Integration via substitution, PLEASE HELP ME?

Question

Find the integral of sqr(4-x^2) using x =2 sin u

help!!!!

answer MUST be in terms of x

Answer 1

x=2sinu
dx = 2cosudu

√(4 - x²) = √[4 - (2sinu)²] = √[4 - 4sin²u] = √[4(1 - sin²u)]

√[4(1 - sin²u)] = √(4cos²u) = 2cosu

The integral using dx=2cosudu is

∫√(4 - x²)dx = ∫(2cosu)(2cosudu)

4∫cos²udu = 2∫[1+cos(2u)]du using cos²u = ½(1+cos2u)

2∫[1+cos(2u)]du = 2u + sin2u

You now need to solve for x in terms of u and resubstitute and solve the equation.

u=arcsin(½x) so 2u = 2arcsin(½x)

sin²u = ½[1 - sin(2u)] so sin(2u) = 1 - sin²u

But √(4 - x²) = √[4(1 - sin²u)] from above

so sin(2u) = ½√(4 - x²)

So finally

∫√(4 - x²)dx = ½√(4 - x²) + 2arcsin(½x)

Answer 2

I = ∫ √(2 - x).(2 + x).dx
let x = 2 sin u
dx = 2 cos u du
I = ∫ √(2 - 2 sin u).(2 + 2 sin u).2cosu du
I = ∫ √(4 - 4 sin ² u).2cos u.du
I = 4 ∫ √(1 - sin ² u).cos u.du
I = 4 ∫ cos ² u.du
I = 2 ∫ cos 2u + 1.du
I = sin 2u + 2 u + C
I = sin.(2.sin ^(-1) (x/2)) + 2.sin^(-1) (x / 2) + C

Answer 3

∫√(4 - x²)dx

= ∫√[4 - 4sin²u] . 2cosu du

= ∫√4cos²u . 2cosu du

= ∫4cos²u du

= ∫2(1 + cos2u) du

= 2u + sin2u + c

= 2u + 2 sinu cosu + c

= 2arcsin (x/2) + 2(x/2)√[1 - sin²u] + c

= 2arcsin(x/2) + x√(1 - x²) + c