Reading the Completed Square
A completed square reveals which trigonometric identity can simplify a quadratic radical. The decision depends on the signs in the expression, not on the fact that it is quadratic. Once the square is completed, identify whether the radicand is a positive constant minus a square, a positive constant plus a square, or a square minus a positive constant.
The corresponding substitutions come from the identities \(1-\sin^2\theta=\cos^2\theta\), \(1+\tan^2\theta=\sec^2\theta\), and \(\sec^2\theta-1=\tan^2\theta\). The angle interval matters: it determines the sign of the trigonometric factor when the square root is taken.
The Three Forms and Their Substitutions
Choose the substitution after expressing the radicand in one of these forms. In the table, \(a>0\). The indicated intervals make the square-root expressions nonnegative and keep the substitutions on a single branch.
| Completed-square form | Substitution | Square root on the stated interval |
|---|---|---|
| \(a^2-u^2\), with \(-a<u<a\) | \(u=a\sin\theta\), \(-\pi/2<\theta<\pi/2\) | \(\sqrt{a^2-u^2}=a\cos\theta\) |
| \(a^2+u^2\), for real \(u\) | \(u=a\tan\theta\), \(-\pi/2<\theta<\pi/2\) | \(\sqrt{a^2+u^2}=a\sec\theta\) |
| \(u^2-a^2\), with \(u>a\) | \(u=a\sec\theta\), \(0<\theta<\pi/2\) | \(\sqrt{u^2-a^2}=a\tan\theta\) |
For the third form, the domain also has a negative branch, \(u<-a\). On that branch one may use \(u=-a\sec\theta\) with \(0<\theta<\pi/2\); then \(\sqrt{u^2-a^2}=a\tan\theta\). A substitution must describe the branch of the original variable being considered. At the boundary points \(u=\pm a\), the radical is zero, so formulas involving division by the radical are not defined there.
In each case the differential follows by differentiating the substitution. The scale \(a\) is part of the substitution, so it must also appear in \(du\): respectively, \(du=a\cos\theta\,d\theta\), \(du=a\sec^2\theta\,d\theta\), and \(du=a\sec\theta\tan\theta\,d\theta\) on the positive branch.
Why These Choices Work
The substitutions work because they turn the radicand into a square of a trigonometric function. Here is the complete verification. For the difference form, substitute \(u=a\sin\theta\) and use the identity \(1-\sin^2\theta=\cos^2\theta\). Since \(\cos\theta>0\) on the chosen interval, taking the square root gives \(a\cos\theta\), not \(a|\cos\theta|\) left unresolved.
For the sum form, \(1+\tan^2\theta=\sec^2\theta\), and \(\sec\theta>0\) on \((-\pi/2,\pi/2)\). For the exterior form, \(\sec^2\theta-1=\tan^2\theta\), and both \(\sec\theta\) and \(\tan\theta\) are positive on \((0,\pi/2)\). These identities and signs give the following results.
Worked Example: A Constant Minus a Shifted Square
Worked Example: A Constant Minus a Shifted Square
Choose a substitution for the radical \(\sqrt{25-(2x-1)^2}\) and transform the differential quotient \(dx/\sqrt{25-(2x-1)^2}\).
Set \(u=2x-1\). The radicand is \(25-u^2\), so it has the form \(a^2-u^2\) with \(a=5\). The original radicand is nonnegative when \(-5\leq u\leq5\), or equivalently \(-2\leq x\leq3\). Use the sine substitution \(u=5\sin\theta\), with \(-\pi/2<\theta<\pi/2\).
On this angle interval, \(\cos\theta>0\), so the radical becomes \(5\cos\theta\). Therefore, wherever the quotient is defined,
Check the substitution: \(25-(2x-1)^2=25-25\sin^2\theta=25\cos^2\theta\), whose positive square root is \(5\cos\theta\) on the chosen interval. Differentiating \(x=(1+5\sin\theta)/2\) gives \(dx/d\theta=(5/2)\cos\theta\), as used above.
Worked Example: A Positive-Sign Sum
Worked Example: A Positive-Sign Sum
Choose a substitution for \(\sqrt{9+4(x+1)^2}\) and transform \(dx/\sqrt{9+4(x+1)^2}\).
Set \(u=2(x+1)\). Then the radicand is \(9+u^2\), which has the form \(a^2+u^2\) with \(a=3\). This radicand is positive for every real \(x\). Use \(u=3\tan\theta\), with \(-\pi/2<\theta<\pi/2\), so that \(\theta=\arctan(u/3)\) lies on one-to-one branch.
Since \(\sec\theta>0\) on this interval, the radical is \(3\sec\theta\). The transformed quotient is
Check the radical: \(9+4(x+1)^2=9+9\tan^2\theta=9\sec^2\theta\), so its positive square root is \(3\sec\theta\). Differentiating the stated \(x\) gives \(dx/d\theta=(3/2)\sec^2\theta\), confirming the differential.
Worked Example: A Square Minus a Constant
Worked Example: A Square Minus a Constant
Choose substitutions on both domain branches for \(\sqrt{9(x-2)^2-16}\), and transform \(dx/\sqrt{9(x-2)^2-16}\) on the positive branch.
Set \(u=3(x-2)\). The radicand is \(u^2-16\), so \(a=4\). It is nonnegative when \(|u|\geq4\). Thus the two open intervals where the quotient is defined are \(x>10/3\) and \(x<2/3\), found from \(3(x-2)>4\) and \(3(x-2)<-4\).
On \(x>10/3\), use \(u=4\sec\theta\), \(0<\theta<\pi/2\). Solving for \(x\) and differentiating gives
Consequently,
Check: \(u^2-16=16(\sec^2\theta-1)=16\tan^2\theta\), whose positive square root is \(4\tan\theta\) on this interval. Also, differentiating \(x=2+(4/3)\sec\theta\) gives \(dx/d\theta=(4/3)\sec\theta\tan\theta\).
On the other branch, \(x<2/3\), use \(u=-4\sec\theta\) with \(0<\theta<\pi/2\). Then \(x=2-(4/3)\sec\theta\), which indeed satisfies \(x<2/3\). The same identity gives \(\sqrt{u^2-16}=4\tan\theta\). The negative sign in \(dx=-(4/3)\sec\theta\tan\theta\,d\theta\) records the direction in which \(x\) changes as \(\theta\) increases.
Using the Completed Square Without Repeating the Algebra
In an expression such as \(A(x-h)^2+B\), the completed square already identifies the shift \(h\). If \(A>0\), set \(u=\sqrt{A}(x-h)\); if \(A<0\), set \(u=\sqrt{-A}(x-h)\). These choices convert the squared term into \(u^2\) or \(-u^2\), respectively. Then compare the constant term and the squared term to the three standard forms.
A positive factor can be extracted without changing which substitution matches. For example, \(\lambda(a^2-u^2)\), where \(\lambda>0\), has square root \(\sqrt{\lambda}\sqrt{a^2-u^2}\). The substitution is still \(u=a\sin\theta\); the factor \(\sqrt{\lambda}\) remains outside the trigonometric identity.
Identify the shifted variable and the coefficient multiplying its square.
Rescale the shifted variable so its squared term is \(u^2\), and identify the positive constant \(a\).
Choose sine for \(a^2-u^2\), tangent for \(a^2+u^2\), or secant for \(u^2-a^2\).
Confirm that the original radicand is nonnegative and that the chosen angle interval gives the correct square-root sign.
A Common Mistake: Choosing by Shape Alone
A substitution that contains the right trigonometric function can still be wrong if it does not match the signs. For example, the incorrect choice \(u=5\tan\theta\) for \(25-u^2\) gives
which does not simplify to a square of a basic trigonometric function by the identity \(1+\tan^2\theta=\sec^2\theta\). The correction is to match the difference form \(a^2-u^2\) with \(u=a\sin\theta\); then \(25-u^2=25\cos^2\theta\). Also, if one uses a substitution outside an interval where the relevant trigonometric factor is positive, the square root may be an absolute value rather than the factor itself. State the interval before simplifying the radical.
Why the Match Matters
Choosing from the completed square makes the substitution systematic: the algebra determines the form, the form determines the trigonometric function, and the domain determines the angle branch. This prevents sign errors and preserves the scale factors in the differential. The same checks are useful whenever a radical appears inside a more complicated integrand: first identify the completed-square form, then check the branch before transforming the rest of the expression.
Check Your Understanding
For each question, identify the form first and then state the substitution and any needed domain restriction.
- Which substitution matches \(\sqrt{36-(x-4)^2}\), and what interval for \(\theta\) makes the square root \(6\cos\theta\)?
- For \(\sqrt{7+9(x+2)^2}\), what normalized variable \(u\) and value of \(a\) give the sum form?
- For \(\sqrt{4(x-1)^2-25}\), what are the two domain branches in \(x\), and which substitution applies on the branch \(x>7/2\)?
- Why is \(\sqrt{a^2+u^2}=a\sec\theta\) valid for \(u=a\tan\theta\) on \(-\pi/2<\theta<\pi/2\)?
- What is the mistake in using \(u=a\tan\theta\) for a radicand \(a^2-u^2\)?
