The value of $\int\limits_0^{π/2}\frac{\tan^{7} x}{\cot^7 x + \tan^7 x}dx$ is

$π/4$

The correct answer is Option (3) → $π/4$

Consider the integral: $I = \int_0^{\pi/2} \frac{\tan^7 x}{\cot^7 x + \tan^7 x} \, dx$

Use symmetry property: $\int_0^{\pi/2} f(x) \, dx = \int_0^{\pi/2} f\left(\frac{\pi}{2}-x\right) \, dx$

Let $x \to \frac{\pi}{2}-x$: $\tan\left(\frac{\pi}{2}-x\right) = \cot x$, $\cot\left(\frac{\pi}{2}-x\right) = \tan x$

Then: $I = \int_0^{\pi/2} \frac{\cot^7 x}{\tan^7 x + \cot^7 x} \, dx$

Add the two forms:

$2I = \int_0^{\pi/2} \frac{\tan^7 x + \cot^7 x}{\tan^7 x + \cot^7 x} \, dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}$

Thus: $I = \frac{\pi}{4}$