Evaluate the int (4 - x^2)^(3/2) from 0 to 1

The expression under the radical, 4 - x^2, suggests the following substitution:
x = 2sin(u) ==> dx = 2cos(u) du.

Our new limits of integration are:
(a) Lower bound: x = 0 ==> sin(u) = 0 ==> u = 0
(b) Upper bound: x = 1 ==> sin(u) = 1/2 ==> u = π/6.

Applying these substitutions and bound changes gives:
∫ (4 - x^2)^(3/2) dx (from x=0 to 1)
= ∫ 2cos(u)[4 - 4sin^2(u)]^(3/2) du (from u=0 to π/6)
= 16 ∫ cos(u)[1 - sin^2(u)]^(3/2) du (from u=0 to π/6), by pulling out constants
= 16 ∫ cos^4(u) du (from u=0 to π/6), since cos^2(u) = 1 - sin^2(u).

In order to integrate ∫ cos^4(u) du, we need to use the double-angle formula for cosine:
cos(2u) = 2cos^2(u) - 1 ==> cos^2(u) = [1 + cos(2u)]/2. (*)

So:
cos^4(u) = [cos^2(u)]^2
= [1 + cos(2u)]^2/4, by applying *(*)
= [cos^2(2u) + 2cos(2u) + 1]/4, by expanding
= [(1/2)cos(4u) + 2cos(2u) + 3/2]/4, by applying (*) again
= (1/8)cos(4u) + (1/2)cos(2u) + 3/8.

Therefore, the integral in question equals:
16 ∫ cos^4(u) du (from u=0 to π/6)
= 16 ∫ [(1/8)cos(4u) + (1/2)cos(2u) + 3/8] du (from u=0 to π/6)
= 2 ∫ [cos(4u) + 4cos(2u) + 3] du (from u=0 to π/6)
= [(1/2)sin(4u) + 2sin(2u) + 3u] (evaluated from u=0 to π/6)
= (√3/4 + √3 + π/2) - (0 + 0 + 0)
= 5√3/4 + π/2
= (5√3 + 2π)/4.

I hope this helps!