Evaluate `int(1)/(4sin^(2) x + 9 cos^(2) x) dx`

Answer

To evaluate the integral \( I = \int \frac{1}{4 \sin^2 x + 9 \cos^2 x} \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{1}{4 \sin^2 x + 9 \cos^2 x} \, dx \]
To simplify the expression, we can multiply the numerator and the denominator by \( \sec^2 x \):
\[ I = \int \frac{\sec^2 x}{4 \sin^2 x \sec^2 x + 9 \cos^2 x \sec^2 x} \, dx \]
Since \( \sec^2 x = 1 + \tan^2 x \), we can rewrite the integral as:
\[ I = \int \frac{\sec^2 x}{4 \tan^2 x + 9} \, dx \]

### Step 2: Factor out constants
Next, we factor out the constant from the denominator:
\[ I = \int \frac{1}{4} \cdot \frac{\sec^2 x}{\tan^2 x + \frac{9}{4}} \, dx \]
This simplifies to:
\[ I = \frac{1}{4} \int \frac{\sec^2 x}{\tan^2 x + \left(\frac{3}{2}\right)^2} \, dx \]

### Step 3: Use substitution
Now, we will use the substitution \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \), or \( dx = \frac{dt}{\sec^2 x} \):
\[ I = \frac{1}{4} \int \frac{1}{t^2 + \left(\frac{3}{2}\right)^2} \, dt \]

### Step 4: Integrate using the arctangent formula
The integral \( \int \frac{1}{t^2 + a^2} \, dt \) is known to be \( \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C \):
\[ I = \frac{1}{4} \cdot \frac{1}{\frac{3}{2}} \tan^{-1} \left( \frac{t}{\frac{3}{2}} \right) + C \]
This simplifies to:
\[ I = \frac{1}{6} \tan^{-1} \left( \frac{2t}{3} \right) + C \]

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

### Final Answer
Thus, the final result of the integral is:
\[ I = \frac{1}{6} \tan^{-1} \left( \frac{2 \tan x}{3} \right) + C \]