`int_(pi//4)^(3pi//4)(1)/(1+cosx)dx=`

Options

A. `2`
B. `6`
C. `5`
D. `3`

Explanation

To solve the integral

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1 + \cos x} \, dx, \]

we will follow these steps:

### Step 1: Rewrite the integrand
We can use the identity \(1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)\). Thus, we can rewrite the integrand:

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

### Step 2: Change the variable
Let \(u = \frac{x}{2}\). Then, \(dx = 2 \, du\). The limits change as follows:
- When \(x = \frac{\pi}{4}\), \(u = \frac{\pi}{8}\).
- When \(x = \frac{3\pi}{4}\), \(u = \frac{3\pi}{8}\).

Now, the integral becomes:

\[ I = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sec^2(u) \, du. \]

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

\[ I = \left[ \tan(u) \right]_{\frac{\pi}{8}}^{\frac{3\pi}{8}}. \]

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

\[ I = \tan\left(\frac{3\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right). \]

### Step 5: Use the tangent addition formula
Using the tangent addition formula, we can express \(\tan\left(\frac{3\pi}{8}\right)\) and \(\tan\left(\frac{\pi}{8}\right)\):

\[ \tan\left(\frac{3\pi}{8}\right) = \tan\left(\frac{\pi}{4} + \frac{\pi}{8}\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{8}\right)}{1 - \tan\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{8}\right)} = \frac{1 + \tan\left(\frac{\pi}{8}\right)}{1 - \tan\left(\frac{\pi}{8}\right)}. \]

### Step 6: Substitute and simplify
Let \(t = \tan\left(\frac{\pi}{8}\right)\). Then:

\[ I = \frac{1 + t}{1 - t} - t = \frac{1 + t - t(1 - t)}{1 - t} = \frac{1 + t - t + t^2}{1 - t} = \frac{1 + t^2}{1 - t}. \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{1 + \tan^2\left(\frac{\pi}{8}\right)}{1 - \tan\left(\frac{\pi}{8}\right)}. \]