∫ _0^(1/2)xdx/√(1-x²)
the half should be at the top of this ∫ and the zero at the bottom.
To evaluate the integral
∫ (0 to 1/2) x dx / √(1 - x²),
we can use a trigonometric substitution. Let's substitute x = sinθ.
When x = sinθ, we can find dx by differentiating both sides with respect to θ:
dx = cosθ dθ.
Next, we need to find the limits of integration in terms of θ.
When x = 0, sinθ = 0, so θ = 0.
When x = 1/2, sinθ = 1/2. By considering the unit circle, we know that sin(π/6) = 1/2. Therefore, θ = π/6.
Now, substitute x = sinθ and dx = cosθ dθ into the integral:
∫ (0 to 1/2) x dx / √(1 - x²)
= ∫ (0 to π/6) sinθ cosθ dθ / √(1 - sin²θ)
= ∫ (0 to π/6) sinθ cosθ dθ / cosθ
= ∫ (0 to π/6) sinθ dθ.
Using the identity ∫ sinθ dθ = -cosθ + C, we can now integrate:
∫ (0 to π/6) sinθ dθ
= [-cosθ] (0 to π/6)
= -cos(π/6) + cos(0)
= -√3/2 + 1
= 1 - √3/2.
Therefore, the value of the integral ∫ (0 to 1/2) x dx / √(1 - x²) is 1 - √3/2.