Question

If ${\int }_0^{\infty }\frac{\sin x}{x}dx=\frac{\pi }{2}$, then ${\int }_0^{\infty }\frac{{\sin }^{3}x}{x}dx$ is equal to

• A. $\pi /2$
• B. $\pi /4$
• C. $\pi /6$
• D. $3\pi /2$

Answer 1

The correct option is A $\pi /2$

We know, ${\sin }^{3}x=\frac{3}{4}\sin x-\frac{1}{4}\sin 3x$

$\Rightarrow {\int }_0^{\infty }\frac{{\sin }^{3}x}{x}dx$

$=\frac{3}{4}{\int }_0^{\infty }\frac{\sin x}{x}dx-\frac{1}{4}{\int }_0^{\infty }\frac{\sin 3x}{x}dx$

$=\frac{3}{4}{\int }_0^{\infty }\frac{\sin x}{x}dx-\frac{1}{4}{\int }_0^{\infty }\frac{\sin u}{u}du(u=3x)$

$=\frac{3}{4}\frac{\pi }{2}-\frac{1}{4}\frac{\pi }{2}=\frac{\pi }{4}$