Why the Radicand Guides the Substitution
A trigonometric substitution is chosen to turn a quadratic expression under a square root into a trigonometric identity. The useful choice depends on the signs in that quadratic expression. Matching those signs correctly can simplify later work; choosing a substitution without checking its range can introduce an incorrect sign or even leave the real domain.
In this tutorial, \(a\) is a positive real number, and each square root means the principal, nonnegative square root. The three standard forms are \(a^2-x^2\), \(a^2+x^2\), and \(x^2-a^2\). Their domains and substitutions are related, but they are not interchangeable.
The Three Standard Matches
| Radicand | Real domain for \(x\) | Substitution and useful range | Transformed square root |
|---|---|---|---|
| \(a^2-x^2\) | \(-a\le x\le a\) | \(x=a\sin\theta,\;-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\) | \(a\cos\theta\) |
| \(a^2+x^2\) | All real \(x\) | \(x=a\tan\theta,\;-\frac{\pi}{2}<\theta<\frac{\pi}{2}\) | \(a\sec\theta\) |
| \(x^2-a^2\), \(x\ge a\) | \(x\ge a\) | \(x=a\sec\theta,\;0\le\theta<\frac{\pi}{2}\) | \(a\tan\theta\) |
| \(x^2-a^2\), \(x\le-a\) | \(x\le-a\) | \(x=-a\sec\theta,\;0\le\theta<\frac{\pi}{2}\) | \(a\tan\theta\) |
The domains in the table follow by requiring the radicand to be nonnegative. For example, \(a^2-x^2\ge0\) means \(x^2\le a^2\), or \(-a\le x\le a\). The condition \(x^2-a^2\ge0\) means \(|x|\ge a\), which has two separate branches. The radicand \(a^2+x^2\) is positive for every real \(x\).
The chosen intervals are not decorative. They ensure that the expression obtained after taking a square root has the stated sign. In particular, a square root cannot be replaced by a trigonometric expression that might be negative on the chosen range.
Why These Substitutions Work
The three matches come directly from identities. The Pythagorean identities are \(\sin^2\theta+\cos^2\theta=1\), \(1+\tan^2\theta=\sec^2\theta\), and \(\sec^2\theta-1=\tan^2\theta\). Substitution turns each radicand into a constant times a square of a trigonometric function.
Taking square roots gives \(a|\cos\theta|\), \(a|\sec\theta|\), or \(a|\tan\theta|\), respectively, because \(a>0\) and \(\sqrt{u^2}=|u|\). On the listed ranges, the first cosine is nonnegative, the second secant is positive, and the tangent in the positive-\(x\) secant branch is nonnegative. For the negative-\(x\) branch, \(x=-a\sec\theta\) with \(0\le\theta<\frac{\pi}{2}\) still gives \(x^2-a^2=a^2\tan^2\theta\), and its square root is again \(a\tan\theta\). This proves the table’s transformed-root formulas.
The differentials also follow by differentiating the substitutions. They are useful when the substitution is later applied inside an integral, although choosing and checking the substitution is the main task here.
Worked Examples
Worked Example: Matching \(\sqrt{49-x^2}\) to a Sine Substitution
Here \(a^2=49\), so \(a=7\). The radicand is nonnegative when \(49-x^2\ge0\), which gives \(x^2\le49\), or \(-7\le x\le7\). The sine match is \(x=7\sin\theta\) with \(-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}\).
On this interval, \(\cos\theta\ge0\). Substituting into the radicand gives
Therefore \(\sqrt{49-x^2}=7|\cos\theta|=7\cos\theta\) on the selected range. The differential is \(dx=7\cos\theta\,d\theta\). As a check, substituting \(x=7\sin\theta\) back into the original radicand produces \(49\cos^2\theta\), and its nonnegative square root on this interval is exactly \(7\cos\theta\).
Worked Example: Matching \(\sqrt{9+x^2}\) to a Tangent Substitution
Here \(a=3\). Since \(9+x^2>0\) for every real \(x\), \(x\) has no restriction from the radicand. Choose \(x=3\tan\theta\) with \(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\). Tangent maps this interval onto all real numbers, so it covers every possible \(x\).
Substitute and simplify:
Throughout this interval \(\cos\theta>0\), so \(\sec\theta>0\). Hence \(\sqrt{9+x^2}=3\sec\theta\), not merely \(3|\sec\theta|\). Differentiating the substitution gives \(dx=3\sec^2\theta\,d\theta\). The check is direct: \(3\sec\theta\) is positive and its square is \(9\sec^2\theta\), the transformed radicand.
Worked Example: Matching \(\sqrt{x^2-16}\) on the Positive Branch
The radicand is nonnegative for \(x^2\ge16\), so this example considers the branch \(x\ge4\). Set \(x=4\sec\theta\), where \(0\le\theta<\frac{\pi}{2}\). On this interval, \(\sec\theta\ge1\), so the substitution gives \(x\ge4\) as required.
The radicand becomes
Since \(\tan\theta\ge0\) on the chosen interval, \(\sqrt{x^2-16}=4\tan\theta\). The differential is \(dx=4\sec\theta\tan\theta\,d\theta\). To check the endpoint and sign, \(\theta=0\) gives \(x=4\) and the transformed root \(0\), as in the original expression. For \(0<\theta<\frac{\pi}{2}\), both the original and transformed roots are positive.
Worked Example: Matching \(\sqrt{x^2-9}\) on the Negative Branch
The negative branch of the domain is \(x\le-3\). To cover it with an angle in the first quadrant, use \(x=-3\sec\theta\), with \(0\le\theta<\frac{\pi}{2}\). Since \(\sec\theta\ge1\), this gives \(x\le-3\).
Squaring removes the minus sign, so the radicand transforms as follows:
Thus \(\sqrt{x^2-9}=3\tan\theta\), because \(\tan\theta\ge0\). Here \(dx=-3\sec\theta\tan\theta\,d\theta\). As a numerical check, take \(\theta=\frac{\pi}{3}\). Then \(\sec\theta=2\), so \(x=-6\), and the original root is \(\sqrt{(-6)^2-9}=\sqrt{27}=3\sqrt3\). The transformed root is \(3\tan(\frac{\pi}{3})=3\sqrt3\), in agreement.
Choosing a Branch for \(x^2-a^2\)
The expression \(x^2-a^2\) has two disconnected parts in its real domain: \(x\ge a\) and \(x\le-a\). A single one-to-one angle interval for \(x=a\sec\theta\) is commonly chosen to describe the positive branch. The negative branch can be handled separately with \(x=-a\sec\theta\). This keeps the angle in the first quadrant and makes the signs easy to read.
It is also possible to use \(x=a\sec\theta\) for negative \(x\) by choosing \(\theta\) in the second quadrant. But then \(\tan\theta<0\), and the root is \(a|\tan\theta|\), not \(a\tan\theta\). The two-branch convention in the table avoids that extra sign adjustment. At \(x=a\) or \(x=-a\), the radicand is zero; these endpoints correspond to \(\theta=0\) in the respective branch choices.
A Common Sign Error
A frequent mistake is to use \(\sqrt{u^2}=u\) without checking that \(u\ge0\). For example, take \(a=3\), \(x=-5\), and \(\sqrt{x^2-9}\). If one uses \(x=3\sec\theta\) with \(\theta\) in the second quadrant, then \(\sec\theta=-5/3\) and \(\tan\theta<0\). The transformed identity gives \(x^2-9=9\tan^2\theta\), but the following calculation is wrong:
A principal square root cannot be negative. The correction is
Indeed, \(x=-5\) gives \(\sqrt{25-9}=4\), while \(\tan\theta=-4/3\), so \(-3\tan\theta=4\). Alternatively, use the negative-branch substitution \(x=-3\sec\theta\) with \(0\le\theta<\frac{\pi}{2}\); then the transformed root is \(3\tan\theta\) with no absolute-value ambiguity.
A Practical Selection Process
Before making a substitution, identify the exact quadratic pattern and check its real domain. If the coefficients do not initially match one of the standard forms, first identify the scaled variable or factor that does. Then select an angle range that represents the intended part of the \(x\)-domain and controls the sign of the trigonometric factor under the square root.
Find the \(x\)-values for which the expression under the square root is nonnegative.
Use sine for \(a^2-x^2\), tangent for \(a^2+x^2\), and secant for \(x^2-a^2\).
Ensure the substitution covers the required real \(x\)-values and fixes the signs of the trigonometric expressions.
Substitute into the radicand, use the identity, and take the nonnegative square root with the chosen range in mind.
Why the Match Matters
These substitutions are useful because they replace a quadratic radicand with a trigonometric square, making the square root easier to express. The matching rule also provides a quick way to detect an unsuitable choice: if the algebra does not produce the relevant Pythagorean identity, or the chosen angle range gives the wrong sign, the substitution needs to be reconsidered.
Keep the distinction between the algebraic identity and the square-root statement clear. For instance, \(x^2-a^2=a^2\tan^2\theta\) is an identity after a secant substitution. It implies \(\sqrt{x^2-a^2}=a|\tan\theta|\) in general; a more specific expression such as \(a\tan\theta\) is valid only on a range where \(\tan\theta\ge0\). The domain and branch are part of the method, not optional details.
Check Your Understanding
For each question, state the real domain and explain how the angle range controls the sign of the transformed square root.
- Which substitution and range are suitable for \(\sqrt{25-x^2}\)?
- Which substitution represents all real \(x\)-values for \(\sqrt{4+x^2}\)?
- For \(\sqrt{x^2-36}\), what substitution covers the branch \(x\ge6\), and what substitution covers \(x\le-6\), using first-quadrant angles?
- Why does \(x=a\sec\theta\) give \(\sqrt{x^2-a^2}=a|\tan\theta|\) before a range is specified?
- For \(\sqrt{16-4x^2}\), identify a scaled sine substitution and its corresponding range for \(\theta\).
