$\int_0^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$ is equal to

Let $I = \int_0^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$ ...(1)

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Using the King Property for definite integrals:

$\int_0^{a} f(x)dx = \int_0^{a} f(a-x)dx$

For $a = \frac{\pi}{2}$:

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

Using the complementary angle identities:

$\sin\left(\frac{\pi}{2}-x\right) = \cos x$

$\cos\left(\frac{\pi}{2}-x\right) = \sin x$

This gives:

$I = \int_0^{\pi/2} \frac{\cos^8 x}{\cos^8 x + \sin^8 x} dx$ ...(2)

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Adding equations (1) and (2):

$I + I = \int_0^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} dx + \int_0^{\pi/2} \frac{\cos^8 x}{\sin^8 x + \cos^8 x} dx$

$2I = \int_0^{\pi/2} \frac{\sin^8 x + \cos^8 x}{\sin^8 x + \cos^8 x} dx$

$2I = \int_0^{\pi/2} 1 \, dx$

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$2I = [x]_0^{\pi/2}$

$2I = \frac{\pi}{2} - 0$

$2I = \frac{\pi}{2}$

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

Therefore, $\int_0^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} dx = \boxed{\frac{\pi}{4}}$