Why the Bounds Matter
For an indefinite integral, a trigonometric substitution changes the integrand and introduces a new variable. For a definite integral, it also changes the limits. The transformed limits are not optional: they record which part of the trigonometric function corresponds to the original interval.
For radicals of the form \(\sqrt{a^2-x^2}\), the sine substitution is especially useful. It simplifies the radical on a chosen interval, but only if the interval and angle range are matched correctly. This is important because a square root is nonnegative, whereas a cosine can be negative on some intervals.
Transforming the Limits and the Radical
On \([-\pi/2,\pi/2]\), the sine function is increasing and the cosine is nonnegative. Therefore \(x=a\sin\theta\) gives
The last equality uses \(a>0\) and \(\cos\theta\geq 0\) on the chosen angle interval. If the original limits are \(u\) and \(v\), then \(u=a\sin\alpha\) and \(v=a\sin\beta\), where
Because sine is increasing on this interval, \(u\leq v\) implies \(\alpha\leq\beta\). Thus the substitution preserves the order of the limits. If an original endpoint equals \(a\) or \(-a\), its angle is respectively \(\pi/2\) or \(-\pi/2\).
Two Useful Definite-Integral Formulas
The substitution immediately gives a formula for the reciprocal square root. Provided \(-a\leq u<v\leq a\), the integral is understood as improper if either endpoint is \(a\) or \(-a\).
For the square root itself, substitution gives \(a^2\cos^2\theta\,d\theta\). Using \(\cos^2\theta=(1+\cos 2\theta)/2\), an antiderivative in \(\theta\) is \(\frac{a^2}{2}(\theta+\sin\theta\cos\theta)\). Hence
These formulas apply when the interval lies within \([-a,a]\), the real domain of the radical. The reciprocal formula also requires interpreting an endpoint singularity as an improper integral when necessary. The singularity at \(x=\pm a\) is integrable: after substitution it becomes a finite angular interval with integrand \(1\).
Why the Substitution Works
The change of variables is justified by matching both the integrand and the interval. On the selected angle range, \(x=a\sin\theta\) is differentiable and strictly increasing, so it maps \([\alpha,\beta]\) to \([u,v]\). The usual change-of-variables rule then replaces \(dx\) by \(a\cos\theta\,d\theta\), with the new bounds corresponding to the old ones.
For an endpoint \(x=c\), solve \(c=a\sin\theta\) using \(\theta=\arcsin(c/a)\) in \([-\pi/2,\pi/2]\).
Replace \(x\), \(dx\), and \(\sqrt{a^2-x^2}\), not just the radical.
Keep the transformed limits through the evaluation, then check the value against the original integral where practical.
Worked Integrals
Worked Example: Integrating a Square Root on a Short Interval
Evaluate \(\int_0^{\sqrt{3}}\sqrt{4-x^2}\,dx\). Here \(a=2\), and both limits lie in \([-2,2]\). The angle limits are \(\alpha=\arcsin(0/2)=0\) and \(\beta=\arcsin(\sqrt{3}/2)=\pi/3\).
Set \(x=2\sin\theta\). Then \(dx=2\cos\theta\,d\theta\), and on \(0\leq\theta\leq\pi/3\), \(\sqrt{4-x^2}=2\cos\theta\). Therefore
As a check, the antiderivative in \(x\) from the same substitution is \(\frac{x}{2}\sqrt{4-x^2}+2\arcsin(x/2)\). Differentiating gives \(\frac{4-2x^2}{2\sqrt{4-x^2}}+\frac{2}{\sqrt{4-x^2}}=\frac{8-2x^2}{2\sqrt{4-x^2}}=\sqrt{4-x^2}\). At \(x=\sqrt{3}\) its value is \(\frac{\sqrt3}{2}+2\cdot\frac{\pi}{3}=\frac{\sqrt3}{2}+\frac{2\pi}{3}\), and at \(x=0\) it is \(0\). The difference is \(\frac{\sqrt3}{2}+\frac{2\pi}{3}\), matching the transformed evaluation.
Worked Example: Reciprocal Square Root with Interior Endpoints
Evaluate \(\int_{-1}^{1}\frac{dx}{\sqrt{9-x^2}}\). Here \(a=3\). The transformed endpoints are \(\alpha=\arcsin(-1/3)\) and \(\beta=\arcsin(1/3)\). Since arcsine is odd, \(\alpha=-\arcsin(1/3)\).
Use \(x=3\sin\theta\). Then \(dx=3\cos\theta\,d\theta\) and the denominator is \(3\cos\theta\), because the angle limits lie in \([-\pi/2,\pi/2]\). Thus
To check, differentiate \(F(x)=\arcsin(x/3)\). Its derivative is \(F'(x)=\frac{1/3}{\sqrt{1-x^2/9}}=\frac{1}{\sqrt{9-x^2}}\). Therefore the original integral is \(F(1)-F(-1)=\arcsin(1/3)-\arcsin(-1/3)=2\arcsin(1/3)\), as obtained.
Worked Example: A Reciprocal Radical with an Endpoint at \(a\)
Evaluate \(\int_0^2\frac{dx}{\sqrt{4-x^2}}\). The integrand is undefined at \(x=2\), so this is an improper integral. Write it as a limit: \(\lim_{b\to2^-}\int_0^b\frac{dx}{\sqrt{4-x^2}}\).
For \(0\leq b<2\), set \(x=2\sin\theta\). The lower angle is \(0\), and the upper angle is \(\arcsin(b/2)\). The transformed integral is
Taking the endpoint limit gives \(\lim_{b\to2^-}\arcsin(b/2)=\pi/2\). Thus the improper integral converges and equals \(\pi/2\). This also checks the endpoint rule: \(x=2\) corresponds to \(\theta=\pi/2\), and the transformed integrand is \(1\) up to that angle.
A Common Bounds Mistake
A frequent error is to change variables in the integrand but leave the original numerical bounds in place. For example, in the integral \(\int_0^1 dx/\sqrt{4-x^2}\), the substitution \(x=2\sin\theta\) does not turn the limits into \(0\) and \(1\). Those numbers are \(x\)-values, not angle values.
Correction: The upper angle solves \(1=2\sin\beta\), so \(\beta=\arcsin(1/2)=\pi/6\). The integral is \(\int_0^{\pi/6}d\theta=\pi/6\).
The correction can also be checked in the original variable: an antiderivative is \(\arcsin(x/2)\), and its change from \(0\) to \(1\) is \(\arcsin(1/2)-\arcsin(0)=\pi/6\). The incorrect value \(1\) came from treating an \(x\)-coordinate as an angle.
Choosing and Checking the Angle Interval
The condition \(-a\leq x\leq a\) is not merely a convenient restriction. It is exactly where \(a^2-x^2\geq0\), so the square root is real. Within that interval, the principal inverse sine gives angles in \([-\pi/2,\pi/2]\), where the replacement \(\sqrt{a^2-x^2}=a\cos\theta\) is valid.
If an integral is written with limits outside \([-a,a]\), it is not a real-valued integral of this radical across the entire stated interval. If the bounds are reversed but still lie in the domain, the usual orientation rule applies: reversing the limits changes the sign. When an endpoint is \(\pm a\), distinguish the square-root integrand, which is defined and continuous there, from the reciprocal integrand, which requires an improper-limit interpretation.
These checks make definite trigonometric substitution reliable: verify the real interval, find each endpoint angle, and use the same angle range when simplifying the radical. A correct transformed integrand with incorrect bounds still gives an incorrect definite integral.
Check Your Understanding
For each question, identify the transformed limits before evaluating or explaining the integral.
- Under \(x=5\sin\theta\), what are the angle limits corresponding to \(x=0\) and \(x=5/2\) on the principal interval?
- Why is \(\sqrt{a^2-x^2}=a\cos\theta\) valid for \(\theta\in[-\pi/2,\pi/2]\), but not automatically valid for every angle?
- Evaluate \(\int_0^{a/2} dx/\sqrt{a^2-x^2}\) for \(a>0\), showing the angle limits.
- What makes \(\int_0^a dx/\sqrt{a^2-x^2}\) improper, and why does it converge?
- For a real integral involving \(\sqrt{a^2-x^2}\), what restriction must the original interval satisfy?
