Question

Let $f(x)=\frac{\sin x}{x}$, then $\int\limits_0^{\pi / 2} f(x) f\left(\frac{\pi}{2}-x\right) d x=$

• A. $\frac{2}{\pi} \int\limits_0^\pi f(x) d x$
• B. $\int\limits_0^\pi f(x) d x$
• C. $\pi \int\limits_0^\pi f(x) d x$
• D. $\frac{1}{\pi} \int\limits_0^\pi f(x) d x$

Answer 1

Answer: (A)

$I=\int\limits_0^{\pi / 2} \frac{\sin x}{x} \frac{\cos x}{\left(\frac{\pi}{2}-x\right)} d x$
$\pi I =\int\limits_0^{\pi / 2} \sin 2 x\left[\frac{1}{x}+\frac{1}{\pi / 2-x}\right] d x$
$=\int\limits_0^{\pi / 2} \frac{\sin 2 x}{x} d x+\int\limits_0^{\pi / 2} \frac{\sin 2 x}{\pi / 2-x} d x$
$=\int\limits_0^{\pi / 2} \frac{\sin 2 x}{x} d x+\int\limits_0^{\pi / 2} \frac{\sin 2 x}{x} d x=4 \int\limits_0^{\pi / 2} \frac{\sin 2 x}{2 x} d x$
$\frac{\pi I }{2}=\int\limits_0^\pi \frac{\sin t }{ t } dt$
[Put $2 x = t$ ]
$I =\frac{2}{\pi} \int\limits_0^\pi \frac{\sin x }{ x } dx$