$\int\limits_{3π/8}^{π/8}\frac{\tan^{2025} x}{\tan^{2025} x + \cot^{2025} x}dx$ is equal to

$-\frac{\pi}{8}$

The correct answer is Option (2) → $-\frac{\pi}{8}$

Given

$I=\displaystyle\int_{3\pi/8}^{\pi/8}\frac{\tan^{2025}x}{\tan^{2025}x+\cot^{2025}x}\,dx$

Reverse limits:

$I=-\displaystyle\int_{\pi/8}^{3\pi/8}\frac{\tan^{2025}x}{\tan^{2025}x+\cot^{2025}x}\,dx$

Let $n=2025$ (odd) and $f(x)=\frac{\tan^{n}x}{\tan^{n}x+\cot^{n}x}$. Using $\tan\!\big(\frac{\pi}{2}-x\big)=\cot x$ gives

$f(x)+f\!\big(\frac{\pi}{2}-x\big)=1$

On $[\frac{\pi}{8},\frac{3\pi}{8}]$ the substitution $x\mapsto \frac{\pi}{2}-x$ maps the interval onto itself, hence

$2\displaystyle\int_{\pi/8}^{3\pi/8}f(x)\,dx=\displaystyle\int_{\pi/8}^{3\pi/8}1\,dx=\frac{\pi}{4}$

Therefore $\displaystyle\int_{\pi/8}^{3\pi/8}f(x)\,dx=\frac{\pi}{8}$ and

$I=-\frac{\pi}{8}$