Question

If ${\int }_0^{\infty }\frac{sinx}{x}dx=\frac{\pi }{2},then{\int }_0^{\infty }\frac{si{n}^{3}x}{x}dx$ equals

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

Answer 1

Answer: (D)

$\frac{\pi }{4}$

As ${\int }_0^{\infty }\frac{sinx}{x}=\frac{\pi }{2}$
$I={\int }_0^{\infty }\frac{sin3x}{x}dx=\frac{1}{4}{\int }_0^{\infty }\left(\frac{3sinx-sin3x}{x}\right)dx$
$=\frac{3}{4}{\int }_0^{\infty }\frac{sinx}{x}dx-\frac{1}{4}{\int }_0^{\infty }\frac{\sin 3x}{x}dx$
For ${\int }_0^{\infty }\frac{\sin 3x}{x}dx$
Putiing $3x=t\Rightarrow 3dx=dt$ and using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(t\right)dt$
We get ${\int }_0^{\infty }\frac{\sin 3x}{x}dx=3\frac{\pi }{2}$
$\therefore I=\frac{3}{4}\times \frac{\pi }{2}-\frac{3}{4}\times \frac{\pi }{2}\times \frac{1}{3}=\frac{3\pi -\pi }{8}=\frac{2\pi }{8}=\frac{\pi }{4}$