Q113 / 160 · AP EAMCET 2023 (19 May Shift 1) · Mathematics
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$1$
- Correct answerC$0$
- D$\frac{\pi}{2}$
Solution
$S_n=\int_0^{\pi / 2} \frac{\sin (2 n-1) x}{\sin x} d x \quad$ (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
$$
\begin{aligned}
\Rightarrow \quad S_{n+1}-S_n & =\int_0^{\pi / 2} \frac{[\sin (2 n+1) x-\sin (2 n-1) x]}{\sin x} d x \\
\Rightarrow \quad S_{n+1}-S_n & =\int_0^{\pi / 2} \frac{2 \cos 2 n x \cdot \sin x}{\sin x} d x \\
& \left\{\because \sin C-\sin \mathrm{D}=2 \cos \frac{C+D}{2} \cdot \sin \frac{C-D}{2}\right\}
\end{aligned}
$$
$\begin{aligned} \Rightarrow \quad S_{n+1}-S_n & =\frac{1}{n}[\sin 2 n x]_0^{\pi / 2}=\frac{1}{n}[\sin n \pi-\sin 0] \\ & =0\{\text { Since } n \text { is an integer }\}\end{aligned}$
Hence, the correct answer is option: 'C'.