question about integration?

∫√(1-x²/4)dx

Let x = 2sinθ
dx = 2cosθdθ
x² = 4sin²θ
x²/4 = sin²θ
1-x²/4 = cos²θ
√(1-x²/4) = cosθ

∫√(1-x²/4)dx
= ∫cosθ.2cosθdθ
= ∫2cos²θdθ

cos(2θ) = 2cos²θ - 1
2cos²θ = (1+cos(2θ))

∫2cos²θdθ
= ∫(1+cos(2θ))dθ
= θ + sin(2θ)/2 + c
= θ + sinθcosθ + c
= sinֿ¹(x/2) + (x/2)√(1-x²/4) + c

Let sin(y) = x/2

Then cos(y)dy = dx/2, so dx = 2cos(y)dy

Then using sin^2(y) + cos^2(y) = 1:
cos(y) = sqrt(1 - sin^2(y)) = sqrt(1 - x^2/4)

Now replace dx with 2cos(y)dy and replace sqrt(1 - x^2/4) with cos(y):

Int[sqrt(1-x^2/4)]dx = Int[cos(y)]2cos(y)dy = 2*Int[cos^2(y)]dy

Using the power reducing formula: cos^2(y) = [1 + cos(2y)]/2

2*Int[cos^2(y)]dy = Int[1 + cos(2y)]dy

This simplifies to y + (1/2)sin(2y) + C

Plugging in y = arcsin(x/2) gives:

arcsin(x/2) + (1/2)sin[2arcsin(x/2)] + C

which can be simplified to:
arcsin(x/2) + (x/2)*sqrt(1-x^2/4) + C

pie?