Question

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

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

Answer 1

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