Question

Let $ f $ be the function $ [-\pi, \pi] $ given by $ f(0) = 9 \,$ and $ f(x) = \sin (\frac {9x}{2})/\sin(\frac {x}{2}) $ for $ x \neq 0 $ .
The value of $ \frac{2}{\pi} \int\limits^{\pi}_{-\pi} $ $ f(x) dx $ is

• A. 0
• B. 4
• C. 8
• D. None of these

Answer 1

Answer: (B)

Given, $f \left(0\right)=9$ and $f \left(x\right)=\frac{\sin\left(\frac{9x}{2}\right)}{\sin\frac{x}{2}}$
Now, $f \left(-x\right)=\frac{\sin\left(-\frac{9x}{2}\right)}{\sin \left(-\frac{x}{2}\right)} =\frac{-\sin \left(\frac{9x}{2}\right)}{-\sin \left(\frac{x}{2}\right)}=f \left(x\right)$
Hence, $f\left(x\right)$ is an even function
$\therefore I=\frac{2}{\pi} \int\limits_{-\pi}^{\pi}f \left(x\right)dx $
$=\frac{2}{\pi}\times2 \int\limits^{\pi}_{0} f \left(x\right)dx$
$=\frac{4}{\pi} \int\limits_{0}^{\pi} \frac{\sin \frac{9x}{2}}{\sin \frac{x}{2}}dx$
$=\frac{4}{\pi} \int\limits_{0}^{\pi} \frac{\sin\left(4x+\frac{x}{2}\right)}{\sin \frac{x}{2}} dx $
$=\frac{4}{\pi} \int\limits_{0}^{\pi} \frac{\sin\,4x\,\cos \frac{x}{2}+\sin \frac{x}{2} \cos \,4x}{\sin \frac{x}{2}}dx$
$=\frac{4}{\pi} \left[\int\limits_{0}^{\pi}\left(\sin\,4x\,\cot \frac{x}{2} dx\right)\right]+0$
$\left[\begin{matrix}\because \int\limits_{0}^{\pi}\cos\,4x\,dx=\left[\frac{\sin\,4x}{4}\right]_{0}^{\pi}=\frac{1}{4}\left(\sin\,4\pi-\sin\right)=\frac{1}{4}\left[0-0\right]=0\end{matrix}\right]=-\frac{1}{4}\left[0-0\right]=0$
$\therefore I=\frac{4}{\pi}\int\limits_{0}^{\pi} \sin\,4x\,\cot \frac{x}{2} dx \ldots\left(i\right)$
$=\frac{4}{\pi} \int\limits_{0}^{\pi} \sin \left(4\pi-4x\right)\cot \left(\frac{\pi-x}{2}\right)dx$
$\Rightarrow I=\frac{4}{\pi} \int\limits_{0}^{\pi} -\sin \,4x \, \tan \frac{x}{2} dx \ldots\left(ii\right)$
On adding Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$2I=\frac{4}{\pi} \left[\int\limits_{0}^{\pi} \sin\,4x\left(cot \frac{x}{2}-\tan\frac{x}{2}\right)dx\right]$
$=\frac{4}{\pi}\left[\int\limits_{0}^{\pi}\sin\,4x \frac{\left(\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}\right)}{\sin \frac{x}{2}\cos \frac{x}{2}} dx\right]$
$=\frac{4}{\pi}\left[\int\limits_{0}^{\pi}\frac{\sin\,4x\,\cos\,x}{\frac{1}{2}\sin\,x}dx\right]$
$=\frac{8}{\pi} \left[\int\limits_{0}^{\pi} \frac{2\,\sin\,2x\,\cos\,2x\,\cos\,x}{\sin\,x}dx\right]$
$=\frac{16}{\pi}\left[\int\limits_{0}^{\pi}\frac{2\,\sin\,x\,\cos\,x\,\cos\,2x\,\cos\,x}{\sin\,x}dx\right]$
$=\frac{32}{\pi} \int\limits_{0}^{\pi}\cos^{2}\,x\, \cos\,2x\,dx$
$=\frac{32}{\pi}\int\limits_{0}^{\pi} \left(\frac{\cos\,2x+1}{2}\right)\cos\,2x\,dx$
$=\frac{16}{\pi}\left[\int\limits_{0}^{\pi}\cos^{2}\,2x\,dx+\int\limits_{0}^{\pi}\cos\,2x\,dx\right]$
$=\frac{16}{\pi} \left[\int\limits_{0}^{\pi} \frac{\cos\,4x+1}{2}dx+\int\limits_{0}^{\pi} \cos\,2x\,dx\right]$
$=\frac{16}{\pi}\left[\frac{1}{2}\left[\frac{\sin\,4x}{4}+x\right]_{0}^{x}+\left[\frac{\sin\,2x}{2}\right]_{0}^{x}\right]$
$=\frac{16}{\pi}\left[\frac{1}{2}\left[0+\pi-0-0\right]+\left[\frac{0-0}{2}\right]\right]=\frac{16}{\pi}\left[\frac{\pi}{2}\right]$
$2I \Rightarrow 8$
$\Rightarrow I=4$