Evaluate ∫ (1 - x^2)^(1/2) dx. What is the result?

Evaluate ∫ (1 - x^2)^(1/2) dx. What is the result?

• A. (1/2)(x)(1 - x^2)^(1/2) + (1/2)arcsin(x) + C
• B. (1/2)(x)(1 + x^2)^(1/2) + arcsin(x) + C
• C. (1/3)(x)(1 - x^2)^(1/2) + C
• D. (1/2)(1 - x^2)^(3/2) + C

Answer: (A)

The integral ∫ (1 - x^2)^(1/2) dx requires a method of integration that often involves trigonometric substitution due to its square root structure. To evaluate this integral, we can use the substitution x = sin(θ), which then transforms the expression inside the square root: 1 - x^2 = 1 - sin^2(θ) = cos^2(θ), leading to (1 - x^2)^(1/2) = cos(θ). Furthermore, dx becomes cos(θ) dθ. Thus, the integral can be rewritten as: ∫ cos(θ) cos(θ) dθ = ∫ cos^2(θ) dθ. Using the identity cos^2(θ) = (1 + cos(2θ))/2, the integral becomes: ∫ (1 + cos(2θ))/2 dθ = (1/2) ∫ dθ + (1/2) ∫ cos(2θ) dθ. The first integral yields (1/2)θ, and the second integral yields (1/4)sin(2θ) + C when evaluated. Combining these results gives: (1

The integral ∫ (1 - x^2)^(1/2) dx requires a method of integration that often involves trigonometric substitution due to its square root structure.

To evaluate this integral, we can use the substitution x = sin(θ), which then transforms the expression inside the square root:

1 - x^2 = 1 - sin^2(θ) = cos^2(θ), leading to (1 - x^2)^(1/2) = cos(θ). Furthermore, dx becomes cos(θ) dθ. Thus, the integral can be rewritten as:

∫ cos(θ) cos(θ) dθ = ∫ cos^2(θ) dθ.

Using the identity cos^2(θ) = (1 + cos(2θ))/2, the integral becomes:

∫ (1 + cos(2θ))/2 dθ = (1/2) ∫ dθ + (1/2) ∫ cos(2θ) dθ.

The first integral yields (1/2)θ, and the second integral yields (1/4)sin(2θ) + C when evaluated. Combining these results gives:

(1