Evaluate the following Integrals.
` int (dx)/((2 sin x+ 3 cos x)^(2))`

Answer

To evaluate the integral

\[ I = \int \frac{dx}{(2 \sin x + 3 \cos x)^2} \]

we will follow these steps:

### Step 1: Multiply by \(\sec^2 x\)

We multiply both the numerator and the denominator by \(\sec^2 x\):

\[ I = \int \frac{\sec^2 x \, dx}{(2 \sin x + 3 \cos x)^2 \sec^2 x} \]

### Step 2: Rewrite the integral

Using the identity \(\sec x = \frac{1}{\cos x}\), we rewrite the integral:

\[ I = \int \frac{\sec^2 x \, dx}{(2 \tan x + 3)^2} \]

### Step 3: Substitution

Let \(t = 2 \tan x + 3\). Then, we differentiate \(t\):

\[ \frac{dt}{dx} = 2 \sec^2 x \implies dt = 2 \sec^2 x \, dx \implies dx = \frac{dt}{2 \sec^2 x} \]

### Step 4: Substitute in the integral

Now substitute \(dx\) in the integral:

\[ I = \int \frac{\sec^2 x \cdot \frac{dt}{2 \sec^2 x}}{t^2} = \int \frac{dt}{2 t^2} \]

### Step 5: Integrate

The integral simplifies to:

\[ I = \frac{1}{2} \int t^{-2} \, dt \]

Integrating gives:

\[ I = \frac{1}{2} \left(-\frac{1}{t}\right) + C = -\frac{1}{2t} + C \]

### Step 6: Substitute back for \(t\)

Substituting back \(t = 2 \tan x + 3\):

\[ I = -\frac{1}{2(2 \tan x + 3)} + C \]

### Final Answer

Thus, the evaluated integral is:

\[ I = -\frac{1}{2(2 \tan x + 3)} + C \]

---