Question

If \( f(t) = \int_0^t \tan^{2n-1}(x) \, dx \), \( n \in \mathbb{N} \), then \( f(t + \pi) =\)

• A. \( f(t) f(\pi) \)
• B. \( f(t) - f(\pi) \)
• C. \( f(t) + f(\pi) \)
• D. \( \frac{f(t)}{f(\pi)} \)

Hint:

Use the periodicity of trigonometric functions (e.g., \(\tan(x + \pi) = \tan x\)) to simplify definite integrals over shifted intervals.

Answer 1

The Correct Option is C

Given: \[ f(t) = \int_0^t \tan^{2n-1}(x) \, dx \] Compute: \[ f(t + \pi) = \int_0^{t + \pi} \tan^{2n-1}(x) \, dx \] Split the integral: \[ f(t + \pi) = \int_0^\pi \tan^{2n-1}(x) \, dx + \int_\pi^{t + \pi} \tan^{2n-1}(x) \, dx \] The first integral is \( f(\pi) \). For the second, substitute \( u = x - \pi \), so \( du = dx \), with limits from \( x = \pi \) to \( x = t + \pi \), or \( u = 0 \) to \( u = t \): \[ \int_\pi^{t + \pi} \tan^{2n-1}(x) \, dx = \int_0^t \tan^{2n-1}(u + \pi) \, dx \] Since \(\tan(u + \pi) = \tan u\), and \(\tan^{2n-1}(u + \pi) = \tan^{2n-1}(u)\): \[ \int_0^t \tan^{2n-1}(u + \pi) \, du = \int_0^t \tan^{2n-1}(u) \, du = f(t) \] Thus: \[ f(t + \pi) = f(\pi) + f(t) \] Option (3) is correct. Options (1), (2), and (4) do not satisfy the relation.