Question

lf ${\int }_0^{\infty }\frac{\sin x}{x}dx=\frac{\pi }{2}$, then ${\int }_0^{\infty }\frac{{\sin }^{3}x}{x}dx$ equals

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

${\int }_0^{\infty }\frac{\sin x}{x}dx=\frac{\pi }{2}$
${\int }_0^{\infty }\frac{{\sin }^{3}x}{x}dx$
$\sin 3x=3\sin x-4{\sin }^{3}x$
${\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}$
${\sin }^{3}x=\frac{1}{4}{\int }_0^{\infty }\frac{3\sin x}{x}dx-{\int }_0^{\infty }\frac{\sin 3x}{3x}d\left(3x\right)$
$=\frac{1}{4}(\frac{3\pi }{2}-\frac{\pi }{2})=\frac{\pi }{4}$
Hence, option D.