AP Calculus BC

Integral of sqrt(1-x^2): The Circle Answer

The integral of the square root of 1 minus x squared is x times the square root of 1 minus x squared, plus arcsin x, all over 2, plus C. The technique that produces it is trigonometric substitution, which is beyond AP, but the definite integral appears on AP as the area of a circular region.

∫1−x2 dx=x1−x2+arcsin⁡x2+C\int \sqrt{1-x^{2}}\,dx = \frac{x\sqrt{1-x^{2}} + \arcsin x}{2} + C

Recognise the geometry instead

The graph of y=1−x2y = \sqrt{1-x^{2}} is the upper half of the unit circle, so a definite integral over [−1,1][-1, 1] is just the area of a semicircle. That is the AP-level route, and it needs no antiderivative at all.

∫−111−x2 dx=π2\int_{-1}^{1}\sqrt{1-x^{2}}\,dx = \frac{\pi}{2}

Geometry beats technique here

When an integrand is a circle, a semicircle, or a triangle, computing the area directly is faster and fully rigorous. AP free response uses this constantly on graph-defined functions.

Where the closed form comes from

The substitution x=sin⁡θx = \sin\theta turns the radical into cos⁡θ\cos\theta and the integral into ∫cos⁡2θ dθ\int \cos^{2}\theta\,d\theta, which the power reducing identity finishes. Trigonometric substitution is NOT on the AP syllabus, so the formula is given here for completeness rather than as a required method.

Checking it by differentiation is on the syllabus, and it works: the product rule and the arcsine derivative combine to give back 1−x2\sqrt{1-x^{2}}.

Common mistakes

  • Writing 1−x2=1−x\sqrt{1-x^{2}} = 1 - x. At x=0.6x = 0.6 that claims 0.40.4 where the true value is 0.80.8.
  • Reaching for the antiderivative when the interval makes it a recognisable circular area.
  • Forgetting the domain restriction −1≤x≤1-1 \le x \le 1.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of sqrt(1-x^2)?

It is x1−x2+arcsin⁡x2+C\frac{x\sqrt{1-x^{2}} + \arcsin x}{2} + C.

Is this on the AP exam?

The antiderivative technique is not; trigonometric substitution is outside the syllabus. The definite integral does appear, as the area of a circular region.

What is the integral from -1 to 1?

π2\frac{\pi}{2}, the area of the upper half of the unit circle.