`int_(-pi//4)^(pi//4) ( dx)/( 1+cos 2x) ` is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the integral \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{1 + \cos 2x} \), we can follow these steps:

### Step 1: Simplify the integrand
We can use the trigonometric identity:
\[ 1 + \cos 2x = 2 \cos^2 x \]
Thus, we can rewrite the integral as:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{2 \cos^2 x} \]

### Step 2: Factor out the constant
We can factor out the constant \( \frac{1}{2} \) from the integral:
\[ I = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \, dx \]

### Step 3: Integrate \( \sec^2 x \)
The integral of \( \sec^2 x \) is:
\[ \int \sec^2 x \, dx = \tan x \]
Thus, we have:
\[ I = \frac{1}{2} \left[ \tan x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \]

### Step 4: Evaluate the limits
Now we will evaluate the limits:
\[ I = \frac{1}{2} \left( \tan\left(\frac{\pi}{4}\right) - \tan\left(-\frac{\pi}{4}\right) \right) \]
We know that:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \quad \text{and} \quad \tan\left(-\frac{\pi}{4}\right) = -1 \]
Thus:
\[ I = \frac{1}{2} \left( 1 - (-1) \right) = \frac{1}{2} \left( 1 + 1 \right) = \frac{1}{2} \cdot 2 = 1 \]

### Final Answer
The value of the integral is:
\[ \boxed{1} \]