equation 2)

Adding these two equations:
\[ 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{1}{1 + \tan x} + \frac{\tan x}{\tan x + 1} \right) \, dx \]
The integrand simplifies as follows:
\[ \frac{1}{1 + \tan x} + \frac{\tan x}{\tan x + 1} = \frac{1 + \tan x}{1 + \tan x} = 1 \]
Thus,
\[ 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx \]

### Step 5: Evaluate the Integral
Now, we can evaluate the integral:
\[ 2I = \left[ x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{6} \]
So,
\[ I = \frac{\pi}{12} \]

### Final Answer
The value of the integral is:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \tan x} \, dx = \frac{\pi}{12} \]