Introduction
A reciprocal square root can lead to an inverse-sine antiderivative when the expression under the root has the form of a positive constant minus a square. The important details are the coefficient of that square and the interval on which the radicand is positive.
This tutorial develops the formula for \(a^2-u^2\), then applies it to bounded radicands written as \(C-B(x-h)^2\). In each case, the formula is valid on the open interval where the radicand is positive. The endpoints make the denominator zero, so they are not part of the domain of the integrand.
The Inverse-Sine Antiderivative
For \(a>0\), the expression \(a^2-u^2\) is positive exactly when \(-a<u<a\). On that interval, the inverse sine \(\arcsin(u/a)\) is differentiable. Its derivative gives the basic formula.
The numerator condition matters. The formula directly integrates \(u'(x)/\sqrt{a^2-u(x)^2}\), not \(1/\sqrt{a^2-u(x)^2}\) for every possible function \(u\). If \(u\) is affine, its derivative is constant, so that constant can be accounted for explicitly.
Why the Formula Works
Here is a derivation from the derivative of the inverse sine. The principal inverse sine takes values in \([-\pi/2,\pi/2]\). For an input strictly between \(-1\) and \(1\), its cosine is positive; this allows the square root in the derivative to be taken as the positive root.
For \(|z|<1\), let \(y=\arcsin z\). Then \(\sin y=z\), and implicit differentiation gives \(\cos y\,dy/dz=1\). Since \(y\in(-\pi/2,\pi/2)\), \(\cos y=\sqrt{1-z^2}>0\). Thus \(dy/dz=1/\sqrt{1-z^2}\).
Set \(z=u(x)/a\), where \(a>0\). The chain rule gives \[ \frac{d}{dx}\arcsin\left(\frac{u}{a}\right) = \frac{u'/a}{\sqrt{1-u^2/a^2}}. \]
Because \(a>0\), \(\sqrt{1-u^2/a^2}=\sqrt{a^2-u^2}/a\). Therefore the derivative is \(u'/\sqrt{a^2-u^2}\), as claimed.
This calculation also explains the domain requirement. At \(u=\pm a\), the denominator in the derivative formula is zero, and the original reciprocal-square-root integrand is undefined. The antiderivative formula applies on intervals where \(|u|<a\), not at those boundary points.
Affine Expressions Inside the Square Root
Suppose \(u(x)=kx+m\), where \(k\ne 0\). Then \(u'(x)=k\), so the formula above gives an antiderivative after dividing by \(k\). The domain is found by solving \(-a<kx+m<a\); its interval endpoints depend on the sign of \(k\).
The coefficient is \(1/k\), including its sign. Indeed, differentiating the right-hand side multiplies the derivative of the inverse sine by \(k\), which cancels the factor \(1/k\). Replacing \(k\) by \(|k|\) without considering the orientation of the affine substitution would generally give the wrong derivative.
Worked Example: An Affine Expression with a Nonunit Slope
Evaluate \[ \int \frac{dx}{\sqrt{25-9x^2}}. \]
Write the radicand as \(5^2-(3x)^2\). Here \(a=5\), \(u=3x\), and \(u'=3\). The formula with the required scale factor gives \[ \int \frac{dx}{\sqrt{25-9x^2}} = \frac{1}{3}\arcsin\left(\frac{3x}{5}\right)+C. \]
For the real integrand, \(25-9x^2>0\), so \(9x^2<25\), or \(-5/3<x<5/3\). To check the answer, differentiate: \[ \frac{d}{dx}\left[\frac{1}{3}\arcsin\left(\frac{3x}{5}\right)\right] = \frac{1}{3}\cdot\frac{3/5}{\sqrt{1-9x^2/25}} = \frac{1}{5\sqrt{1-9x^2/25}} = \frac{1}{\sqrt{25-9x^2}}. \] The derivative matches the integrand throughout the stated interval.
Applying the Formula to a Bounded Completed Square
Let \(B>0\) and \(C>0\). For the bounded form \(C-B(x-h)^2\), positivity requires \(B(x-h)^2<C\). Equivalently, \[ |x-h|<\sqrt{\frac{C}{B}}. \] Thus the radicand is positive only on the bounded interval \(h-\sqrt{C/B}<x<h+\sqrt{C/B}\). Its two endpoints make the radicand zero and are excluded from the integrand's domain.
Factor out \(C\) to match the inverse-sine pattern. The resulting antiderivative includes a factor \(1/\sqrt B\); retaining that factor is essential.
The letter \(C_0\) here denotes the constant of integration; \(C\) in the radicand is the positive parameter. To verify the factor, differentiate the proposed antiderivative. The derivative of the inverse-sine input contributes \(\sqrt{B}/\sqrt{C}\), while the derivative of the inverse sine contributes the reciprocal square root of one minus the input squared. Together with \(1/\sqrt B\), these factors simplify to the original integrand.
Worked Example: A Shifted Bounded Form
Evaluate \[ \int \frac{dx}{\sqrt{16-4(x-1)^2}}. \]
Identify \(C=16\), \(B=4\), and \(h=1\). The real domain follows from \[ 16-4(x-1)^2>0 \quad\Longleftrightarrow\quad (x-1)^2<4 \quad\Longleftrightarrow\quad -1<x<3. \] The completed-square formula gives \[ \int \frac{dx}{\sqrt{16-4(x-1)^2}} = \frac{1}{2}\arcsin\left(\frac{x-1}{2}\right)+C_0. \]
Check by differentiation: \[ \frac{d}{dx}\left[\frac{1}{2}\arcsin\left(\frac{x-1}{2}\right)\right] = \frac{1}{2}\cdot\frac{1/2}{\sqrt{1-(x-1)^2/4}} = \frac{1}{4\sqrt{1-(x-1)^2/4}} = \frac{1}{\sqrt{16-4(x-1)^2}}. \] This agrees with the integrand for \(-1<x<3\).
Completing the Square Before Integrating
A quadratic may not initially appear as a constant minus a square. When completing the square produces \(C-B(x-h)^2\) with both \(B\) and \(C\) positive, the completed-square formula applies on the interval where that expression is positive. This use of completing the square is specific to the bounded form: not every quadratic radicand has this shape or a bounded real domain.
Worked Example: A Quadratic Radicand
Evaluate \[ \int \frac{dx}{\sqrt{-2x^2+8x+6}}. \]
First factor \(-2\) from the quadratic and linear terms: \[ -2x^2+8x+6=-2(x^2-4x)+6. \] Complete the square inside the parentheses: \[ x^2-4x=(x-2)^2-4. \] Substitute this expression and simplify: \[ -2\big((x-2)^2-4\big)+6 = -2(x-2)^2+8+6 = 14-2(x-2)^2. \] Thus \(C=14\), \(B=2\), and \(h=2\). The radicand is positive when \[ 14-2(x-2)^2>0 \quad\Longleftrightarrow\quad (x-2)^2<7 \quad\Longleftrightarrow\quad 2-\sqrt7<x<2+\sqrt7. \] Apply the formula: \[ \int \frac{dx}{\sqrt{-2x^2+8x+6}} = \frac{1}{\sqrt2}\arcsin\left(\frac{x-2}{\sqrt7}\right)+C_0. \]
For a direct check, differentiate: \[ \frac{d}{dx}\left[\frac{1}{\sqrt2}\arcsin\left(\frac{x-2}{\sqrt7}\right)\right] = \frac{1}{\sqrt2}\cdot\frac{1/\sqrt7}{\sqrt{1-(x-2)^2/7}} = \frac{1}{\sqrt{14}\sqrt{1-(x-2)^2/7}} = \frac{1}{\sqrt{14-2(x-2)^2}}. \] Since \(14-2(x-2)^2=-2x^2+8x+6\), this is the original integrand on the stated interval.
A Common Scale-Factor Mistake
A frequent error is to recognize the inverse-sine pattern but omit the coefficient created by the inner function. For example, for \(\sqrt{25-9x^2}\), writing \(\arcsin(3x/5)\) as the antiderivative omits the factor \(1/3\). Differentiation exposes the error: \[ \frac{d}{dx}\arcsin\left(\frac{3x}{5}\right) = \frac{3}{\sqrt{25-9x^2}}, \] which is three times the required integrand. The correction is to divide by \(3\): \[ \frac{d}{dx}\left[\frac13\arcsin\left(\frac{3x}{5}\right)\right] = \frac{1}{\sqrt{25-9x^2}}. \]
It is also an error to ignore the domain when writing the result. The inverse-sine expression is real for inputs in \([-1,1]\), but the reciprocal square root requires a strictly positive radicand. Therefore, for an indefinite integral here, specify an interval on which the radicand is positive; do not include its zero endpoints.
Checklist for These Integrals
For a reciprocal square root of a bounded form, proceed in this order:
- Rewrite the radicand as \(a^2-u(x)^2\), or complete the square to obtain \(C-B(x-h)^2\).
- Determine the interval where the radicand is strictly positive.
- Identify the derivative of the inner expression and include the corresponding scale factor.
- Differentiate the proposed antiderivative to verify that it reproduces the integrand on that interval.
The domain step and the derivative check are part of the calculation, not optional additions. They confirm both that the real inverse-sine expression is defined and that its scale is correct.
Check Your Understanding
Use the inverse-sine formula, and state the interval where each integrand is real.
- Find an antiderivative of \(1/\sqrt{36-16x^2}\).
- Find an antiderivative of \(1/\sqrt{9-(x+2)^2}\).
- Complete the square in \(-x^2+6x+7\), then find an antiderivative of \(1/\sqrt{-x^2+6x+7}\).
- Explain why the antiderivative formula does not make the integrand defined at a point where the radicand is zero.
- Differentiate \(\frac{1}{4}\arcsin(4x/7)\) and identify the reciprocal square root it produces.
