Evaluate `int _(0)^(pi//2)(dx)/(1+ cos x)`

Answer

To evaluate the integral

\[ I = \int_0^{\frac{\pi}{2}} \frac{dx}{1 + \cos x} \]

we can use a trigonometric identity to simplify the integrand.

### Step 1: Use the trigonometric identity
We know that

\[ \cos x = 2 \cos^2\left(\frac{x}{2}\right) - 1 \]

This implies that

\[ 1 + \cos x = 1 + (2 \cos^2\left(\frac{x}{2}\right) - 1) = 2 \cos^2\left(\frac{x}{2}\right) \]

Thus, we can rewrite the integral:

\[ I = \int_0^{\frac{\pi}{2}} \frac{dx}{2 \cos^2\left(\frac{x}{2}\right)} = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sec^2\left(\frac{x}{2}\right) dx \]

### Step 2: Make a substitution
Let

\[ u = \frac{x}{2} \quad \Rightarrow \quad dx = 2 du \]

When \( x = 0 \), \( u = 0 \) and when \( x = \frac{\pi}{2} \), \( u = \frac{\pi}{4} \). Thus, the integral becomes:

\[ I = \frac{1}{2} \int_0^{\frac{\pi}{4}} \sec^2(u) \cdot 2 du = \int_0^{\frac{\pi}{4}} \sec^2(u) du \]

### Step 3: Integrate
The integral of \( \sec^2(u) \) is:

\[ \int \sec^2(u) du = \tan(u) + C \]

So we have:

\[ I = \left[ \tan(u) \right]_0^{\frac{\pi}{4}} = \tan\left(\frac{\pi}{4}\right) - \tan(0) \]

### Step 4: Evaluate the limits
We know that:

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

Thus:

\[ I = 1 - 0 = 1 \]

### Final Answer
The value of the integral is:

\[ \boxed{1} \]