Q116 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics
If $\cos x+\cos 2 x+\ldots+\cos n x=\frac{A(x)}{2 \sin x / 2}$, then $\int_0^\pi A(x) d x=$
- A$\frac{n^2}{n+1}$
- Correct answerB$\frac{-4 n}{2 n+1}$
- C$\frac{2 n}{2 n+1}$
- D$\frac{-n}{2 n+1}$
Solution
$\begin{aligned} & \text { Given, } \cos x+\cos 2 x+\ldots+\cos n x=\frac{A(x)}{2 \sin x / 2} \\ & \therefore \quad A(x)=2 \cos \left(x+\frac{(n-1)}{2} x\right) \sin \frac{n x}{2} \\ & A(x)=2 \cos \left(\frac{n+1}{2}\right) x \sin \frac{n x}{2} \\ & A(x)=\sin \left(\frac{2 n+1}{2}\right) x-\sin \frac{x}{2} \\ & \int_0^\pi A(x) d x=\int_0^\pi\left(\sin \left(\frac{2 n+1}{2}\right) x-\sin \frac{x}{2}\right) d x \\ & =\left[-\frac{2}{2 n+1} \cos \left(\frac{2 n+1}{2}\right) x+2 \cos \frac{x}{2}\right]_0^\pi \\ & =\left(\frac{-2}{2 n+1} \cos \left(\frac{2 n+1}{2}\right) \pi+2 \cos \frac{\pi}{2}\right) \\ & -\left(\frac{-2}{2 n+1} \cos 0+2 \cos 0\right) \\ & =0-\left[\frac{-2}{2 n+1}+2\right]=-\frac{4 n}{2 n+1} \\ & \end{aligned}$
Hence, the correct answer is option: 'B'.