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

Answer: (A)

$\frac{2}{\pi }{\int }_0^{\pi }f\left(x\right)dx$

$l={\int }_0^{n/2}\frac{\sin x}{x}\frac{\cos x}{(\frac{\pi }{2}-x)}dx$

$\pi ={\int }_0^{\pi /2}\sin 2x[\frac{1}{x}+\frac{1}{\pi /2-x}]dx$

$={\int }_0^{\pi /2}\frac{\sin 2x}{x}dx+{\int }_0^{\pi /2}\frac{\sin 2x}{\pi /2-x}dx$

$={\int }_0^{\pi /2}\frac{\sin 2x}{x}dx+{\int }_0^{\pi /2}\frac{\sin 2x}{x}dx=4{\int }_0^{\pi /2}\frac{\sin 2x}{2x}dx$

$\frac{\pi }{2}={\int }_0^{\pi }\frac{\sin t}{t}dt$ [Substitute $2x=t$]

$l=\frac{2}{\pi }{\int }_0^{\pi }\frac{\sin x}{x}dx$