Evaluate:`int \ (sin^2x cos^2x)/(sin^5x+cos^3x sin^2x + sin^3x cos^2x + cos^5x)^2 \ dx`

Options

A. `(-1)/(1+cot^(3)x)+C`
B. `(1)/(3(1+tan^(3)x))+C`
C. `(-1)/(3(1+tan^(3)x))+C`
D. `(1)/(1+cot^(3)x)+C`

Explanation

To evaluate the integral

\[ I = \int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} \, dx, \]

we will follow these steps:

### Step 1: Simplify the Integral
We start by rewriting the integral \(I\):

\[ I = \int \frac{\sin^2 x \cos^2 x}{(\sin^5 x + \cos^3 x \sin^2 x + \sin^3 x \cos^2 x + \cos^5 x)^2} \, dx. \]

### Step 2: Divide by \(\cos^{10} x\)
To simplify the expression, we divide both the numerator and denominator by \(\cos^{10} x\):

\[ I = \int \frac{\frac{\sin^2 x \cos^2 x}{\cos^{10} x}}{\left(\frac{\sin^5 x}{\cos^{10} x} + \frac{\cos^3 x \sin^2 x}{\cos^{10} x} + \frac{\sin^3 x \cos^2 x}{\cos^{10} x} + \frac{\cos^5 x}{\cos^{10} x}\right)^2} \, dx. \]

This gives us:

\[ I = \int \frac{\tan^2 x}{(\tan^5 x + \tan^3 x + \tan^2 x + 1)^2} \sec^6 x \, dx. \]

### Step 3: Factor the Denominator
Next, we can factor the denominator:

\[ \tan^5 x + \tan^3 x + \tan^2 x + 1 = (\tan^2 x + 1)(\tan^3 x + 1). \]

Thus, we have:

\[ I = \int \frac{\tan^2 x \sec^6 x}{(\tan^2 x + 1)^2 (\tan^3 x + 1)^2} \, dx. \]

### Step 4: Substitute \(t = \tan^3 x + 1\)
Let \(t = \tan^3 x + 1\). Then, we differentiate:

\[ dt = 3\tan^2 x \sec^2 x \, dx \implies dx = \frac{dt}{3\tan^2 x \sec^2 x}. \]

### Step 5: Substitute in the Integral
Substituting \(dx\) into the integral gives:

\[ I = \int \frac{\tan^2 x \sec^6 x}{(\tan^2 x + 1)^2 t^2} \cdot \frac{dt}{3\tan^2 x \sec^2 x}. \]

This simplifies to:

\[ I = \frac{1}{3} \int \frac{\sec^4 x}{(\tan^2 x + 1)^2 t^2} \, dt. \]

### Step 6: Evaluate the Integral
Now, we know that \(\tan^2 x + 1 = \sec^2 x\), so we can rewrite the integral:

\[ I = \frac{1}{3} \int \frac{1}{t^2} \, dt. \]

The integral of \(t^{-2}\) is:

\[ \int t^{-2} \, dt = -\frac{1}{t} + C. \]

### Step 7: Substitute Back
Substituting back for \(t\):

\[ I = -\frac{1}{3} \cdot \frac{1}{\tan^3 x + 1} + C. \]

### Final Answer
Thus, the evaluated integral is:

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