`int(1)/(sin^(2)x-4 cos^(2)x)dx`

Answer

To solve the integral \( I = \int \frac{1}{\sin^2 x - 4 \cos^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{1}{\sin^2 x - 4 \cos^2 x} \, dx \]

### Step 2: Divide numerator and denominator by \(\cos^2 x\)
To simplify the expression, we divide both the numerator and the denominator by \(\cos^2 x\):
\[ I = \int \frac{1/\cos^2 x}{\frac{\sin^2 x}{\cos^2 x} - 4} \, dx = \int \frac{\sec^2 x}{\tan^2 x - 4} \, dx \]

### Step 3: Substitute \( \tan x = t \)
Let \( t = \tan x \). Then, the derivative \( dt = \sec^2 x \, dx \), which means \( dx = \frac{dt}{\sec^2 x} \). Substituting this into the integral gives:
\[ I = \int \frac{1}{t^2 - 4} \, dt \]

### Step 4: Recognize the integral form
The expression \( t^2 - 4 \) can be factored as \( (t - 2)(t + 2) \). We can use the formula for the integral:
\[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln \left| \frac{x - a}{x + a} \right| + C \]
where \( a = 2 \).

### Step 5: Apply the formula
Applying the formula:
\[ I = \frac{1}{2 \cdot 2} \ln \left| \frac{t - 2}{t + 2} \right| + C = \frac{1}{4} \ln \left| \frac{t - 2}{t + 2} \right| + C \]

### Step 6: Substitute back \( t = \tan x \)
Now, substituting back \( t = \tan x \):
\[ I = \frac{1}{4} \ln \left| \frac{\tan x - 2}{\tan x + 2} \right| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{1}{4} \ln \left| \frac{\tan x - 2}{\tan x + 2} \right| + C \]