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 $x\ne 0$ .
The value of $\frac{2}{\pi }\underset{-\pi }{\overset{\pi }{\int }}$ $f(x)dx$ 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
$\therefore I=\frac{2}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f\left(x\right)dx$
$=\frac{2}{\pi }\times 2\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx$
$=\frac{4}{\pi }\underset{0}{\overset{\pi }{\int }}\frac{\sin \frac{9x}{2}}{\sin \frac{x}{2}}dx$
$=\frac{4}{\pi }\underset{0}{\overset{\pi }{\int }}\frac{\sin \left(4x+\frac{x}{2}\right)}{\sin \frac{x}{2}}dx$
$=\frac{4}{\pi }\underset{0}{\overset{\pi }{\int }}\frac{\sin 4x\cos \frac{x}{2}+\sin \frac{x}{2}\cos 4x}{\sin \frac{x}{2}}dx$
$=\frac{4}{\pi }\left[\underset{0}{\overset{\pi }{\int }}\left(\sin 4x\cot \frac{x}{2}dx\right)\right]+0$
$\left[\begin{array}{c}\because \underset{0}{\overset{\pi }{\int }}\cos 4xdx={\left[\frac{\sin 4x}{4}\right]}_0^{\pi }=\frac{1}{4}\left(\sin 4\pi -\sin \right)=\frac{1}{4}\left[0-0\right]=0\end{array}\right]=-\frac{1}{4}\left[0-0\right]=0$
$\therefore I=\frac{4}{\pi }\underset{0}{\overset{\pi }{\int }}\sin 4x\cot \frac{x}{2}dx\dots \left(i\right)$
$=\frac{4}{\pi }\underset{0}{\overset{\pi }{\int }}\sin \left(4\pi -4x\right)\cot \left(\frac{\pi -x}{2}\right)dx$
$\Rightarrow I=\frac{4}{\pi }\underset{0}{\overset{\pi }{\int }}-\sin 4x\tan \frac{x}{2}dx\dots \left(ii\right)$
On adding Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$2I=\frac{4}{\pi }\left[\underset{0}{\overset{\pi }{\int }}\sin 4x\left(cot\frac{x}{2}-\tan \frac{x}{2}\right)dx\right]$
$=\frac{4}{\pi }\left[\underset{0}{\overset{\pi }{\int }}\sin 4x\frac{\left({\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}\right)}{\sin \frac{x}{2}\cos \frac{x}{2}}dx\right]$
$=\frac{4}{\pi }\left[\underset{0}{\overset{\pi }{\int }}\frac{\sin 4x\cos x}{\frac{1}{2}\sin x}dx\right]$
$=\frac{8}{\pi }\left[\underset{0}{\overset{\pi }{\int }}\frac{2\sin 2x\cos 2x\cos x}{\sin x}dx\right]$
$=\frac{16}{\pi }\left[\underset{0}{\overset{\pi }{\int }}\frac{2\sin x\cos x\cos 2x\cos x}{\sin x}dx\right]$
$=\frac{32}{\pi }\underset{0}{\overset{\pi }{\int }}{\cos }^{2}x\cos 2xdx$
$=\frac{32}{\pi }\underset{0}{\overset{\pi }{\int }}\left(\frac{\cos 2x+1}{2}\right)\cos 2xdx$
$=\frac{16}{\pi }\left[\underset{0}{\overset{\pi }{\int }}{\cos }^{2}2xdx+\underset{0}{\overset{\pi }{\int }}\cos 2xdx\right]$
$=\frac{16}{\pi }\left[\underset{0}{\overset{\pi }{\int }}\frac{\cos 4x+1}{2}dx+\underset{0}{\overset{\pi }{\int }}\cos 2xdx\right]$
$=\frac{16}{\pi }\left[\frac{1}{2}{\left[\frac{\sin 4x}{4}+x\right]}_0^x+{\left[\frac{\sin 2x}{2}\right]}_0^x\right]$
$=\frac{16}{\pi }\left[\frac{1}{2}\left[0+\pi -0-0\right]+\left[\frac{0-0}{2}\right]\right]=\frac{16}{\pi }\left[\frac{\pi }{2}\right]$
$2I\Rightarrow 8$
$\Rightarrow I=4$