If $$I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x+ \cos^{\frac{3}{2} x}}dx$$, then $$\int_{0}^{21}\frac{x\sin x \cos x}{\sin^{4} x+\cos^{4} x}dx$$ equals :

• Step 1: Use the property $$\int_0^a f(x) dx = \int_0^a f(a-x) dx$$.

$$J = \int_0^{\pi/2} \frac{(\frac{\pi}{2}-x) \sin(\frac{\pi}{2}-x) \cos(\frac{\pi}{2}-x)}{\sin^4(\frac{\pi}{2}-x) + \cos^4(\frac{\pi}{2}-x)} dx = \int_0^{\pi/2} \frac{(\frac{\pi}{2}-x) \cos x \sin x}{\cos^4 x + \sin^4 x} dx$$

• Step 2: Add the two forms of J.

$$2J = \frac{\pi}{2} \int_0^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$$

• Step 3: Solve the remaining integral.

Divide numerator and denominator by $$\cos^4 x$$:

$$\int \frac{\tan x \sec^2 x}{\tan^4 x + 1} dx$$

Let $$u = \tan^2 x, du = 2\tan x \sec^2 x dx$$. Limits change from $$0$$ to $$\infty$$.

$$2J = \frac{\pi}{2} \cdot \frac{1}{2} \int_0^\infty \frac{du}{u^2 + 1} = \frac{\pi}{4} [\tan^{-1} u]_0^\infty = \frac{\pi}{4} \cdot \frac{\pi}{2} = \frac{\pi^2}{8}$$

$$J = \frac{\pi^2}{16}$$

Correct Option: C ($$\pi^2/16$$)