`int_(0)^(pi//4) (1)/(1+cos 2x)dx`

Answer

To solve the integral

\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{1 + \cos 2x} \, dx, \]

we can follow these steps:

### Step 1: Simplify the integrand
We know that

\[ \cos 2x = 2\cos^2 x - 1. \]

Thus,

\[ 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x. \]

Therefore, we can rewrite the integrand:

\[ \frac{1}{1 + \cos 2x} = \frac{1}{2\cos^2 x}. \]

### Step 2: Rewrite the integral
Now, we can express the integral as:

\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{2\cos^2 x} \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx. \]

### Step 3: Integrate
The integral of \(\sec^2 x\) is \(\tan x\). Thus, we have:

\[ \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx = \frac{1}{2} \left[ \tan x \right]_{0}^{\frac{\pi}{4}}. \]

### Step 4: Evaluate the limits
Now we evaluate the limits:

\[ = \frac{1}{2} \left( \tan\left(\frac{\pi}{4}\right) - \tan(0) \right). \]

We know that:

\[ \tan\left(\frac{\pi}{4}\right) = 1 \quad \text{and} \quad \tan(0) = 0. \]

Thus, we have:

\[ = \frac{1}{2} (1 - 0) = \frac{1}{2}. \]

### Final Answer
Therefore, the value of the integral is:

\[ \int_{0}^{\frac{\pi}{4}} \frac{1}{1 + \cos 2x} \, dx = \frac{1}{2}. \]

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