MCQ
$\int\limits_0^\pi {\frac{{\sin \left( {2n + 1} \right)\,\frac{x}{2}}}{{\sin \frac{x}{2}}}\,\,dx = .......} $
- A$0$
- B$\frac{\pi }{2}$
- ✓$\pi $
- D$2\pi $
✓
Answer
Correct option: C.
$\pi $
$\sin(2n+1)\frac{x}{2}=\sin(2n+1)\frac{x}{2}$
$-\sin(2n-1)\frac{x}{2}+\sin(2n-1)\frac{x}{2}$
$-\sin(2n-3)\frac{x}{2}+...+\sin\frac{3x}{2}-\sin\frac{x}{2}$
$=2\cos\ n x.\sin\frac{x}{2}+2\cos(n-1)x.\sin\frac{x}{2}+....+2 \cos \ x \sin\frac{x}{2}+\sin\frac{x}{2}$
$I=2\int^{\pi}_0\left(\cos \ n \ x+\cos(n-1)x+...+\cos x+\frac{1}{2}\right)dx$
$=2\int^{\pi}_0\left(\cos\ n \ x+\cos(n-1)x+...+\cos x+\frac{1}{2}\right)dx$
$=2\left[\frac{\sin \ nx}{n}+\frac{\sin(n-1)x}{n-1}+.....+\sin x\right]^{\pi}_0+2\left[\frac{x}{2}\right]^{\pi}_0$
$=2(0)+2\left[\frac{\pi}{2}+0\right]$
$-\sin(2n-1)\frac{x}{2}+\sin(2n-1)\frac{x}{2}$
$-\sin(2n-3)\frac{x}{2}+...+\sin\frac{3x}{2}-\sin\frac{x}{2}$
$=2\cos\ n x.\sin\frac{x}{2}+2\cos(n-1)x.\sin\frac{x}{2}+....+2 \cos \ x \sin\frac{x}{2}+\sin\frac{x}{2}$
$I=2\int^{\pi}_0\left(\cos \ n \ x+\cos(n-1)x+...+\cos x+\frac{1}{2}\right)dx$
$=2\int^{\pi}_0\left(\cos\ n \ x+\cos(n-1)x+...+\cos x+\frac{1}{2}\right)dx$
$=2\left[\frac{\sin \ nx}{n}+\frac{\sin(n-1)x}{n-1}+.....+\sin x\right]^{\pi}_0+2\left[\frac{x}{2}\right]^{\pi}_0$
$=2(0)+2\left[\frac{\pi}{2}+0\right]$
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