∫ _0^(1/2)xdx/√(1-x²)

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

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)

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 -

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

To solve this integral, you can use a trigonometric substitution. Let's set x = sin(t), where t is the angle in the first quadrant such that sin(t) = x.

Differentiating both sides with respect to t, we have dx = cos(t) dt.

Now, let's substitute these values into the integral:

∫ (0 to 1/2) xdx/√(1-x²)
= ∫ (0 to π/6) sin(t)(cos(t) dt / √(1-sin²(t)))

Since sin(t) = x, we can substitute this in as well:

= ∫ (0 to π/6) xcos(t) dt / √(1-x²)

Next, we can use the trigonometric identity: cos²(t) = 1 - sin²(t)

Rearranging, we have sin²(t) = 1 - cos²(t)

Substituting this into the integral, we get:

= ∫ (0 to π/6) x√(1 - cos²(t)) dt / √(1-x²)

Rewriting the integral in terms of x:

= ∫ (0 to π/6) x√(sin²(t)) dt / √(1-x²)
= ∫ (0 to π/6) xsin(t) dt / √(1-x²)

Since we have x = sin(t), we can substitute this value into the integral:

= ∫ (0 to π/6) (sin(t))(sin(t)) dt / √(1-sin²(t))
= ∫ (0 to π/6) (sin²(t)) dt / √(1-sin²(t))

Now, we have a simpler integral to evaluate. We can solve it by using a trigonometric identity:

∫ (0 to π/6) (sin²(t)) dt = ∫ (0 to π/6) (1 - cos²(t)) dt
= ∫ (0 to π/6) dt - ∫ (0 to π/6) (cos²(t)) dt

Simplifying further:

= t|_(0 to π/6) - [ (∫ (0 to π/6) (1 + cos(2t))/2 dt ]
= π/6 - 1/2 ( [t + (sin(2t))/2] | _(0 to π/6) )
= π/6 - 1/2 ( (π/6) + (sin(2π/6))/2 - (0 + (sin(0))/2) )
= π/6 - 1/2 (π/6 + 1/2)

By simplifying further, we get:

= π/6 - π/12 - 1/4
= π/12 - 1/4

So, the value of the integral ∫ (0 to 1/2) xdx/√(1-x²) is π/12 - 1/4.