First you have to know its domain, which you can find to be -8 ≤ 0 ≤ 8 as the function neither exists for values of x less than -8 nor for values of x greater than 8.
∫{ √(64-x²)/16 }dx, limits from -8 to 8
= (1/16)[ ∫√(64-x²)dx]
Since the function is even,
= (2/16)[ ∫√(64-x²)dx]
= (1/8) ∫√(64-x²)dx
There is a formula that,
∫√(a²-x²)dx = (x/2)√(a²-x²) + a²/2[sin^-1 (x/a)] + c
Here a=8.
Substituting these in the above formula.
= (1/8){ (x/2)√(64-x²) + 64/2[sin^-1 (x/8)] }
= (1/8){ (x/2)√(64-x²) + 32[sin^-1 (x/8)] }
Applying the limits,
= (1/8){ (8/2)√(64-64) + 32[sin^-1 (8/8)] - (0/2)√(64-0) - 32[sin^-1 (0/8)] }
= (1/8){0+ 32[sin^-1 (1)] - 0 - 32[sin^-1 (0)] }
= (1/8){ 32[sin^-1 (1)] }
= (32/8)(π/2)
= 4π/2
= 2π