AP Calculus AB and BC

Integral of 1/sqrt(9-x^2): Answer and Proof

The integral of 1/sqrt(9-x^2) is arcsin(x/3) + C. The integrand matches the standard form 1 over the square root of a^2 minus x^2 with a = 3, whose antiderivative is arcsin(x/a) with no coefficient in front. Differentiating arcsin(x/3) returns the integrand.

∫19−x2 dx=arcsin⁡x3+C\int \frac{1}{\sqrt{9-x^2}}\,dx = \arcsin\frac{x}{3} + C

Matching the arcsine form

A square root of a constant minus x2x^2 is the arcsine pattern. The constant under the root names a2a^2, so a2=9a^2 = 9 and a=3a = 3.

∫1a2−x2 dx=arcsin⁡xa+C\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \arcsin\frac{x}{a} + C
∫19−x2 dx=arcsin⁡x3+C\int \frac{1}{\sqrt{9-x^2}}\,dx = \arcsin\frac{x}{3} + C

Substitution shows why no coefficient survives. With x=3ux = 3u and dx=3 dudx = 3\,du, the root becomes 31−u23\sqrt{1-u^2}, and the two threes cancel.

∫3 du31−u2=∫du1−u2=arcsin⁡u+C=arcsin⁡x3+C\int \frac{3\,du}{3\sqrt{1-u^2}} = \int \frac{du}{\sqrt{1-u^2}} = \arcsin u + C = \arcsin\frac{x}{3} + C

Checking by differentiating

Use ddxarcsin⁡u=u′1−u2\frac{d}{dx}\arcsin u = \frac{u'}{\sqrt{1-u^2}} with u=x3u = \frac{x}{3} and u′=13u' = \frac{1}{3}.

ddxarcsin⁡x3=131−x29=13139−x2=19−x2\frac{d}{dx}\arcsin\frac{x}{3} = \frac{\frac{1}{3}}{\sqrt{1-\frac{x^2}{9}}} = \frac{\frac{1}{3}}{\frac{1}{3}\sqrt{9-x^2}} = \frac{1}{\sqrt{9-x^2}}

Pulling 13\frac{1}{3} out of the root cancels the 13\frac{1}{3} on top, which is why the arcsine form needs no leading fraction.

The mistakes students make

  • Writing 13arcsin⁡x3\frac{1}{3}\arcsin\frac{x}{3} by copying the 1a\frac{1}{a} from the arctangent formula. The square root already supplies that factor, so arcsine takes none.
  • Using a=9a = 9 and answering arcsin⁡x9\arcsin\frac{x}{9}. The number under the root is a2a^2.
  • Confusing the pattern with ∫19−x2 dx\int\frac{1}{9-x^2}\,dx, which has no root and is a partial fractions problem instead.
  • Ignoring the domain. The integrand is defined only for −3<x<3-3 < x < 3, so limits of integration outside that interval make the integral improper or undefined.

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Frequently asked questions

Why is there no 1/3 in front, when the arctangent version has a 1/a?

The square root holds a2a^2, so factoring it out produces an aa that cancels the 1a\frac{1}{a} from the chain rule. In the arctangent form the denominator has a2a^2 with no root, and nothing cancels.

What is the integral of 1/sqrt(9-x^2) from 0 to 3/2?

∫03/219−x2 dx=[arcsin⁡x3]03/2=arcsin⁡12=π6\int_0^{3/2}\frac{1}{\sqrt{9-x^2}}\,dx = \left[\arcsin\frac{x}{3}\right]_0^{3/2} = \arcsin\frac{1}{2} = \frac{\pi}{6}.

Is arccos(x/3) also an antiderivative?

It differs by a sign, so no. Since arcsin⁡x3+arccos⁡x3=π2\arcsin\frac{x}{3} + \arccos\frac{x}{3} = \frac{\pi}{2}, the function −arccos⁡x3-\arccos\frac{x}{3} is an antiderivative, differing from arcsin⁡x3\arcsin\frac{x}{3} only by a constant.