Why the Angle Interval Matters
A trigonometric substitution can turn a quadratic radical into a simpler trigonometric expression. But the substitution is not only an equation between variables. It also includes a choice of angle interval. That choice controls whether the trigonometric function is one-to-one and, crucially, whether a cosine or tangent is positive or negative.
For example, \(\cos^2\theta\) determines \(|\cos\theta|\), not the sign of \(\cos\theta\). Thus the identity \(\sqrt{\cos^2\theta}=|\cos\theta|\) does not automatically permit replacing the square root by \(\cos\theta\). The replacement is valid only on an interval where \(\cos\theta\geq 0\).
Domain and Branch
A substitution such as \(x=a\sin\theta\) or \(x=a\sec\theta\), where \(a>0\), can have more than one angle corresponding to the same value of \(x\). A branch choice is a restriction of \(\theta\) to an interval on which the chosen trigonometric function is one-to-one and has the required signs.
The interval for \(x\) matters too. The radicand must be nonnegative for a real square root, and a secant substitution for \(x^2-a^2\) must be restricted to either \(x\geq a\) or \(x\leq -a\) if the substitution is to be inverted consistently.
| Radical | Substitution and branch | Square-root identity on that branch |
|---|---|---|
| \(\sqrt{a^2-x^2}\), \(-a\leq x\leq a\) | \(x=a\sin\theta\), \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\) | \(\sqrt{a^2-x^2}=a\cos\theta\) |
| \(\sqrt{a^2+x^2}\), \(x\in\mathbb R\) | \(x=a\tan\theta\), \(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\) | \(\sqrt{a^2+x^2}=a\sec\theta\) |
| \(\sqrt{x^2-a^2}\), \(x\geq a\) | \(x=a\sec\theta\), \(0\leq\theta<\frac{\pi}{2}\) | \(\sqrt{x^2-a^2}=a\tan\theta\) |
| \(\sqrt{x^2-a^2}\), \(x\leq -a\) | \(x=-a\sec\theta\), \(0\leq\theta<\frac{\pi}{2}\) | \(\sqrt{x^2-a^2}=a\tan\theta\) |
For differentiable changes of variable, the open interiors of these intervals are the usual working domains. The endpoints may be included when the algebraic identity makes sense there, but the substitution can have zero derivative at an endpoint or need separate treatment there.
Sine Substitution: Keep Cosine Nonnegative
For \(a>0\), the real domain of \(\sqrt{a^2-x^2}\) is \([-a,a]\). Use \(x=a\sin\theta\) with \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\), on which sine is one-to-one and cosine is nonnegative. Then \(dx=a\cos\theta\,d\theta\), and
The restriction on \(\theta\) is what justifies the final equality. If an angle outside this interval is used, the identity \(\sqrt{a^2\cos^2\theta}=a\cos\theta\) may be false. To return to \(x\), use the principal inverse sine: \(\theta=\arcsin(x/a)\), whose values lie in \([-\frac{\pi}{2},\frac{\pi}{2}]\).
Worked Example: Choosing the Sine Branch
Use a sine substitution to represent \(\sqrt{25-x^2}\) at \(x=3\), and check the value using the chosen branch.
Here \(a=5\), and \(x=3\) is in the allowed domain \([-5,5]\). Set \(x=5\sin\theta\) with \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\). Then \(\sin\theta=3/5\), so the principal angle has \(\cos\theta=4/5\), which is nonnegative.
Check by direct substitution into the original radical: \(\sqrt{25-3^2}=\sqrt{16}=4\). Also, \(5\sin\theta=5(3/5)=3=x\), so the selected angle recovers the original variable.
Tangent Substitution: The Principal Interval Covers Every Real \(x\)
For \(a>0\), set \(x=a\tan\theta\) and restrict \(\theta\) to \((-\frac{\pi}{2},\frac{\pi}{2})\). Tangent is one-to-one on this interval and takes every real value, so every \(x\in\mathbb R\) corresponds to exactly one angle there: \(\theta=\arctan(x/a)\). On this interval \(\cos\theta>0\), hence \(\sec\theta>0\). Using \(1+\tan^2\theta=\sec^2\theta\),
Here \(dx=a\sec^2\theta\,d\theta\). The positivity of secant is essential: the square root is positive, so the expression that replaces it must also be positive.
Worked Example: Recovering a Tangent-Branch Angle
Let \(x=-4\) and use \(x=3\tan\theta\) on \((-\frac{\pi}{2},\frac{\pi}{2})\). Find the radical \(\sqrt{9+x^2}\) in terms of \(\theta\), then check the reconstructed value.
The substitution gives \(\tan\theta=-4/3\). Since \(\theta\) is in the principal interval, its cosine is positive. A right triangle with opposite side \(-4\), adjacent side \(3\), and hypotenuse \(5\) gives \(\sec\theta=5/3\). Therefore,
Directly, \(\sqrt{9+(-4)^2}=\sqrt{25}=5\), as required. The substitution also recovers \(x=3\tan\theta=3(-4/3)=-4\). The positive value of secant is consistent with the positive square root.
Secant Substitution: Split the Positive and Negative Domains
The real domain of \(\sqrt{x^2-a^2}\) is \((-\infty,-a]\cup[a,\infty)\). It consists of two separate intervals, so it is useful to handle the two signs of \(x\) separately.
For \(x\geq a\), choose \(x=a\sec\theta\) with \(0\leq\theta<\frac{\pi}{2}\). Secant is one-to-one on this interval, \(\sec\theta\geq1\), and \(\tan\theta\geq0\). Thus
For \(x\leq -a\), use \(x=-a\sec\theta\) on the same angle interval. Then \(x\leq-a\), \(dx=-a\sec\theta\tan\theta\,d\theta\), and \(x^2=a^2\sec^2\theta\), so again \(\sqrt{x^2-a^2}=a\tan\theta\). The minus sign in \(x=-a\sec\theta\) is what selects the negative \(x\)-branch; it does not make the square root negative.
One could instead use \(x=a\sec\theta\) with angles in \((\frac{\pi}{2},\pi]\) for negative \(x\), but then \(\tan\theta\leq0\) and the radical is \(-a\tan\theta\), not \(a\tan\theta\). Either convention can work if its signs are followed consistently. The two-branch convention in the table keeps \(\tan\theta\geq0\) in both cases.
Worked Example: The Negative Secant Branch
Take \(a=5\) and \(x=-13\). Use the negative secant branch and verify the radical and the original variable.
Set \(x=-5\sec\theta\) with \(0\leq\theta<\frac{\pi}{2}\). Then \(\sec\theta=13/5\). A right triangle has hypotenuse \(13\), adjacent side \(5\), and opposite side \(12\), since \(13^2-5^2=169-25=144=12^2\). Thus \(\tan\theta=12/5\), which is positive on the chosen branch.
Checking in the original expression gives \(\sqrt{(-13)^2-25}=\sqrt{169-25}=\sqrt{144}=12\). Checking the back-substitution gives \(x=-5\sec\theta=-5(13/5)=-13\). Both the sign of \(x\) and the nonnegative sign of the radical are preserved.
Why the Branch Conditions Work
The sign issue in these substitutions follows from a basic identity. For any real \(u\), \(\sqrt{u^2}=|u|\), because the square root is defined to be nonnegative. Therefore, a step from \(\sqrt{u^2}\) to \(u\) requires the additional condition \(u\geq0\); if \(u\leq0\), the result is \(-u\).
For the sine substitution, \(u=a\cos\theta\), and the principal sine branch ensures \(u\geq0\). For the tangent substitution, \(u=a\sec\theta\), and the chosen interval ensures \(u>0\). For the secant substitutions, \(u=a\tan\theta\), and the first-quadrant angle interval ensures \(u\geq0\). This is the complete reason the radical formulas in the table hold.
Identify where the radicand is nonnegative, and whether that domain has separate pieces.
Use an interval that covers the relevant \(x\)-values without assigning multiple angles to one \(x\).
Use \(\sqrt{u^2}=|u|\), then use the interval to determine whether \(|u|=u\) or \(|u|=-u\).
Use the corresponding inverse trigonometric value or a right triangle, retaining the sign specified by the original domain.
A Common Branch Mistake
A frequent error is to use the right trigonometric identity but an angle outside the branch that makes the needed expression nonnegative. For example, when \(x=3\) and \(a=5\), the equation \(x=5\sin\theta\) permits both \(\theta=\arcsin(3/5)\) and the supplementary angle \(\theta=\pi-\arcsin(3/5)\). On the second choice, \(\cos\theta=-4/5\).
The incorrect calculation is
A square root cannot equal \(-4\). The error is not in the identity \(\sin^2\theta+\cos^2\theta=1\); it is in treating \(\sqrt{\cos^2\theta}\) as \(\cos\theta\) without checking its sign. The correction is either to restrict to the principal sine branch, where \(\cos\theta\geq0\), or to retain the absolute value \(5|\cos\theta|\). The principal branch is usually simpler because it also gives the direct inverse \(\theta=\arcsin(x/5)\).
Why Consistent Back-Substitution Matters
The purpose of choosing a branch is not merely to simplify a radical in the middle of a calculation. It also makes the return from \(\theta\) to \(x\) unambiguous. For sine, the principal inverse gives \(\theta=\arcsin(x/a)\). For tangent, it gives \(\theta=\arctan(x/a)\). For secant, the sign of \(x\) must be retained: the positive branch has \(x=a\sec\theta\), while the negative branch has \(x=-a\sec\theta\).
When using a right triangle to recover a trigonometric expression, its side lengths are nonnegative. The original variable’s sign is therefore not encoded by the triangle alone; for a negative secant branch it must be supplied by the minus sign in \(x=-a\sec\theta\). Keeping that sign separate prevents a positive length such as \(\sqrt{x^2-a^2}\) from being confused with the possibly negative variable \(x\).
Check Your Understanding
For each question, state the relevant branch and explain how it determines the sign.
- For \(x=a\sin\theta\), why does the interval \([-\frac{\pi}{2},\frac{\pi}{2}]\) justify writing \(\sqrt{a^2-x^2}=a\cos\theta\)?
- For \(x=a\tan\theta\), what interval gives exactly one angle for each real \(x\), and why is \(\sec\theta\) positive there?
- Why must the domain for a secant substitution of \(\sqrt{x^2-a^2}\) be treated as separate positive and negative branches?
- If \(x=-a\sec\theta\) with \(0\leq\theta<\frac{\pi}{2}\), what are the signs of \(x\), \(dx/d\theta\), and \(\sqrt{x^2-a^2}\) in the interior?
- Explain the error in replacing \(\sqrt{\cos^2\theta}\) by \(\cos\theta\) for an unrestricted angle.
