Why Branches Matter
A substitution involving a square root is not reversible merely because algebra produces a formula for the original variable. Squaring an equation can discard its sign information, and the resulting formula may describe parameter values that do not belong to the original substitution. The correct parameter interval is part of the substitution.
For an Euler substitution, tracking that interval does several jobs at once. It tells us whether the inverse formula is defined, whether it returns the intended interval of the original variable, and whether the reconstructed square root has the correct sign. These checks are especially important when the radical is real on two separate intervals or when a denominator vanishes at a boundary.
Squaring Requires a Sign Check
Write the quadratic under the radical as \(Q(x)=ax^2+bx+c\), and consider only values of \(x\) for which \(Q(x)\geq 0\). One common Euler substitution, used when \(a>0\), is \(t=\sqrt{Q(x)}+\sqrt{a}\,x\). Rearranging it gives \(\sqrt{Q(x)}=t-\sqrt{a}\,x\). Squaring this equation produces an algebraic relation, but squaring alone cannot ensure that \(t-\sqrt{a}\,x\) is nonnegative. That sign condition must still be checked.
In general, if \(u=\sqrt{Q(x)}\), then \(u\geq 0\). An equation derived by squaring \(u=v\) gives \(u^2=v^2\), which implies \(u=|v|\), not necessarily \(u=v\). Thus a proposed inverse must satisfy both the squared equation and the original sign requirement.
A General Reversibility Check
For the substitution \(t=\sqrt{Q(x)}+\sqrt{a}\,x\), where \(a>0\), expand the square of \(t-\sqrt{a}\,x=\sqrt{Q(x)}\). This gives
Cancel \(ax^2\) and collect the terms involving \(x\):
Consequently, when \(2\sqrt{a}\,t+b\ne 0\), the candidate inverse is \(x=(t^2-c)/(2\sqrt{a}\,t+b)\). This algebra gives a candidate, not yet a complete inverse: the denominator must be nonzero on the chosen parameter interval, and the unsquared relation \(\sqrt{Q(x)}=t-\sqrt{a}\,x\) must hold there.
Specify where \(Q(x)\geq 0\), and choose one interval at a time if that set is disconnected.
Use the original substitution, including its sign, to determine which values of \(t\) actually occur.
Exclude any parameter values that make its denominator zero, and verify that the inverse returns values in the chosen \(x\)-interval.
Substitute the inverse into the original, unsquared equation and confirm the nonnegative square-root value.
Worked Examples
Worked Example: A Positive-Range Parameter
Use \(t=\sqrt{x^2+1}+x\) and determine its range and inverse for \(x\in\mathbb{R}\). Then verify the radical branch.
Rearrange the original equation as \(\sqrt{x^2+1}=t-x\), and square:
Cancel \(x^2\), so \(1=t^2-2tx\). If \(t\ne 0\), solve for \(x\):
Now recover the radical from the original, unsquared equation:
This expression is positive precisely when \(t>0\). For every \(x\in\mathbb{R}\), \(\sqrt{x^2+1}>|x|\), so \(t=\sqrt{x^2+1}+x>0\). Conversely, for \(t>0\), the inverse formula gives a real \(x\), and the recovered radical \((t^2+1)/(2t)\) is positive. Its square is \(x^2+1\), as required. Thus the range is \(t\in(0,\infty)\), and the inverse is valid throughout that range.
As a direct composition check, substituting the inverse and reconstructed radical into the substitution gives
The denominator \(2t\) does not vanish on \(t>0\), so there is no excluded parameter value within the range.
Worked Example: A Substitution That Extends Across Zero
Let \(Q(x)=x^2+3\) and use \(t=(\sqrt{x^2+3}-\sqrt{3})/x\) for \(x\ne 0\). Find the range, extend the substitution at \(x=0\), and verify its inverse.
Rationalizing the numerator gives
The right-hand expression tends to \(0\) as \(x\) tends to \(0\), so define \(t=0\) at \(x=0\). From the original relation \(tx+\sqrt{3}=\sqrt{x^2+3}\), squaring gives
and hence \(x[(t^2-1)x+2\sqrt{3}\,t]=0\). For \(x\ne0\), the inverse is \(x=2\sqrt{3}\,t/(1-t^2)\), provided \(t^2\ne1\). At \(x=0\), this same formula gives \(x=0\) when \(t=0\).
To see the parameter range and check the branch, take any \(-1<t<1\) and use the inverse. Its denominator is positive, and the reconstructed radical is
The square of this expression equals \(x^2+3\): both are \(\displaystyle 3(1+t^2)^2/(1-t^2)^2\). Thus it is the required square root. Differentiating the inverse gives \(dx/dt=2\sqrt{3}(1+t^2)/(1-t^2)^2>0\) on \((-1,1)\). Also, as \(t\) tends to \(-1\) from above, \(x\) tends to \(-\infty\), and as \(t\) tends to \(1\) from below, \(x\) tends to \(+\infty\). The inverse therefore covers \(\mathbb{R}\) once, and the parameter range is \((-1,1)\).
Finally, substituting the inverse and recovered radical into the original relation reproduces \(t\), since
when \(t\ne0\); the extended definition handles \(t=0\). The inverse denominator \(1-t^2\) vanishes at \(t=\pm1\), but neither endpoint belongs to the range.
Worked Example: Separate Branches for a Real-Root Quadratic
For \(Q(x)=x^2-1\), use \(t=\sqrt{x^2-1}/(x-1)\) on the intervals where it is defined. Determine the parameter ranges and verify the inverse on each branch.
The radical is real for \(x\leq-1\) or \(x\geq1\), but the substitution is undefined at \(x=1\). Squaring \(t(x-1)=\sqrt{x^2-1}\) gives \(t^2(x-1)^2=x^2-1\). Since \(x^2-1=(x-1)(x+1)\), when \(x\ne1\) and \(t^2\ne1\), cancellation and rearrangement yield
On \(x<-1\), both \(x-1\) and \(\sqrt{x^2-1}\) are respectively negative and nonnegative, so \(t<0\). Moreover, \(t^2=(x+1)/(x-1)\) lies in \([0,1)\), giving \(-1<t\leq0\). The endpoint \(x=-1\) corresponds to \(t=0\); as \(x\) tends to \(-\infty\), \(t\) tends to \(-1\), which is not attained.
On \(x>1\), \(t>0\) and \(t^2=(x+1)/(x-1)>1\), so \(t>1\). The value \(t=1\) is not attained at any finite \(x\), and it makes the inverse denominator zero. Therefore this branch has parameter range \((1,\infty)\).
Check the radical using the inverse:
For \(-1<t\leq0\), the numerator and denominator in the last fraction are both nonpositive, so the result is nonnegative. For \(t>1\), both are positive. Its square equals \(x^2-1\), since
Thus \(t(x-1)=\sqrt{x^2-1}\) on both parameter ranges. The two ranges remain separate: the inverse has a pole at \(t=-1\) and \(t=1\), neither of which belongs to either range.
A Common Branch Error
A frequent mistake is to keep the inverse formula but discard the range restriction found from the original substitution. In the first example, someone might use \(x=(t^2-1)/(2t)\) for every nonzero real \(t\). Taking \(t=-1\) gives \(x=0\), and the squared equation is satisfied because \(t^2=1\). But the original substitution at \(x=0\) gives \(t=\sqrt{1}+0=1\), not \(-1\). In fact, the proposed recovered radical is \((t^2+1)/(2t)=-1\), which cannot equal \(\sqrt{x^2+1}=1\).
Why Reversibility Matters in Integration
An Euler substitution is used to replace an algebraic radical by rational expressions in a parameter. That replacement is useful only on a stated branch where the substitution and its inverse agree. If the original variable ranges across a point where the formula is undefined, or across two intervals with different parameter ranges, one must treat the intervals separately or use a suitable limiting argument at an endpoint.
When an antiderivative is obtained in the parameter, converting it back to \(x\) also uses the inverse formula. The branch checks establish which values of the parameter are allowed and that the radical in the resulting expression has its original meaning. They also prevent apparent poles in the inverse from being mistaken for valid points of the substitution. A denominator zero outside the range is irrelevant; a denominator zero inside the proposed range means the inverse formula cannot be used there without further analysis.
Check Your Understanding
For each question, keep track of the original unsquared relation as well as any equation obtained by squaring.
- For \(t=\sqrt{x^2+4}+x\), what sign must \(t\) have, and why?
- In the substitution \(t=(\sqrt{x^2+3}-\sqrt{3})/x\), why is it legitimate to define a value at \(x=0\) by using the rationalized expression?
- For \(t=\sqrt{x^2-1}/(x-1)\), why must the domains \(x<-1\) and \(x>1\) be tracked separately?
- What two checks are required after squaring an equation involving \(\sqrt{Q(x)}\)?
- If an inverse denominator vanishes at a parameter value, what must be determined before deciding whether that value causes a problem?
