Managing Poles in the Half-Angle Substitution

Why poles require care

The substitution $t=\tan(x/2)$ turns trigonometric expressions into rational expressions in $t$. But $t$ is not defined at every real value of $x$, and a single formula for $t$ does not describe an entire interval of arbitrary length in a one-to-one way. Those facts matter especially for definite integrals: the transformed limits must follow the branch actually traversed by $x$.

There are two different kinds of trouble to watch for. First, $t$ is undefined at certain values of $x$, even when the original integrand is perfectly finite there. Second, the transformed rational expression may have a denominator that vanishes at a finite value of $t$. That zero may represent a genuine singularity of the original integrand, or it may cancel after simplification. These cases must not be treated as though they were the same.

Key idea: The half-angle parameter has poles at odd multiples of $\pi$. Split an interval there when necessary, track the parameter limits on each piece, and check finite transformed denominator zeros against the original integrand.

Branches and parameter limits

The basic identities are

\sin x=\frac{2t}{1+t^2}, \qquad \cos x=\frac{1-t^2}{1+t^2}, \qquad dx=\frac{2\,dt}{1+t^2}, \qquad t=\tan\left(\frac{x}{2}\right).

The parameter is undefined when $x/2=\pi/2+k\pi$, or equivalently when $x=(2k+1)\pi$, where $k\in\mathbb Z$. On each open interval between consecutive such points, $t$ is continuous and strictly increasing: wherever it is defined,

\frac{dt}{dx}=\frac12\sec^2\left(\frac{x}{2}\right)>0.

More precisely, on the branch $((2k-1)\pi,(2k+1)\pi)$, the parameter ranges from $-\infty$ to $+\infty$. Its inverse on that branch is

x=2k\pi+2\arctan t,

where $\arctan t$ takes values in $(-\pi/2,\pi/2)$. Thus an endpoint must be converted using the branch containing the interval, not by applying a single inverse formula regardless of the $x$-values.

For example, on the branch $(\pi,3\pi)$, $x=2\pi+2\arctan t$. Therefore $x=3\pi/2$ corresponds to $t=-1$, not $t=1$. The tangent function repeats its values, but the increasing parameter on a particular branch still determines the correct direction and limits.

When a definite integral crosses a pole

If an interval crosses $x=(2k+1)\pi$, the substitution cannot be applied across the whole interval as one ordinary change of variable. Split the original interval at that point. The left-hand piece approaches $t=+\infty$; the right-hand piece begins at $t=-\infty$. If the original integrand is finite at the split point, these are two improper parameter integrals whose values are added.

Definition: An integral split at an interior point $c$ is interpreted as the sum of the one-sided improper integrals, if both limits exist finitely: $$ \int_a^b f(x)\,dx = \lim_{u\to c^-}\int_a^u f(x)\,dx + \lim_{v\to c^+}\int_v^b f(x)\,dx. $$ If either one-sided integral diverges, the ordinary improper integral does not converge. A cancellation between divergent sides does not make it converge.

This definition is important here even when $f$ is continuous at $c$: it prescribes how the change of variable is performed. The two parameter tails must be evaluated separately, with their own limits.

Worked Example: Crossing a coordinate pole where the integrand is finite

Evaluate $\displaystyle\int_0^{2\pi}\frac{dx}{2+\cos x}$. The original denominator is at least $1$, so the integrand is continuous throughout the interval, including at $x=\pi$. The parameter, however, is undefined there. Split at $x=\pi$.

Using the half-angle identities, the transformed integrand is

\frac{dx}{2+\cos x} = \frac{\frac{2\,dt}{1+t^2}} {2+\frac{1-t^2}{1+t^2}} = \frac{2\,dt}{3+t^2}.

On $0\le x<\pi$, $t$ increases from $0$ to $+\infty$. On $\pi<x\le2\pi$, it increases from $-\infty$ to $0$. Therefore

\int_0^{2\pi}\frac{dx}{2+\cos x} = \int_0^\infty\frac{2\,dt}{t^2+3} + \int_{-\infty}^0\frac{2\,dt}{t^2+3}.

An antiderivative in $t$ is $\frac{2}{\sqrt3}\arctan(t/\sqrt3)$, since its derivative is $2/(t^2+3)$. Evaluating the two limits separately gives

\begin{aligned} \int_0^\infty\frac{2\,dt}{t^2+3} &= \frac{2}{\sqrt3}\left(\frac{\pi}{2}-0\right) =\frac{\pi}{\sqrt3},\\ \int_{-\infty}^0\frac{2\,dt}{t^2+3} &= \frac{2}{\sqrt3}\left(0-\left(-\frac{\pi}{2}\right)\right) =\frac{\pi}{\sqrt3}. \end{aligned}

Adding the two pieces gives $\boxed{2\pi/\sqrt3}$. The antiderivative check confirms the transformed integrand, and the limits show that both improper pieces converge. In particular, the left and right sides of $x=\pi$ correspond to different infinite ends of the parameter line.

Zeros in a transformed denominator

After substitution and simplification, a rational integrand may have a denominator that vanishes at some finite $t=t_0$. On a branch centered at $2k\pi$, that parameter value corresponds to $x=2k\pi+2\arctan(t_0)$. Check the original integrand at that $x$. If its denominator vanishes and no cancellation removes the singularity, the original integral must be treated as improper there. If factors cancel, simplify and inspect the resulting expression before deciding whether a singularity remains.

Conversely, a pole in $t$ at $t=\infty$ is not automatically a singularity of the original integrand. Example 1 showed that a change-of-variable pole can occur at an entirely regular point of the original integrand. It is the behavior of the original function, or equivalently the convergence of the correctly transformed one-sided integrals, that determines whether the integral exists.

Worked Example: A finite transformed zero that gives divergence

Consider $\displaystyle\int_0^{\pi/2}\frac{dx}{1-\sin x}$. The denominator of the original integrand tends to zero at the upper endpoint. Under the substitution, $t$ increases from $0$ to $1$, and

\frac{dx}{1-\sin x} = \frac{\frac{2\,dt}{1+t^2}} {1-\frac{2t}{1+t^2}} = \frac{2\,dt}{(t-1)^2}.

The transformed denominator vanishes at $t=1$, which corresponds to $x=\pi/2$. Hence the integral is improper. For $0\le b<1$,

\int_0^b\frac{2\,dt}{(t-1)^2} = \left[-\frac{2}{t-1}\right]_0^b = -\frac{2}{b-1}-2.

As $b\to1^-$, $-2/(b-1)\to+\infty$, so the integral diverges to $+\infty$. This agrees with the original integrand: $1-\sin x$ tends to zero as $x\to\pi/2^-$, while the numerator remains positive. The finite denominator zero is a genuine singularity, not a coordinate pole.

A branch example without crossing a pole

It is possible for an interval to lie wholly on a branch other than $(-\pi,\pi)$. The branch index affects how $x$ is recovered, even though the trigonometric identities in $t$ are unchanged.

Worked Example: Using the branch centered at $2\pi$

Evaluate $\displaystyle\int_{3\pi/2}^{5\pi/2}\frac{dx}{2+\cos x}$. This interval is contained in the branch $(\pi,3\pi)$, whose inverse relation is $x=2\pi+2\arctan t$. Its parameter endpoints are

t(3\pi/2)=\tan(3\pi/4)=-1, \qquad t(5\pi/2)=\tan(5\pi/4)=1.

As in the first example, the transformed integrand is $2/(t^2+3)$. Thus

\begin{aligned} \int_{3\pi/2}^{5\pi/2}\frac{dx}{2+\cos x} &=\int_{-1}^1\frac{2\,dt}{t^2+3}\\ &=\left[\frac{2}{\sqrt3}\arctan\left(\frac{t}{\sqrt3}\right)\right]_{-1}^1\\ &=\frac{4}{\sqrt3}\arctan\left(\frac{1}{\sqrt3}\right) =\frac{2\pi}{3\sqrt3}. \end{aligned}

Differentiating the displayed antiderivative gives $2/(t^2+3)$, and the endpoint values are $t=-1$ and $t=1$, in increasing order. The result is therefore $\boxed{2\pi/(3\sqrt3)}$. This interval does not cross a pole of the substitution, but identifying its branch prevents assigning incorrect parameter limits.

Why the split procedure works

The justification is the ordinary change-of-variable rule applied only where the substitution is defined and one-to-one, followed by one-sided limits at the boundaries. On every branch interval, $t$ is differentiable with positive derivative, so the usual definite-integral substitution applies to any closed subinterval contained in that branch. The full improper integral is then the limit of those subinterval integrals.

Choose branch intervals.
Mark every point $x=(2k+1)\pi$ inside the original interval. These are precisely where $t$ is undefined.

Transform each piece.
On each piece, use the trigonometric identities and $dx=2\,dt/(1+t^2)$. The parameter is monotone there, so the endpoint order is preserved.

Take one-sided limits.
At a split point, use $+\infty$ on the left and $-\infty$ on the right. Evaluate each improper integral separately.

Check finite denominator zeros.
Translate each finite zero back to $x$, then determine whether the original integrand has a genuine singularity or whether the apparent singularity cancels.

Because the rule is applied on each branch before taking limits, it does not assume that $t$ remains defined or one-to-one across a pole. If the original integrand is continuous at such a point, the split integrals represent the two sides of an ordinary finite integral. If the original integrand is singular there, convergence must be established separately on both sides.

Common mistake: using the same parameter endpoint twice

A tempting but incorrect calculation is to substitute into $\int_0^{2\pi} dx/(2+\cos x)$ and claim that the parameter starts and ends at $t=0$, so the transformed integral has equal limits and equals zero. This calculation is wrong:

\text{Incorrect:}\qquad \int_0^{2\pi}\frac{dx}{2+\cos x} \overset{\text{wrong}}{=} \int_0^0\frac{2\,dt}{t^2+3} =0.

The parameter does equal $0$ at both endpoints, but it does not travel directly from $0$ back to $0$. It increases without bound as $x\to\pi^-$, then restarts from $-\infty$ as $x\to\pi^+$. The correct calculation is the sum of the two separate integrals from $0$ to $+\infty$ and from $-\infty$ to $0$, which equals $2\pi/\sqrt3$. Equal endpoint values do not justify a single substitution across a point where the substitution is undefined.

Check Your Understanding

Use branch intervals and one-sided limits where needed. For each question, decide whether a transformed denominator zero represents a genuine singularity.

At which values of $x$ is $t=\tan(x/2)$ undefined?
On the branch $(\pi,3\pi)$, what are the parameter limits for $x$ from $3\pi/2$ to $5\pi/2$?
Why must an interval crossing $x=\pi$ be split when using the half-angle substitution?
For $\int_0^{\pi/2} dx/(1+\sin x)$, the transformed denominator is proportional to $(t+1)^2$. Does its zero lie in the parameter interval for this integral?
Why is it not valid to infer that an integral is zero merely because its parameter has equal values at the two original endpoints?

Managing Poles in the Half-Angle Substitution

What You'll Learn

Why poles require care

Branches and parameter limits

When a definite integral crosses a pole

Worked Example: Crossing a coordinate pole where the integrand is finite

Worked Example: Crossing a coordinate pole where the integrand is finite

Zeros in a transformed denominator

Worked Example: A finite transformed zero that gives divergence

Worked Example: A finite transformed zero that gives divergence

A branch example without crossing a pole

Worked Example: Using the branch centered at \(2\pi\)

Why the split procedure works

Common mistake: using the same parameter endpoint twice

Check Your Understanding

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Vol. 2: Solutions Manual

Managing Poles in the Half-Angle Substitution

What You'll Learn

Why poles require care

Branches and parameter limits

When a definite integral crosses a pole

Worked Example: Crossing a coordinate pole where the integrand is finite

Worked Example: Crossing a coordinate pole where the integrand is finite

Zeros in a transformed denominator

Worked Example: A finite transformed zero that gives divergence

Worked Example: A finite transformed zero that gives divergence

A branch example without crossing a pole

Worked Example: Using the branch centered at \(2\pi\)

Why the split procedure works

Common mistake: using the same parameter endpoint twice

Check Your Understanding

Test yourself: Calculus II final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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