`int(1)/(3sin^(2) x+4 cos^(2) x)dx`

Answer

To solve the integral \( \int \frac{1}{3 \sin^2 x + 4 \cos^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the Denominator
We start with the integral:
\[ \int \frac{1}{3 \sin^2 x + 4 \cos^2 x} \, dx \]
To simplify the denominator, we can multiply and divide by \( \cos^2 x \):
\[ = \int \frac{\cos^2 x}{3 \sin^2 x \cos^2 x + 4 \cos^4 x} \, dx \]

### Step 2: Substitute \( \tan x \)
Now we can use the substitution \( t = \tan x \), which gives us \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2} \). Also, we know that \( \sin^2 x = \frac{t^2}{1+t^2} \) and \( \cos^2 x = \frac{1}{1+t^2} \).

Substituting these into the integral, we have:
\[ = \int \frac{\frac{1}{1+t^2}}{3 \frac{t^2}{(1+t^2)} + 4 \frac{1}{(1+t^2)^2}} \cdot \frac{dt}{1+t^2} \]
This simplifies to:
\[ = \int \frac{1}{3t^2 + 4} \, dt \]

### Step 3: Factor Out Constants
Next, we factor out the constant:
\[ = \frac{1}{3} \int \frac{1}{t^2 + \frac{4}{3}} \, dt \]

### Step 4: Use the Integration Formula
We can now use the integration formula:
\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
In our case, \( a^2 = \frac{4}{3} \) so \( a = \frac{2}{\sqrt{3}} \).

Thus, we have:
\[ = \frac{1}{3} \cdot \frac{1}{\frac{2}{\sqrt{3}}} \tan^{-1} \left( \frac{t}{\frac{2}{\sqrt{3}}} \right) + C \]
This simplifies to:
\[ = \frac{\sqrt{3}}{6} \tan^{-1} \left( \frac{\sqrt{3} t}{2} \right) + C \]

### Step 5: Substitute Back for \( t \)
Now, we substitute back \( t = \tan x \):
\[ = \frac{\sqrt{3}}{6} \tan^{-1} \left( \frac{\sqrt{3} \tan x}{2} \right) + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{1}{3 \sin^2 x + 4 \cos^2 x} \, dx = \frac{\sqrt{3}}{6} \tan^{-1} \left( \frac{\sqrt{3} \tan x}{2} \right) + C \]