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AP EAMCET · Maths · Definite Integration

If \(S_n=\int_0^{\frac{\pi}{2}} \frac{\sin (2 n-1) x}{\sin x} d x\) and \(n\) is an integer, then \(\mathrm{S}_{\mathrm{n}+1}-\mathrm{S}_{\mathrm{n}}=\)

  1. A \(-\frac{\pi}{2}\)
  2. B \(1\)
  3. C \(0\)
  4. D \(\frac{\pi}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0\)

Step-by-step Solution

Detailed explanation

\(S_n=\int_0^{\pi / 2} \frac{\sin (2 n-1) x}{\sin x} d x \quad\) (given) Hence \(S_{n+1}=\int_0^{\pi / 2} \frac{\sin (2 n+1) x}{\sin x} d x\) \(\{\because\) Replacing 'n' by \((n+1)\}\) Therefore…