$$\int_0^{\pi/4} \frac{\cos^2 x \sin^2 x}{(\cos^3 x + \sin^3 x)^2} dx$$ is equal to

Let $$I = \displaystyle\int_{0}^{\pi/4} \frac{\cos^{2}x \,\sin^{2}x}{\bigl(\cos^{3}x+\sin^{3}x\bigr)^{2}}\,dx$$.

Case 1: Convert the integral to a single-variable rational form using $$t=\tan x$$.
For $$t=\tan x$$ we have $$\sin x = \frac{t}{\sqrt{1+t^{2}}},\qquad \cos x = \frac{1}{\sqrt{1+t^{2}}},\qquad dx = \frac{dt}{1+t^{2}}.$$ When $$x$$ goes from $$0$$ to $$\pi/4$$, $$t$$ goes from $$0$$ to $$1$$.

Compute the numerator:
$$\cos^{2}x\,\sin^{2}x = \frac{1}{1+t^{2}}\cdot\frac{t^{2}}{1+t^{2}} = \frac{t^{2}}{(1+t^{2})^{2}}.$$ Compute the denominator:
$$\cos^{3}x + \sin^{3}x = \frac{1}{(1+t^{2})^{3/2}} + \frac{t^{3}}{(1+t^{2})^{3/2}} = \frac{1+t^{3}}{(1+t^{2})^{3/2}}.$$ Therefore $$\bigl(\cos^{3}x+\sin^{3}x\bigr)^{2} = \frac{(1+t^{3})^{2}}{(1+t^{2})^{3}}.$$ Putting these into the integrand gives $$\frac{\cos^{2}x\sin^{2}x}{(\cos^{3}x+\sin^{3}x)^{2}} = \frac{t^{2}/(1+t^{2})^{2}}{(1+t^{3})^{2}/(1+t^{2})^{3}} = \frac{t^{2}(1+t^{2})}{(1+t^{3})^{2}}.$$ Multiplying by $$dx=\dfrac{dt}{1+t^{2}}$$, $$I = \int_{0}^{1} \frac{t^{2}(1+t^{2})}{(1+t^{3})^{2}}\cdot\frac{dt}{1+t^{2}} = \int_{0}^{1} \frac{t^{2}}{(1+t^{3})^{2}}\,dt.$$

Case 2: Evaluate the rational integral by substituting $$u=t^{3}$$.
Put $$u = t^{3}\; \Longrightarrow\; du = 3t^{2}\,dt \; \Longrightarrow\; t^{2}\,dt = \frac{du}{3}.$$ As $$t$$ varies from $$0$$ to $$1$$, $$u$$ also varies from $$0$$ to $$1$$. Hence $$I = \int_{0}^{1} \frac{t^{2}}{(1+t^{3})^{2}}\,dt = \frac13\int_{0}^{1} \frac{du}{(1+u)^{2}}.$$

Case 3: Integrate the simple power function.
Recall $$\displaystyle\int (1+u)^{-2}\,du = -\frac{1}{1+u}+C.$$ Therefore $$\frac13\int_{0}^{1} \frac{du}{(1+u)^{2}} = \frac13\Bigl[-\frac{1}{1+u}\Bigr]_{0}^{1} = \frac13\Bigl(-\frac12 + 1\Bigr) = \frac13\cdot\frac12 = \frac16.$$

Hence $$I = \dfrac16$$.

The correct option is Option A (1/6).