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 \(\mathrm{S}_{\mathrm{n}+1}-\mathrm{S}_{\mathrm{n}}=\)
- A \(-\frac{\pi}{2}\)
- B \(1\)
- C \(0\)
- D \(\frac{\pi}{2}\)
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…
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