Evaluate $\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$

$\frac{\pi}{4}$

The correct answer is Option (2) → $\frac{\pi}{4}$

Let $I = \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx \quad \dots (1)$

Then, by $P_4$

$I = \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^4 (\frac{\pi}{2} - x)}{\sin^4 (\frac{\pi}{2} - x) + \cos^4 (\frac{\pi}{2} - x)} \, dx = \int\limits_{0}^{\frac{\pi}{2}} \frac{\cos^4 x}{\cos^4 x + \sin^4 x} \, dx \quad \dots (2)$

Adding (1) and (2), we get

$2I = \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} \, dx = \int\limits_{0}^{\frac{\pi}{2}} dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$

Hence $I = \frac{\pi}{4}$