`int(dx)/(cos x + sqrt(3) sin x)` equals :

Options

A. `1/2 "log tan"(x/2 + pi/12) + C`
B. `1/2 "log tan"(x/2 - pi/12) + C`
C. `"log tan"(x/2 + pi/12) + C`
D. `"log tan"(x/2 +- pi/12) + C`

Answer

To solve the integral

\[ \int \frac{dx}{\cos x + \sqrt{3} \sin x} \]

we will follow these steps:

### Step 1: Multiply numerator and denominator by \( \frac{1}{2} \)

We start by multiplying both the numerator and the denominator by \( \frac{1}{2} \):

\[ \int \frac{dx}{\cos x + \sqrt{3} \sin x} = \int \frac{\frac{1}{2} dx}{\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x} \]

### Step 2: Recognize the denominator as a sine function

Notice that \( \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x \) can be expressed in terms of sine. We can rewrite it as:

\[ \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x = \sin\left(\frac{\pi}{6}\right) \cos x + \cos\left(\frac{\pi}{6}\right) \sin x \]

This is of the form \( \sin(a + b) \) where \( a = x \) and \( b = \frac{\pi}{6} \):

\[ = \sin\left(x + \frac{\pi}{6}\right) \]

### Step 3: Substitute into the integral

Now we can rewrite the integral:

\[ \int \frac{\frac{1}{2} dx}{\sin\left(x + \frac{\pi}{6}\right)} = \frac{1}{2} \int \frac{dx}{\sin\left(x + \frac{\pi}{6}\right)} \]

### Step 4: Integrate using the logarithmic identity

The integral of \( \frac{1}{\sin u} \) is \( \log \left| \tan\left(\frac{u}{2}\right) \right| + C \). Therefore, we have:

\[ \frac{1}{2} \int \frac{dx}{\sin\left(x + \frac{\pi}{6}\right)} = \frac{1}{2} \log \left| \tan\left(\frac{x + \frac{\pi}{6}}{2}\right) \right| + C \]

### Step 5: Final answer

Thus, the final answer is:

\[ \frac{1}{2} \log \left| \tan\left(\frac{x + \frac{\pi}{6}}{2}\right) \right| + C \]