Which is the standard antiderivative of sqrt(a^2 - x^2) dx?

Which is the standard antiderivative of sqrt(a^2 - x^2) dx?

• A. (x/2) sqrt(a^2 - x^2) + (a^2/2) arcsin(x/a) + C
• B. (x/2) sqrt(a^2 - x^2) + (a^2/2) arccos(x/a) + C
• C. (x/2) sqrt(a^2 - x^2) + (a^2/2) arctan(x/a) + C
• D. -(x/2) sqrt(a^2 - x^2) - (a^2/2) arcsin(x/a) + C

Answer: (A)

The integral uses a trig substitution: set x = a sin θ, so dx = a cos θ dθ and sqrt(a^2 − x^2) = a cos θ. The integral becomes ∫ a^2 cos^2 θ dθ, which simplifies to (a^2/2)(θ + sin θ cos θ) + C. Substituting back, θ = arcsin(x/a), sin θ = x/a, and cos θ = sqrt(a^2 − x^2)/a, giving (x/2) sqrt(a^2 − x^2) + (a^2/2) arcsin(x/a) + C. This is the standard antiderivative. The form with arccos would not produce the correct derivative to match sqrt(a^2 − x^2). The arctan form also fails to match, and the negative version would differentiate to the negative of the integrand. Therefore the arcsin form is the correct one.

The integral uses a trig substitution: set x = a sin θ, so dx = a cos θ dθ and sqrt(a^2 − x^2) = a cos θ. The integral becomes ∫ a^2 cos^2 θ dθ, which simplifies to (a^2/2)(θ + sin θ cos θ) + C. Substituting back, θ = arcsin(x/a), sin θ = x/a, and cos θ = sqrt(a^2 − x^2)/a, giving (x/2) sqrt(a^2 − x^2) + (a^2/2) arcsin(x/a) + C. This is the standard antiderivative.

The form with arccos would not produce the correct derivative to match sqrt(a^2 − x^2). The arctan form also fails to match, and the negative version would differentiate to the negative of the integrand. Therefore the arcsin form is the correct one.