Evaluate $\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$
Answer & explanation
Correct answer: option 2
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 $I = \frac{\pi}{4}$