Answer: Integral of 1/(1+tan^n x)

Key Idea / Intuition

The substitution $x \mapsto \frac{\pi}{2} - x$ swaps $\tan x$ with $\cot x = 1/\tan x$. When you add the original integral to the substituted version, the two integrands sum to exactly $1$ everywhere on $[0, \pi/2]$. So $2I_n = \pi/2$, regardless of $n$.

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Formal Proof / Solution

Step 1: Apply the King's substitution $x = \frac{\pi}{2} - t$.

Under this substitution, $dx = -dt$, and when $x=0$, $t=\pi/2$; when $x=\pi/2$, $t=0$. Also: $\tan\!\left(\tfrac{\pi}{2} - t\right) = \cot t = \frac{1}{\tan t}$

So: $I_n = \int_{\pi/2}^{0} \frac{1}{1 + \cot^n t}\,(-dt) = \int_0^{\pi/2} \frac{1}{1 + \cot^n t}\,dt$

Step 2: Simplify the new integrand.

$\frac{1}{1 + \cot^n t} = \frac{1}{1 + \dfrac{1}{\tan^n t}} = \frac{\tan^n t}{1 + \tan^n t}$

Step 3: Add the two expressions.

$2I_n = \int_0^{\pi/2} \frac{1}{1+\tan^n x}\,dx + \int_0^{\pi/2} \frac{\tan^n x}{1+\tan^n x}\,dx = \int_0^{\pi/2} \frac{1 + \tan^n x}{1 + \tan^n x}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}$

Step 4: Conclude.

$\boxed{I_n = \frac{\pi}{4}}$

for every $n > 0$, independently of $n$. The result is the same whether $n = 1, 2, 1000$, or $\pi$.

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Why this is surprising: The integrand $\frac{1}{1+\tan^n x}$ changes its shape dramatically as $n$ varies — for large $n$ it becomes nearly a step function jumping at $x = \pi/4$ — yet the area is always exactly $\pi/4$. The symmetry of the interval $[0, \pi/2]$ about its midpoint $\pi/4$ is the hidden reason.