Question

Let $f\left(x\right)=\frac{\sin x}{x}$, then ${\int }_0^{\pi /2}f\left(x\right)f(\frac{\pi }{2}-x)dx=$

• A. $\frac{2}{\pi }{\int }_0^{\pi }f\left(x\right)dx$
• B. ${\int }_0^{\pi }f\left(x\right)dx$
• C. $\pi {\int }_{-0}^{\pi }f\left(x\right)dx$
• D. $\frac{1}{\pi }{\int }_0^{\pi }f\left(x\right)dx$

Answer 1

The correct option is A $\frac{2}{\pi }{\int }_0^{\pi }f\left(x\right)dx$

Given : $f\left(x\right)=\frac{\sin x}{x}{\int }_0^{\frac{\pi }{2}}f\left(x\right).f(\frac{\pi }{2}-x)dx$

$I={\int }_0^{\frac{\pi }{2}}f\left(x\right).f(\frac{\pi }{2}-x)dxI={\int }_0^{\frac{\pi }{2}}\frac{\sin x}{x}.\frac{\sin (\frac{\pi }{2}-x)}{(\frac{\pi }{2}-x)}={\int }_0^{\frac{\pi }{2}}\frac{\sin x\cos xdx}{x(\frac{\pi }{2}-x)}I={\int }_0^{\frac{\pi }{2}}\frac{2\sin x\cos xdx}{x(\pi -2x)}={\int }_0^{\frac{\pi }{2}}\frac{\sin 2xdx}{x(\pi -2x)}Let2x=tx\to 0t\to 02dx=dtx\to \frac{\pi }{2}t\to \pi I={\int }_0^{\pi }\frac{\sin tdt}{\frac{t}{2}[\pi -t]}={\int }_0^{\pi }\frac{\sin tdt}{t[\pi -t]}I=\left[{\int }_0^{\pi }\sin t[\frac{1}{t}+\frac{1}{(\pi -t)}]dt\right]I=\left[{\int }_0^{\pi }\frac{\sin tdt}{t}+{\int }_0^{\pi }\frac{\sin t}{(\pi -t)}dt\right]I=\left[{\int }_0^{\pi }\frac{\sin tdt}{t}+{\int }_0^{\pi }\frac{\sin tdt}{t}\right][{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx]I=2{\int }_0^{\pi }\frac{\sin tdt}{t}=2{\int }_0^{\pi }f\left(x\right)dx$