Question

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

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

Answer 1

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

${\sin }^{3}x=\frac{3}{4}\sin x-\frac{1}{4}\sin 3x\therefore {\int }_0^{\infty }\frac{{\sin }^{3}x}{x}\mathrm{dx}=\frac{3}{4}{\int }_0^{\infty }\frac{\sin x}{x}\mathrm{dx}-\frac{1}{4}{\int }_0^{\infty }\frac{\sin 3x}{x}\mathrm{dx}=\frac{3}{4}\cdot \frac{\pi }{2}-\frac{1}{4}{\int }_0^{\infty }\frac{\sin t}{t}\mathrm{dt}\left(\mathrm{Let}3x=t\Rightarrow 3\mathrm{dx}=\mathrm{dt}\Rightarrow \mathrm{dx}=\frac{\mathrm{dt}}{3}\right]=\frac{3}{4}\cdot \frac{\pi }{2}-\frac{1}{4}\cdot \frac{\pi }{2}=\frac{\pi }{4}$