If we apply the substitution t = tan x/2 to the integral

If we apply the substitution $t = \tan \frac{x}{2}$ to the integral $\int_0^{2\pi} \frac{dx}{4-3\cos x}$, we get: $$ \int_0^{2\pi} \frac{dx}{4-3\cos x} = \int_0^0 \frac{2dt}{(1+t^2)\left(4-3\cdot\frac{1-t^2}{1+t^2}\right)} = 0 $$ However, the integrand $f(x) = \frac{1}{4-3\cos x}$ is non-negative! Where is the error?

This is an important example illustrating a common error when using the tangent half-angle substitution over the interval $[0, 2\pi]$.

Identifying the error

The error occurs in the limits of integration!

The substitution $t = \tan \frac{x}{2}$ appears to give:

• When $x = 0$: $t = \tan 0 = 0$
• When $x = 2\pi$: $t = \tan \pi = 0$

This leads to $\int_0^0 (\ldots) dt = 0$, which is incorrect!

Why is this wrong? The tangent half-angle substitution $t = \tan \frac{x}{2}$ has a discontinuity at $x = \pi$ because $\tan \frac{\pi}{2}$ is undefined. Therefore, this substitution is not valid over the entire interval $[0, 2\pi]$.

Understanding the discontinuity

The function $t = \tan \frac{x}{2}$ has the following behavior:

• As $x \to \pi^-$: $\frac{x}{2} \to \frac{\pi}{2}^-$, so $t = \tan \frac{x}{2} \to +\infty$
• As $x \to \pi^+$: $\frac{x}{2} \to \frac{\pi}{2}^+$, so $t = \tan \frac{x}{2} \to -\infty$

The substitution "jumps" from $+\infty$ to $-\infty$ at $x = \pi$, so we cannot use it directly over $[0, 2\pi]$.

Correct approach: Split the integral

We must split the integral at $x = \pi$:

\[\int_0^{2\pi} \frac{dx}{4-3\cos x} = \int_0^\pi \frac{dx}{4-3\cos x} + \int_\pi^{2\pi} \frac{dx}{4-3\cos x}\]

For the first integral $\int_0^\pi \frac{dx}{4-3\cos x}$:

Using $t = \tan \frac{x}{2}$:

• $\cos x = \frac{1-t^2}{1+t^2}$
• $dx = \frac{2dt}{1+t^2}$
• When $x = 0$: $t = 0$
• When $x = \pi$: $t = \tan \frac{\pi}{2} \to +\infty$

\[\int_0^\pi \frac{dx}{4-3\cos x} = \int_0^{+\infty} \frac{2dt}{(1+t^2)\left(4-3\cdot\frac{1-t^2}{1+t^2}\right)}\]

Simplifying the denominator:

\[4 - 3\cos x = 4 - 3\cdot\frac{1-t^2}{1+t^2} = \frac{4(1+t^2) - 3(1-t^2)}{1+t^2} = \frac{4 + 4t^2 - 3 + 3t^2}{1+t^2} = \frac{1 + 7t^2}{1+t^2}\]

Therefore:

\[\int_0^{+\infty} \frac{2dt}{(1+t^2) \cdot \frac{1+7t^2}{1+t^2}} = \int_0^{+\infty} \frac{2dt}{1+7t^2}\]

For the second integral $\int_\pi^{2\pi} \frac{dx}{4-3\cos x}$:

Using the same substitution:

• When $x = \pi$: $t \to +\infty$ (from the right, we can think of starting at $-\infty$)
• When $x = 2\pi$: $t = \tan \pi = 0$

Actually, it's easier to use the substitution $u = x - \pi$, so $x \in [\pi, 2\pi]$ becomes $u \in [0, \pi]$.

Then:

\[\int_\pi^{2\pi} \frac{dx}{4-3\cos x} = \int_0^\pi \frac{du}{4-3\cos(u+\pi)} = \int_0^\pi \frac{du}{4-3(-\cos u)} = \int_0^\pi \frac{du}{4+3\cos u}\]

Wait, let me reconsider. By symmetry, since $\cos(2\pi - x) = \cos x$:

\[\int_\pi^{2\pi} \frac{dx}{4-3\cos x} = \int_0^\pi \frac{dx}{4-3\cos x}\]

So both integrals are equal!

Evaluating the integral

We need to evaluate:

\[I = \int_0^{+\infty} \frac{2dt}{1+7t^2} = 2\int_0^{+\infty} \frac{dt}{1+7t^2}\]

Let $u = \sqrt{7}t$, so $du = \sqrt{7}dt$:

\[I = 2\int_0^{+\infty} \frac{1}{\sqrt{7}} \cdot \frac{du}{1+u^2} = \frac{2}{\sqrt{7}} \left[\arctan u\right]_0^{+\infty} = \frac{2}{\sqrt{7}} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{7}} = \frac{\pi\sqrt{7}}{7}\]

Final answer

Since the integral splits into two equal parts:

\[\int_0^{2\pi} \frac{dx}{4-3\cos x} = 2 \cdot \frac{\pi\sqrt{7}}{7} = \frac{2\pi\sqrt{7}}{7}\]

Verification

Let's verify this makes sense:

• The integrand $\frac{1}{4-3\cos x}$ is indeed always positive since $-1 \leq \cos x \leq 1$, so $1 \leq 4-3\cos x \leq 7$
• The integral value is $\frac{2\pi\sqrt{7}}{7} \approx 2.36 > 0$, which is consistent
• The answer is positive, as expected for a positive integrand

Summary of the error

1. What went wrong: The substitution $t = \tan \frac{x}{2}$ was applied over the entire interval $[0, 2\pi]$ without recognizing that it has a discontinuity at $x = \pi$.
2. Why it matters: The function $\tan \frac{x}{2}$ is not continuous on $[0, 2\pi]$. At $x = \pi$, we have $\tan \frac{\pi}{2}$, which is undefined.
3. General principle: When using the tangent half-angle substitution $t = \tan \frac{x}{2}$, it is only valid on intervals where $\frac{x}{2} \neq \frac{\pi}{2} + n\pi$ (i.e., $x \neq \pi + 2n\pi$). For integrals over $[0, 2\pi]$, we must split at $x = \pi$.
4. The incorrect reasoning: The limits appeared to be the same ($0$ to $0$) because $\tan 0 = \tan \pi = 0$, but this ignores the discontinuity in between.

Correct answer:

\[\boxed{\int_0^{2\pi} \frac{dx}{4-3\cos x} = \frac{2\pi\sqrt{7}}{7}}\]

Key lesson: Always check the domain of your substitution function! The tangent half-angle substitution is only valid on intervals of length less than $2\pi$ that don't contain points where $\tan \frac{x}{2}$ is undefined.