`int_(pi//4)^(3pi//4)(dx)/(1+cosx)` is equal to

Options

A. `-2`
B. 2
C. 4
D. `-1`

Explanation

To solve the integral \(\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x}\), we will follow these steps:

### Step 1: Rewrite the integrand
We know that \(1 + \cos x\) can be rewritten using 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: Set up the integral
Now, we can set up the integral with the new expression:
\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{2} \sec^2\left(\frac{x}{2}\right) dx \]

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

Thus, the integral becomes:
\[ \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sec^2(u) \, du \]

### Step 4: Integrate
The integral of \(\sec^2(u)\) is \(\tan(u)\):
\[ \int \sec^2(u) \, du = \tan(u) \]

So, we evaluate:
\[ \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sec^2(u) \, du = \tan\left(\frac{3\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right) \]

### Step 5: Find the values of \(\tan\left(\frac{3\pi}{8}\right)\) and \(\tan\left(\frac{\pi}{8}\right)\)
Using the tangent addition formula:
\[ \tan\left(\frac{3\pi}{8}\right) = \tan\left(\frac{\pi}{4} + \frac{\pi}{8}\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{8}\right)}{1 - \tan\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{8}\right)} \]
Since \(\tan\left(\frac{\pi}{4}\right) = 1\):
\[ \tan\left(\frac{3\pi}{8}\right) = \frac{1 + \tan\left(\frac{\pi}{8}\right)}{1 - \tan\left(\frac{\pi}{8}\right)} \]

### Step 6: Final result
Thus, the final result of the integral is:
\[ \tan\left(\frac{3\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right) = \frac{1 + \tan\left(\frac{\pi}{8}\right)}{1 - \tan\left(\frac{\pi}{8}\right)} - \tan\left(\frac{\pi}{8}\right) \]

This simplifies to:
\[ \frac{1 + \tan\left(\frac{\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right)(1 - \tan\left(\frac{\pi}{8}\right))}{1 - \tan\left(\frac{\pi}{8}\right)} = \frac{1}{1 - \tan\left(\frac{\pi}{8}\right)} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} = \tan\left(\frac{3\pi}{8}\right) - \tan\left(\frac{\pi}{8}\right) \]