`int(1)/(cos^(2) x-3 sin^(2) x)dx`

Answer

To solve the integral \( I = \int \frac{1}{\cos^2 x - 3 \sin^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{\cos^2 x - 3 \sin^2 x} \, dx \]

### Step 2: Divide by \(\cos^2 x\)
Next, we divide the numerator and the denominator by \(\cos^2 x\):
\[ I = \int \frac{1/\cos^2 x}{1 - 3 \tan^2 x} \, dx \]
This simplifies to:
\[ I = \int \frac{\sec^2 x}{1 - 3 \tan^2 x} \, dx \]

### Step 3: Factor Out 3
Now, we can factor out 3 from the denominator:
\[ I = \frac{1}{3} \int \frac{3 \sec^2 x}{1 - 3 \tan^2 x} \, dx \]

### Step 4: Substitute \( t = \tan x \)
Let \( t = \tan x \). Then, we have:
\[ \sec^2 x \, dx = dt \]
Thus, we can rewrite the integral as:
\[ I = \frac{1}{3} \int \frac{dt}{1 - 3t^2} \]

### Step 5: Use the Integral Formula
We can use the formula for the integral:
\[ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a + x}{a - x} \right| + C \]
Here, \( a^2 = \frac{1}{3} \) implies \( a = \frac{1}{\sqrt{3}} \). Thus:
\[ I = \frac{1}{3} \cdot \frac{1}{2 \cdot \frac{1}{\sqrt{3}}} \ln \left| \frac{\frac{1}{\sqrt{3}} + t}{\frac{1}{\sqrt{3}} - t} \right| + C \]

### Step 6: Simplify the Expression
Now, simplifying gives:
\[ I = \frac{1}{2\sqrt{3}} \ln \left| \frac{\frac{1}{\sqrt{3}} + \tan x}{\frac{1}{\sqrt{3}} - \tan x} \right| + C \]

### Step 7: Final Answer
Thus, the final answer is:
\[ I = \frac{1}{2\sqrt{3}} \ln \left| \frac{1 + \sqrt{3} \tan x}{1 - \sqrt{3} \tan x} \right| + C \]