int_(-2)^2sqrt(4-x^2)text()dx evaluate the integral

To evaluate the given integral ∫((-2)^(2)√(4-x^2)) dx, we'll start by simplifying the integrand.

Let's break down the steps:

Step 1: Rewrite the integral. The bounds of integration are not given, so we'll assume it is indefinite.

∫((-2)^(2)√(4-x^2)) dx

Step 2: Simplify the expression within the square root.

√(4-x^2) represents the equation of a circle centered at the origin with a radius of 2. By rewriting it as √(2^2 - x^2), we recognize it as the equation for a semicircle.

Step 3: Change the form of the integral:
∫((-2)^(2)√(4-x^2)) dx = ∫(4√(4-x^2)) dx

Step 4: Make a trigonometric substitution:
Let x = 2sin(θ), where -π/2 ≤ θ ≤ π/2.

Step 5: Calculate dx:
Differentiate x = 2sin(θ) with respect to θ to get dx in terms of dθ:
dx = 2cos(θ) dθ

Step 6: Substitute x and dx in terms of θ:
When x = -2, θ = -(π/2) and when x = 2, θ = π/2.

The integral becomes: ∫(4√(4-(2sin(θ))^2)) (2cos(θ)) dθ from -(π/2) to (π/2)

Step 7: Simplify the integrand:
By using the trigonometric identity sin^2(θ) + cos^2(θ) = 1, we can simplify the expression inside the square root.

√(4-(2sin(θ))^2) = √(4-4sin^2(θ)) = √(4cos^2(θ)) = 2|cos(θ)|

Step 8: Substitute the simplified integrand back into the integral:
∫(4√(4-(2sin(θ))^2)) (2cos(θ)) dθ
= ∫(4)(2|cos(θ)|)(2cos(θ)) dθ

Step 9: Simplify the integrand further:
∫(4)(2|cos(θ)|)(2cos(θ)) dθ = 16|cos(θ)|cos(θ) dθ

Step 10: Evaluate the integral:
∫16|cos(θ)|cos(θ) dθ = 16∫cos^2(θ) dθ

Using the double-angle formula for cosine (cos(2θ) = 2cos^2(θ) - 1), we can transform this into the form of a known integral:

cos^2(θ) = (1 + cos(2θ))/2

Therefore, we have:

∫16|cos(θ)|cos(θ) dθ = 16∫((1 + cos(2θ))/2) dθ

Step 11: Integrate:
∫((1 + cos(2θ))/2) dθ = (1/2)∫(1 + cos(2θ)) dθ
= (1/2)(∫1 dθ + ∫cos(2θ) dθ)
= (1/2)(θ + (sin(2θ)/2))

Step 12: Evaluate the definite integral:
Evaluate the above expression from -π/2 to π/2:

((1/2)(θ + (sin(2θ)/2))) | -π/2 to π/2

Plug in the upper and lower bounds and evaluate:

((1/2)(π/2 + (sin(2(π/2))/2))) - ((1/2)(-(π/2) + (sin(2(-(π/2)))/2)))

Simplifying further:

((1/2)(π/2 + sin(π)/2) - (1/2)(-π/2 + sin(-π)/2))
= ((1/2)(π/2 + 0/2) - (1/2)(-(π/2) + 0/2))
= ((1/2)(π/2) - (1/2)(-π/2))
= (π/4 + π/4)
= π/2

Therefore, the value of the integral ∫((-2)^(2)√(4-x^2)) dx is π/2.