If I_n=∫_0^π/4 tan^n x dx then I_2024 +I_2026 is equal to:

The correct answer is Option (1) → \(\displaystyle\frac{1}{2025}\)

Adding \(I_n\) and \(I_{n+2}\) and using \(1 + \tan^2 x = \sec^2 x\):

    \( \displaystyle I_n + I_{n+2} = \int_{0}^{\pi/4} \tan^n x\,(1 + \tan^2 x)\,dx \)

    \( \displaystyle = \int_{0}^{\pi/4} \tan^n x\,\sec^2 x\,dx \)

Let \(t = \tan x\), so \(dt = \sec^2 x\,dx\); the limits 0 and \(\frac{\pi}{4}\) become 0 and 1:

    \( \displaystyle = \int_{0}^{1} t^n\,dt = \frac{1}{n+1} \)

With n = 2024:

    \( \displaystyle I_{2024} + I_{2026} = \frac{1}{2025} \)

Tip: The reduction result \(I_n + I_{n+2} = \frac{1}{n+1}\) for \(I_n = \int_{0}^{\pi/4}\tan^n x\,dx\) is worth remembering; it answers this type of question in one line.