From a Trigonometric Radical to a Parameter
The substitution \(t=\tan(x/2)\) turns sine and cosine into rational functions of \(t\). A radical involving cosine does not automatically become rational, however: after the substitution, it generally contains a quotient of quadratic expressions under a square root. Seeing that structure clearly helps determine whether the whole transformed integrand is rational, or whether a radical remains.
Throughout, let \(a,b\in\mathbb R\), and consider real \(x\) for which \(a+b\cos x\geq 0\). The substitution is defined when \(x\neq (2k+1)\pi\) for every integer \(k\), because tangent has a pole at those points. On any interval that avoids such a point, it has the familiar differential and trigonometric formulas.
Substituting the cosine formula into the radicand gives
Since \(1+t^2>0\) for every real \(t\), the square root can be written as a quotient of nonnegative square roots wherever the original radicand is nonnegative.
This is the basic half-angle form. The numerator contains a quadratic expression with no linear term, while the denominator contains \(1+t^2\). Depending on \(a\) and \(b\), the numerator may simplify, but the quotient is not generally a rational function.
Domain and Sign Conditions
The formula must be used on a domain where the quantities under the square roots are nonnegative. Because \(1+t^2\) is positive, the condition is exactly
The square root symbol denotes the principal, nonnegative square root. This matters when simplifying a quadratic such as \(t^2\): \(\sqrt{t^2}=|t|\), not \(t\) on the entire real line. On an interval where \(t\geq 0\), that expression does equal \(t\); where \(t\leq 0\), it equals \(-t\).
Worked Example: Rewriting a General Radical
Rewrite \(\sqrt{5+3\cos x}\) in terms of \(t=\tan(x/2)\), and check the result at \(x=0\).
Here \(a=5\), \(b=3\), so \(a+b=8\) and \(a-b=2\). Therefore
For every real \(t\), both quadratic expressions under the square roots are positive, so this expression is defined. At \(x=0\), \(t=\tan 0=0\). The original radical is \(\sqrt{5+3}= \sqrt{8}=2\sqrt{2}\), and the transformed expression is
The two evaluations agree.
When Is the Radical Rational in \(t\)?
A rational function is a quotient of polynomials in \(t\). The expression \(\sqrt{a+b\cos x}\), after substitution, is rational in \(t\) only if the quotient \(\big((a+b)+(a-b)t^2\big)/(1+t^2)\) is the square of a rational function, up to a nonnegative constant factor. For real \(a,b\), this happens in the nontrivial real-valued case precisely when \(b=0\): then the original radicand is the constant \(a\).
Here is a justification of the obstruction when \(b\neq 0\). Work with rational functions over the real numbers, and write \(N(t)=(a+b)+(a-b)t^2\). The polynomial \(1+t^2\) is irreducible over the reals: it has no real root. If \(N(t)\) were divisible by \(1+t^2\), then, because both have degree at most two, \(N\) would have to be a constant multiple of \(1+t^2\). Comparing the constant and \(t^2\) coefficients would require \(a+b=a-b\), hence \(b=0\). Thus, when \(b\neq 0\), the factor \(1+t^2\) does not cancel from \(N(t)/(1+t^2)\).
In a square of a rational function, every irreducible polynomial factor has an even net exponent: squaring doubles every exponent in the numerator and denominator. Here the factor \(1+t^2\) has exponent \(-1\) in the quotient, since it occurs in the denominator and not the numerator. That odd exponent rules out a rational square. This proves the claim. When \(b=0\), the radical is simply the constant \(\sqrt a\), provided \(a\geq 0\); if \(a=0\), it is identically zero.
This conclusion concerns the radical itself. In a full integrand, additional factors may include other radicals that cancel it, or may combine with it to make every factor under a root occur to an even power. But multiplying a nonsquare radical by an ordinary rational function cannot make it rational: if \(R(t)\sqrt{S(t)}\) were rational for a nonzero rational function \(R\), then \(\sqrt{S(t)}\) would also be rational by division by \(R(t)\).
Worked Examples of Simplification and Rationality
Worked Example: The Constant-Radicand Case
Rewrite \(\sqrt{7+0\cos x}\) using the half-angle parameter, including the differential, and check the result at \(x=\pi/3\).
The parameters are \(a=7\), \(b=0\). The radicand is constant, and the transformed expression is
Thus, for an integral containing this factor, the combined radical and differential become
which has a rational function of \(t\) as its variable-dependent factor. To check the radical at \(x=\pi/3\), the original value is \(\sqrt{7+0\cdot\cos(\pi/3)}=\sqrt7\). At this \(x\), \(t=\tan(\pi/6)=1/\sqrt3\). The transformed quotient is
The expression agrees with the original value.
Worked Example: A Coefficient Cancels the Radical
Consider \(\sqrt{2(1+\cos x)}\sec(x/2)\,dx\) on \(-\pi<x<\pi\). Show that its transformed integrand is rational.
On this interval, \(\cos(x/2)>0\), so \(\sqrt{2(1+\cos x)}=2\cos(x/2)\). Also, \(t=\tan(x/2)\), and \(\sec(x/2)=\sqrt{1+t^2}\) on this interval. These identities give
Using \(dx=2\,dt/(1+t^2)\), the complete differential transforms to
This is rational in \(t\). To check the cancellation directly from the general radical formula, set \(a=b=2\). Then its transformed radical is
Multiplying by \(\sec(x/2)=\sqrt{1+t^2}\) gives \(2\), as above. The interval restriction is essential for the stated equality with \(2\cos(x/2)\); it ensures that the half-angle cosine is positive.
As a check at \(x=0\), the original coefficient is \(\sqrt{2(1+1)}\sec 0=2\). Also \(t=0\), so the transformed coefficient before \(dx\) is \(2\), and the differential factor is \(2\,dt\) at \(t=0\). Thus the transformed density is \(4\,dt\), agreeing with \(4/(1+t^2)\,dt\) at \(t=0\).
Worked Example: A Radical Remains Despite the Differential
Transform \(\sqrt{5+3\cos x}\,dx\), and determine whether the result is rational in \(t\).
The radical from the first example is \(\sqrt{2}\sqrt{t^2+4}/\sqrt{t^2+1}\), and \(dx=2\,dt/(1+t^2)\). Therefore
The factor \(2\sqrt2/(1+t^2)\) is rational in \(t\), but the remaining square-root quotient is not. Indeed, here \(b=3\neq0\), so the theorem applies. The differential contributes another rational factor; it does not remove the nonsquare factor from the radical.
At \(x=0\), the original coefficient is \(2\sqrt2\), and \(t=0\). The transformed coefficient, including \(dx/dt\), is
This is twice the original coefficient because the transformed expression is a density with respect to \(dt\): at \(x=0\), \(dx/dt=2\). In fact, the substitution rule requires precisely this factor, so the check confirms both the radical and the differential.
A Practical Test for the Full Integrand
First rewrite the radical using \(t\), then include every other factor and the differential before deciding whether the result is rational. The differential contributes \(2/(1+t^2)\), which is rational, so by itself it cannot change whether the radical is rational. A rational trigonometric coefficient also cannot cancel a nonsquare radical. Cancellation is possible when another algebraic factor supplies the matching square root, as in the preceding example.
Replace \(\cos x\) by \((1-t^2)/(1+t^2)\), and replace other trigonometric factors by their half-angle forms.
Use \(\sqrt{a+b\cos x}=\sqrt{(a+b)+(a-b)t^2}/\sqrt{1+t^2}\) and \(dx=2\,dt/(1+t^2)\).
Check whether another factor cancels the radical. A rational prefactor alone cannot turn a nonsquare rational function under a root into a rational function.
Require the original radicand to be nonnegative and retain absolute values when a square root simplifies to the absolute value of a variable.
The parameter substitution is useful even when a radical remains: it exposes the exact algebraic expression that must be handled next. But it is important not to label an integrand rational merely because its sine and cosine factors have become rational. The radical must also simplify or cancel.
Check Your Understanding
Try each question by writing the transformed expression and checking its domain.
- Use \(t=\tan(x/2)\) to rewrite \(\sqrt{4+2\cos x}\). For which real \(t\) is the radical defined?
- For \(a=6\) and \(b=0\), what does the radical become after substitution, and what factor does \(dx\) contribute?
- Why does a rational function multiplying \(\sqrt{((a+b)+(a-b)t^2)/(1+t^2)}\) not make it rational when \(b\neq0\)?
- On \(-\pi<x<\pi\), simplify \(\sqrt{2(1+\cos x)}\sec(x/2)\), and state why the interval matters.
- When substituting into an integral, why must the factor \(dx=2\,dt/(1+t^2)\) be included before classifying the transformed integrand?
