From 0 to 2π, the integral can be changed into the integral of (2π-x) (sinx) 8.

When the two integrals are added, we can see that

The integral of (sinx)^8, the original integral = π times between 0 and 2π.

(sinx)^8 =4π times the integral from 0 to π/2.

Reuse wallis formula

Original integral = 4 π× (7/8 )× (5/6 )× (3/4 )× (1/2 )× (π/2) = (35/64) π?