Deriving the Sine Substitution for a²−x²

Why a Sine Substitution Fits This Radicand

A square root such as \(\sqrt{a^2-x^2}\) contains a difference of squares. The Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\) has the same structure: rearranging it gives \(1-\sin^2\theta=\cos^2\theta\). This suggests expressing \(x\) as \(a\sin\theta\), so the radicand becomes a multiple of \(\cos^2\theta\).

The algebra alone is not enough. A square root is nonnegative, while \(\cos\theta\) can be negative for some angles. We therefore need an angle range on which the cosine is nonnegative, and we must differentiate the substitution to transform \(dx\) correctly.

Deriving the Substitution from a Right Triangle

First consider \(0

Let \(\theta\) be the acute angle opposite the side of length \(x\). The definition of sine gives \(\sin\theta=x/a\), so \(x=a\sin\theta\). The definition of cosine gives \(\cos\theta=\sqrt{a^2-x^2}/a\), so the radical equals \(a\cos\theta\).

The triangle directly represents values \(0

The identity for the radical also holds on this full interval. Substituting \(x=a\sin\theta\) gives \(a^2-x^2=a^2(1-\sin^2\theta)=a^2\cos^2\theta\). Taking the nonnegative square root gives \(a|\cos\theta|\). Because \(\cos\theta\ge 0\) throughout the selected interval and \(a>0\), this is \(a\cos\theta\).

The Differential and the Integral Transformation

Differentiate \(x=a\sin\theta\) with respect to \(\theta\). Since \(a\) is constant, the chain rule gives \(dx/d\theta=a\cos\theta\), or \(dx=a\cos\theta\,d\theta\). With the radical identity, this means that \(dx\) and \(\sqrt{a^2-x^2}\) have the same factor \(a\cos\theta\) in this angle range.

Consequently, an integral in \(x\) can be rewritten by replacing each occurrence of \(x\) with \(a\sin\theta\), each occurrence of \(\sqrt{a^2-x^2}\) with \(a\cos\theta\), and \(dx\) with \(a\cos\theta\,d\theta\). These replacements come from one substitution and must be made consistently.

For a definite integral, the limits must also be expressed in the new variable. If an endpoint is \(x=c\), its corresponding angle is \(\theta=\arcsin(c/a)\), where \(\arcsin\) denotes the inverse sine taking values in \([-\frac{\pi}{2},\frac{\pi}{2}]\). The original and transformed endpoints represent the same values of the variable, expressed in different coordinates.

Why the Angle Range Makes the Method Work

The chosen interval does more than ensure a convenient sign. It makes \(\theta\mapsto a\sin\theta\) one-to-one, so the substitution has a unique inverse angle for every \(x\in[-a,a]\). In the interior of the interval, \(\cos\theta>0\), and therefore \(dx/d\theta=a\cos\theta>0\). The variable \(x\) increases with \(\theta\), and the substitution can be reversed without ambiguity.

At \(\theta=\pm\frac{\pi}{2}\), the derivative is zero, and those angles correspond to the endpoints \(x=\pm a\), where the radical is zero. The algebraic identities remain valid at the endpoints. When applying a change of variable to an integral that includes an endpoint, the endpoint behavior must be treated in the usual way; the zero derivative there does not justify dividing by \(\cos\theta\) at that endpoint.

Worked Substitutions

A Common Sign Error

A tempting but incomplete calculation is to write \(\sqrt{a^2\cos^2\theta}=a\cos\theta\) for every angle. In general, the square root of a square is an absolute value, not necessarily the original quantity:

For example, at \(\theta=2\pi/3\), \(\cos\theta=-1/2\). The square root is \(a/2\), whereas \(a\cos\theta=-a/2\), which is negative and cannot equal a nonnegative square root. The correction is to restrict \(\theta\) to \([-\frac{\pi}{2},\frac{\pi}{2}]\), where \(\cos\theta\ge0\), and then write \(\sqrt{a^2\cos^2\theta}=a\cos\theta\).

A separate bookkeeping error is to replace the radical but leave \(dx\) unchanged. For instance, under \(x=5\sin\theta\), \(dx\) is \(5\cos\theta\,d\theta\), not \(d\theta\). Omitting that factor changes the integral. Differentiate the substitution explicitly whenever the differential is transformed.

Summary

The sine substitution follows from the right-triangle ratios and the Pythagorean identity. For \(a>0\), the conditions \(x=a\sin\theta\) and \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\) give the exact replacements \(\sqrt{a^2-x^2}=a\cos\theta\) and \(dx=a\cos\theta\,d\theta\). The angle range ensures both that cosine is nonnegative and that each \(x\in[-a,a]\) has a unique corresponding angle. In definite integrals, the bounds must be converted to those corresponding angles as well.