Integrating the Square Root of a²−x²

Why a Radical Needs a Trigonometric Substitution

The integral \(\int\sqrt{a^2-x^2}\,dx\) cannot be done with an algebraic substitution such as \(u=a^2-x^2\), because the factor \(x\) needed for \(du\) is missing. The Pythagorean identity offers a better idea. If \(x=a\sin\theta\), then \(a^2-x^2=a^2-a^2\sin^2\theta=a^2\cos^2\theta\), and the square root becomes a trigonometric function with no radical.

Geometrically, the graph of \(y=\sqrt{a^2-x^2}\) is the upper half of a circle of radius \(a\). Its area from \(0\) to \(a\) is a quarter of a circle, so the answer must reproduce \(\frac{\pi a^2}{4}\). This gives an excellent check on the final formula.

Deriving the Formula

Let \(x=a\sin\theta\) with \(-\frac\pi2\le\theta\le\frac\pi2\). Then \(dx=a\cos\theta\,d\theta\), and \(\sqrt{a^2-x^2}=a\cos\theta\) because \(\cos\theta\ge0\) on this range. Therefore

Integrating gives \(\frac{a^2}{2}\bigl(\theta+\frac{\sin2\theta}{2}\bigr)=\frac{a^2}{2}(\theta+\sin\theta\cos\theta)\), using \(\sin2\theta=2\sin\theta\cos\theta\). Now return to \(x\). From \(x=a\sin\theta\), we have \(\theta=\arcsin\frac xa\), \(\sin\theta=\frac xa\), and, from the right triangle with hypotenuse \(a\) and opposite side \(x\), \(\cos\theta=\frac{\sqrt{a^2-x^2}}{a}\). Hence

Checking by Differentiation

The derivative of \(\frac{a^2}{2}\arcsin\frac xa\) is \(\frac{a^2}{2}\cdot\frac{1}{\sqrt{a^2-x^2}}\). The derivative of \(\frac x2\sqrt{a^2-x^2}\) is \(\frac12\sqrt{a^2-x^2}-\frac{x^2}{2\sqrt{a^2-x^2}}\). Combining over the common denominator \(2\sqrt{a^2-x^2}\), the numerator is \(a^2+(a^2-x^2)-x^2=2(a^2-x^2)\), so the sum equals \(\sqrt{a^2-x^2}\), as required.

Worked Examples

A Reliable Procedure

Common Mistakes

A frequent error is to leave the answer in terms of \(\theta\), or to forget that \(\sin2\theta=2\sin\theta\cos\theta\) introduces both \(x\) and the radical when returned to \(x\). Another is to write \(\sqrt{a^2-x^2}=a-x\) or similar illegal simplifications. The radical of a difference is not a difference of radicals.

A third mistake is dropping the coefficient \(\frac{a^2}{2}\) from the arcsine term; for \(a=3\) it is \(\frac92\), not \(\frac32\).

Quick Reference

Interpreting the Result

The antiderivative has two terms with distinct geometric meanings. The term \(\frac x2\sqrt{a^2-x^2}\) is the area of a right triangle with legs \(x\) and \(\sqrt{a^2-x^2}\), inscribed in the circle. The term \(\frac{a^2}{2}\arcsin\frac xa\) is the area of a circular sector with angle \(\arcsin\frac xa\). Together they give the area of the region under the arc from \(0\) to \(x\), a triangle plus a sector.

A Numerical Spot Check

Compare the area \(1.913\) from the second example with a trapezoid estimate on \([0,1]\). The heights are \(2\) at \(x=0\) and \(\sqrt3\approx1.732\) at \(x=1\), so the estimate is \(\frac12(2+1.732)=1.866\). It is slightly low because the curve is concave and the chord lies below it, which is what the exact answer \(1.913\) shows.

Practice Problems with Answers

Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.

1. Evaluate \(\int_0^2\sqrt{4-x^2}\,dx\). Hint: it is a quarter circle of radius \(2\). Answer: \(\pi\)
2. Evaluate \(\int_0^1\sqrt{4-x^2}\,dx\). Hint: use \(\arcsin\frac12=\frac\pi6\). Answer: \(\frac{\sqrt3}{2}+\frac\pi3\)
3. Find an antiderivative of \(\sqrt{9-4x^2}\). Hint: first let \(u=2x\). Answer: \(\frac x2\sqrt{9-4x^2}+\frac94\arcsin\frac{2x}{3}\)
4. Evaluate \(\int_0^3\sqrt{25-x^2}\,dx\). Hint: \(\sqrt{25-9}=4\). Answer: \(6+\frac{25}{2}\arcsin\frac35\)

Reading the Answer as Triangle Plus Sector

Place the point \((x,\sqrt{a^2-x^2})\) on the circle \(x^2+y^2=a^2\) in the upper half-plane. The region under the arc from \(0\) to \(x\) splits into a right triangle with vertices \((0,0)\), \((x,0)\), and \((x,\sqrt{a^2-x^2})\), and a circular sector from the \(y\)-axis to the ray through that point. The triangle has area \(\frac12x\sqrt{a^2-x^2}\). The sector has central angle \(\arcsin\frac xa\), because the angle is measured from the \(y\)-axis, and radius \(a\), so its area is \(\frac12a^2\arcsin\frac xa\). Adding them reproduces the antiderivative exactly. This argument gives the formula without calculus, and it is a good way to remember it.

Negative Values of x

The formula holds for \(-a\le x\le a\), including negative \(x\). In that range \(\arcsin\frac xa\) is negative, and the triangle term \(\frac x2\sqrt{a^2-x^2}\) is negative too, so the antiderivative is an odd function. The definite integral over a symmetric interval of the even integrand \(\sqrt{a^2-x^2}\) is therefore twice the integral over \([0,a]\). For example, \(\int_{-a}^{a}\sqrt{a^2-x^2}\,dx=2\cdot\frac{\pi a^2}{4}=\frac{\pi a^2}{2}\), the area of a semicircle.

Why the Substitution Range Matters

Restricting \(\theta\) to \([-\frac\pi2,\frac\pi2]\) guarantees that \(\cos\theta\ge0\), so \(\sqrt{\cos^2\theta}=\cos\theta\) with no absolute value. If a different range were used, the sign could flip and the answer would be wrong by a sign. Choosing the principal range of the inverse sine automatically gives the correct branch.

Using the Result as a Building Block

This antiderivative is the starting point for many applications. The arc length of an ellipse, the area of an ellipse, and the volume of a sphere by slicing all lead to integrals of the form \(\int\sqrt{a^2-x^2}\,dx\) or a close variant. For example, the sphere of radius \(a\) has cross-sections of area \(\pi(a^2-x^2)\), which integrates without trigonometry, but the circle area under the arc itself, \(\frac{\pi a^2}{2}\) for the semicircle, relies on the formula derived here. Keep the two-term structure in mind: a radical term that vanishes at \(x=\pm a\), and an arcsine term that supplies the angle.

Where This Fits in the Course

This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Deriving the Sine Substitution for a²−x²,” and the one that follows is “Integrating the Reciprocal Square Root of a²−x².” It is the first of the trigonometric substitutions, built on the Pythagorean identity, and the next tutorials adapt the same idea to the other two radical forms.