$\int\limits_0^{\pi/2}\frac{\sin^8 x}{\sin^8x + \cos^8 x}dx$ is equal to

$\frac{π}{4}$

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

Let the integral be: $\displaystyle I = \int_0^{\frac{\pi}{2}} \frac{\sin^8 x}{\sin^8 x + \cos^8 x}\, dx$

Use the identity: $\displaystyle \int_0^{\frac{\pi}{2}} f(x)\, dx = \int_0^{\frac{\pi}{2}} f\left(\frac{\pi}{2} - x\right)\, dx$

Let $\displaystyle f(x) = \frac{\sin^8 x}{\sin^8 x + \cos^8 x}$ Then $\displaystyle f\left(\frac{\pi}{2} - x\right) = \frac{\cos^8 x}{\cos^8 x + \sin^8 x} = \frac{\cos^8 x}{\sin^8 x + \cos^8 x}$

Now add both:

$\displaystyle I + I = \int_0^{\frac{\pi}{2}} \left( \frac{\sin^8 x}{\sin^8 x + \cos^8 x} + \frac{\cos^8 x}{\sin^8 x + \cos^8 x} \right) dx = \int_0^{\frac{\pi}{2}} 1\, dx = \frac{\pi}{2}$

$\displaystyle \Rightarrow 2I = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4}$