Integrating the Reciprocal Square Root of a²−x²

The Simplest Trigonometric Substitution

The integral \(\int\frac{dx}{\sqrt{a^2-x^2}}\) is the cleanest example of trigonometric substitution. You may already know the answer, because the derivative of \(\arcsin\frac xa\) is \(\frac{1}{\sqrt{a^2-x^2}}\). The substitution \(x=a\sin\theta\) explains why, and it produces the same result by a method that works for all three radical forms.

Deriving the Formula

Let \(x=a\sin\theta\) with \(-\frac\pi2<\theta<\frac\pi2\). Then \(dx=a\cos\theta\,d\theta\) and \(\sqrt{a^2-x^2}=a\cos\theta\), which is positive on this open interval. The integral becomes

The cancellation is complete: the radical is exactly cancelled by the differential, leaving only \(\int d\theta\). No double-angle identity or triangle is needed, because \(\theta=\arcsin\frac xa\) is read directly from the substitution.

Why the Range of θ Matters

The sine function is not one-to-one, so the substitution \(x=a\sin\theta\) is a legitimate change of variable only on an interval where sine is invertible. The standard choice is \(-\frac\pi2<\theta<\frac\pi2\), where sine takes each value in \((-1,1)\) exactly once and \(\cos\theta>0\). This choice gives the principal arcsine, with outputs between \(-\frac\pi2\) and \(\frac\pi2\). If a different interval were used, such as \(\frac\pi2<\theta<\frac{3\pi}{2}\), we would have \(\cos\theta<0\) and the radical would equal \(-a\cos\theta\), which changes the sign of the answer.

A related antiderivative is \(-\arccos\frac xa+C\). It differs from \(\arcsin\frac xa\) by the constant \(\frac\pi2\), because \(\arcsin t+\arccos t=\frac\pi2\). Both are correct; the arcsine version is the standard one.

Checking by Differentiation

By the chain rule, \(\frac{d}{dx}\arcsin\frac xa=\frac{1}{\sqrt{1-(x/a)^2}}\cdot\frac1a=\frac{1}{a\sqrt{1-x^2/a^2}}=\frac{1}{\sqrt{a^2-x^2}}\). The last step uses \(a\sqrt{1-x^2/a^2}=\sqrt{a^2-x^2}\), valid for \(a>0\).

Worked Examples

Endpoints at ±a

At \(x=\pm a\) the integrand is unbounded, but the integral still converges. For instance, \(\int_0^{1}\frac{dx}{\sqrt{1-x^2}}=\lim_{t\to1^-}\arcsin t=\frac\pi2\). The improper integral exists because the singularity \(\frac{1}{\sqrt{1-x}}\) is integrable. For the generalized case, \(\int_0^a\frac{dx}{\sqrt{a^2-x^2}}=\frac\pi2\) for every \(a>0\), reflecting the fact that the integral is the angle swept from \(0\) to \(\frac\pi2\), independent of the radius.

A Reliable Procedure

Common Mistakes

The first mistake is to write \(\arcsin x\) instead of \(\arcsin\frac xa\); the radius must divide \(x\). The second is to forget the scaling factor when the coefficient of \(x^2\) is not \(1\), as in \(\sqrt{9-4x^2}\). The third is to apply the formula to \(\sqrt{a^2+x^2}\) or \(\sqrt{x^2-a^2}\), which lead to different antiderivatives.

Quick Reference

Practice Problems with Answers

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

1. Evaluate \(\int_0^1\frac{dx}{\sqrt{4-x^2}}\). Hint: use \(\arcsin\frac x2\). Answer: \(\frac\pi6\)
2. Evaluate \(\int_0^{3/2}\frac{dx}{\sqrt{9-4x^2}}\). Hint: include the factor \(\frac12\). Answer: \(\frac\pi4\)
3. Find an antiderivative of \(\frac{1}{\sqrt{5-(x-1)^2}}\). Hint: let \(u=x-1\). Answer: \(\arcsin\frac{x-1}{\sqrt5}\)
4. Evaluate \(\int_{-1}^{1}\frac{dx}{\sqrt{2-x^2}}\). Hint: the integrand is even. Answer: \(\frac\pi2\)

Geometric Meaning: Arc Length on a Circle

The integrand \(\frac{1}{\sqrt{1-x^2}}\) is exactly the arc-length element of the unit circle. If \(y=\sqrt{1-x^2}\), then \(y'=-\frac{x}{\sqrt{1-x^2}}\), and \(\sqrt{1+(y')^2}=\sqrt{1+\frac{x^2}{1-x^2}}=\frac{1}{\sqrt{1-x^2}}\). Integrating this element from \(0\) to \(x\) gives the arc length from the point \((0,1)\) to the point \((x,\sqrt{1-x^2})\), which is the angle \(\arcsin x\) measured from the \(y\)-axis. This is why the integral from \(0\) to \(1\) is \(\frac\pi2\): it is the length of a quarter of the unit circle.

Interpreting the Result

The antiderivative \(\arcsin\frac xa\) increases from \(-\frac\pi2\) to \(\frac\pi2\) as \(x\) goes from \(-a\) to \(a\). The slope is smallest at \(x=0\), where it equals \(\frac1a\), and increases without bound as \(x\to\pm a\). A radius larger than \(1\) flattens the curve: the angle changes more slowly with \(x\), by the factor \(\frac1a\). This explains why the scaling factor appears in the antiderivative of \(\frac{1}{\sqrt{9-4x^2}}\) as \(\frac12\), and why the result does not depend on the radius when the interval is scaled with it.

A Numerical Spot Check

Check \(\int_0^1\frac{dx}{\sqrt{4-x^2}}=\frac\pi6\approx0.5236\) by Simpson's rule with three points. The heights are \(0.5\), \(\frac{1}{\sqrt{3.75}}\approx0.5164\), and \(\frac{1}{\sqrt3}\approx0.5774\). Simpson's estimate is \(\frac{0.5}{3}(0.5+4\cdot0.5164+0.5774)\approx0.5238\), which agrees with \(0.5236\) to three digits.

When to Use the Formula Directly

Once the derivation is understood, there is no need to repeat the substitution each time. Treat \(\int\frac{du}{\sqrt{a^2-u^2}}=\arcsin\frac ua\) as a standard form, just as you treat \(\int\frac{du}{u}=\ln|u|\). The only work is to recognize the form: identify the radius \(a\), identify the variable \(u\), and track any constant multiple of \(x\) inside \(u\). The trigonometric substitution is the fallback when the integrand has additional factors that the standard form does not cover, as in the tutorials that follow.

Domain and Continuity

The antiderivative \(\arcsin\frac xa\) is continuous on the closed interval \([-a,a]\) but differentiable only on the open interval \((-a,a)\). A definite integral with endpoints in the open interval is a proper integral. An endpoint at \(\pm a\) makes the integral improper, although it converges. If an interval extends beyond \([-a,a]\), the integrand is not a real number, so the integral is meaningless. Always check the radicand before choosing endpoints; for example, \(\int_0^3\frac{dx}{\sqrt{4-x^2}}\) is undefined because the integrand is not real for \(x>2\).

Where This Fits in the Course

This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Integrating the Square Root of a²−x²,” and the one that follows is “Polynomial Factors Over √(a²−x²).” It is the simplest instance of the sine substitution, with no leftover trigonometric integral, which makes it a good model for how the substitution cancels the radical and leaves only the angle itself as the answer.