Compute ∫ sqrt(9 - x^2) dx. Which expression is correct?

Compute ∫ sqrt(9 - x^2) dx. Which expression is correct?

• A. (x/2) sqrt(9 - x^2) - (9/2) arcsin(x/3) + C
• B. (x/2) sqrt(9 - x^2) + (9/2) arccos(x/3) + C
• C. (x/2) sqrt(9 - x^2) + (9/2) arcsin(x/3) + C
• D. -(x/2) sqrt(9 - x^2) + (9/2) arcsin(x/3) + C

Answer: (C)

Using a trig substitution for the form sqrt(a^2 - x^2) is the natural approach. Let x = a sin θ, so dx = a cos θ dθ and sqrt(a^2 - x^2) = sqrt(a^2 - a^2 sin^2 θ) = a cos θ. The integral becomes ∫ (a cos θ)(a cos θ) dθ = a^2 ∫ cos^2 θ dθ. Since cos^2 θ = (1 + cos 2θ)/2, this is a^2 ∫ (1 + cos 2θ)/2 dθ = (a^2/2) θ + (a^2/4) sin 2θ + C. Convert back: θ = arcsin(x/a) and sin 2θ = 2 sin θ cos θ = 2 (x/a) sqrt(1 - x^2/a^2) = 2x sqrt(a^2 - x^2) / a^2. The second term becomes (a^2/4)·(2x sqrt(a^2 - x^2)/a^2) = (x/2) sqrt(a^2 - x^2). So the antiderivative is (x/2) sqrt(a^2 - x^2) + (a^2/2) arcsin(x/a) + C. Plugging in a = 3 gives (x/2) sqrt(9 - x^2) + (9/2) arcsin(x/3) + C. Equivalent forms using arccos are also valid up to a constant, since arcsin and arccos differ by a constant.

Using a trig substitution for the form sqrt(a^2 - x^2) is the natural approach. Let x = a sin θ, so dx = a cos θ dθ and sqrt(a^2 - x^2) = sqrt(a^2 - a^2 sin^2 θ) = a cos θ. The integral becomes ∫ (a cos θ)(a cos θ) dθ = a^2 ∫ cos^2 θ dθ. Since cos^2 θ = (1 + cos 2θ)/2, this is a^2 ∫ (1 + cos 2θ)/2 dθ = (a^2/2) θ + (a^2/4) sin 2θ + C.

Convert back: θ = arcsin(x/a) and sin 2θ = 2 sin θ cos θ = 2 (x/a) sqrt(1 - x^2/a^2) = 2x sqrt(a^2 - x^2) / a^2. The second term becomes (a^2/4)·(2x sqrt(a^2 - x^2)/a^2) = (x/2) sqrt(a^2 - x^2). So the antiderivative is

(x/2) sqrt(a^2 - x^2) + (a^2/2) arcsin(x/a) + C.

Plugging in a = 3 gives

(x/2) sqrt(9 - x^2) + (9/2) arcsin(x/3) + C.

Equivalent forms using arccos are also valid up to a constant, since arcsin and arccos differ by a constant.