The integral $$\int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$$ is equal to:

We start with the integral: $$\int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$$

First, recall the algebraic identity for the sum of cubes: $$\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x)$$ Since $$\sin^2 x + \cos^2 x = 1$$, this simplifies to: $$\sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x)$$

Therefore, the denominator becomes: $$(\sin^3 x + \cos^3 x)^2 = [(\sin x + \cos x)(1 - \sin x \cos x)]^2 = (\sin x + \cos x)^2 (1 - \sin x \cos x)^2$$

The integral now is: $$\int \frac{\sin^2 x \cos^2 x}{(\sin x + \cos x)^2 (1 - \sin x \cos x)^2} dx$$

Notice that the numerator $$\sin^2 x \cos^2 x = (\sin x \cos x)^2$$, so we write: $$\int \frac{(\sin x \cos x)^2}{(\sin x + \cos x)^2 (1 - \sin x \cos x)^2} dx$$

To simplify, we express everything in terms of $$\tan x$$. Divide both numerator and denominator by $$\cos^4 x$$:

Numerator: $$\frac{\sin^2 x \cos^2 x}{\cos^4 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$$

Denominator: $$\frac{(\sin^3 x + \cos^3 x)^2}{\cos^4 x} = \left( \frac{\sin^3 x}{\cos^3 x} + \frac{\cos^3 x}{\cos^3 x} \right)^2 \cdot \frac{\cos^6 x}{\cos^4 x} = (\tan^3 x + 1)^2 \cos^2 x$$

The integral becomes: $$\int \frac{\tan^2 x}{(\tan^3 x + 1)^2 \cos^2 x} dx$$

We know that $$\frac{1}{\cos^2 x} dx = \sec^2 x dx = d(\tan x)$$. Let $$t = \tan x$$, so $$dt = \sec^2 x dx = \frac{1}{\cos^2 x} dx$$.

Substituting $$t$$ and $$dt$$, the integral simplifies to: $$\int \frac{t^2}{(t^3 + 1)^2} dt$$

Now, solve this integral using substitution. Let $$u = t^3 + 1$$, then $$du = 3t^2 dt$$, which gives $$t^2 dt = \frac{du}{3}$$.

Substituting: $$\int \frac{t^2}{(t^3 + 1)^2} dt = \int \frac{1}{u^2} \cdot \frac{du}{3} = \frac{1}{3} \int u^{-2} du = \frac{1}{3} \cdot \frac{u^{-1}}{-1} + c = -\frac{1}{3u} + c$$

Substitute back $$u = t^3 + 1$$: $$-\frac{1}{3(t^3 + 1)} + c$$

Since $$t = \tan x$$, we have: $$-\frac{1}{3(\tan^3 x + 1)} + c = -\frac{1}{3(1 + \tan^3 x)} + c$$

Comparing with the options:

• Option A: $$\frac{1}{(1+\cot^3 x)} + c$$
• Option B: $$-\frac{1}{3(1+\tan^3 x)} + c$$
• Option C: $$\frac{\sin^3 x}{(1+\cos^3 x)} + c$$
• Option D: $$-\frac{\cos^3 x}{3(1+\sin^3 x)} + c$$

Our result matches option B.

Hence, the correct answer is Option B.