Integrate ∫2xdx/sqrt(1-x^2) from 0 to 1/2 =

Please show how you solved it. Thanks!

∫ 2xdx / √(1-x²) from 0 to 1/2
u = 1-x² → du = -2x dx
x dx = -(1/2) du
= 2 ∫ (-½)/u^½ du
= - ∫ u^-½ du
= -2 u^1/2
……………..1/2
= -2[√(1 – x²)]
………………0
= -2[√(1 – (1/2)²) - √(1 – 0²)]
= -2[√(3/4) – 1]
= -2[½√3 – 1]
= -√3 + 2
= 0.26795

let sqrt(1-x^2)=t
-2x/2sqrt(1-x^2)dx=dt
Now, limit changes too.x->0,t->1 and as x->1/2,t->sqrt3/2
I=int (1 to sqrt3/2) (-2) dt
I=-2[t](1 to sqrt 3/2)
I=-2(sqrt3/2-1)
..plz check this..