AP EAMCET · Maths · Definite Integration
If \(f(t)=\int_0^t \tan ^{(2 n-1)} x d x, n \in N\), then \(f(t+\pi)=\)
- A \(f(t) f(\pi)\)
- B \(\mathrm{f}(\mathrm{t})-\mathrm{f}(\pi)\)
- C \(\mathrm{f}(\mathrm{t})+\mathrm{f}(\pi)\)
- D \(\frac{f(t)}{f(\pi)}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{f}(\mathrm{t})+\mathrm{f}(\pi)\)
Step-by-step Solution
Detailed explanation
\(f(t+\pi) = \int_0^{t+\pi} \tan^{(2n-1)} x dx\) \(f(t+\pi) = \int_0^{\pi} \tan^{(2n-1)} x dx + \int_{\pi}^{t+\pi} \tan^{(2n-1)} x dx\) Let \(y = x - \pi\) in the second integral. Then \(x = y + \pi\) and \(dx = dy\).…
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