MCQ
$\int\limits_0^\pi {\,\frac{{\sin \left( {n + \frac{1}{2}} \right){\rm{ }}x}}{{\sin x}}} \,dx$, $(n \in N)$ equals
- A$n\pi $
- B$(2n + 1)\frac{\pi }{2}$
- ✓$\pi $
- D$0$
✓
Answer
Correct option: C.
$\pi $
c
(c) $2\sin \frac{x}{2}.\left( {\frac{1}{2} + \cos x + \cos 2x + ..... + \cos nx} \right)$
(c) $2\sin \frac{x}{2}.\left( {\frac{1}{2} + \cos x + \cos 2x + ..... + \cos nx} \right)$
$ = \sin \frac{x}{2} + 2\sin \frac{x}{2}\cos x + 2\sin \frac{x}{2}\cos 2x + .... + 2\sin \frac{x}{2}\cos nx$
$ = \sin \frac{x}{2} + \sin \frac{{3x}}{2} - \sin \frac{x}{2} + \sin \frac{{5x}}{2} - \sin \frac{{3x}}{2} + .....$
$ + \sin \left( {n + \frac{1}{2}} \right)x - \sin \left( {n - \frac{1}{2}} \right)x$$ = \sin \left( {n + \frac{1}{2}} \right)x$
$\therefore$ $\frac{1}{2} + \cos x + \cos 2x + ..... + \cos nx = \frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{{2\sin \left( {\frac{x}{2}} \right)}}$
==> $\int_0^\pi {\frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{{\sin \left( {\frac{x}{2}} \right)}}dx} $
$= 2\left( {\int_0^\pi {\frac{1}{2}dx + \int_0^\pi {\cos xdx + ..... + \int_0^\pi {\cos nx\,dx} } } } \right)$
$ = 2\left( {\frac{\pi }{2} + \sin x + ..... + \frac{{\sin nx}}{n}} \right)_0^\pi = \pi $.
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