Question

∫ _0^(1/2)xdx/√(1-x²)

the half should be at the top of this ∫ and the zero at the bottom.

Answer 1

Ah, no problem

We had for the indefinate integral

Integral = -sqrt(1-x^2)
so we need
-sqrt(1-x^2) at x = 1/2
minus
-sqrt(1-x^2) at x = 0
or
-sqrt (1-1/4) + sqrt(1)
=1 - sqrt(3/4)
= 1 -(1/2) sqrt(3)

Answer 2

or we could have done it in t

integral = -cos t
where x = sin t
when x = 0, t = 0 and cos t = 1
when x = 1/2 , t = 30 degrees or pi/6
then cos t = (1/2)sqrt 3
so
(1/2) sqrt(3) - 1

sign difference is due to square root could be + or -

Answer 3

sign ambiguities
-cos pi/6 = -(1/2)sqrt(3)

-(-cos 0) = +1
so
- (1/2) sqrt(3) -1

when sin t = 0, t could be 0 or 180 (pi)
so cos t could be +1 or -1
when sin t = 1/2
t could be pi/6 or 5 pi/6
and cos t could be -(1/2) sqrt(3)
so that sqrt sign makes the signs of the answer tricky