∫_3π/8^π/8(tan^2025 x)/(tan^2025 x + cot^2025 x)dx is equal to

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

Let I be the given integral. Its lower limit is larger than its upper limit, and reversing the limits changes the sign:

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

Here \(\frac{\pi}{8} + \frac{3\pi}{8} = \frac{\pi}{2}\). By the property \(\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx\), x is replaced by \(\frac{\pi}{2} - x\); since \(\tan\left(\frac{\pi}{2} - x\right) = \cot x\),

    \(\displaystyle J = \int_{\pi/8}^{3\pi/8} \frac{\cot^{2025} x}{\cot^{2025} x + \tan^{2025} x}\,dx\)

Adding the two forms of J, the integrands add up to 1:

    \(\displaystyle 2J = \int_{\pi/8}^{3\pi/8} 1\,dx = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{\pi}{4}\)

    \(\displaystyle J = \frac{\pi}{8},\quad I = -\frac{\pi}{8}\)

Tip: The limits here run from 3π/8 down to π/8, so the answer is negative. \(\frac{\pi}{8}\), Option (1), is the value with the limits in the usual order.