Question

If $\underset{0}{\overset{\infty }{\int }}{\left(\frac{\sin x}{x}\right)}^{3}dx=A$ and $\underset{0}{\overset{\infty }{\int }}\left(\frac{x-\sin x}{x^3}\right)dx=\frac{aA}{b}$, where $a$ and $b$ are relative prime then the value of $(a+b)$ equals

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (C)

$I=\underset{0}{\overset{\infty }{\int }}\frac{x-\sin x}{x^3}dx$
$x\to 3tI=\underset{0}{\overset{\infty }{\int }}\frac{3t-\sin 3t}{27t^3}\cdot 3dt$
$I=\underset{0}{\overset{\infty }{\int }}\frac{3t-3\sin t+4{\sin }^{3}t}{9t^3}dt$
$I=\frac{1}{3}\underset{0}{\overset{\infty }{\int }}\frac{t-\sin t}{t^3}dt+\frac{4}{9}\underset{0}{\overset{\infty }{\int }}\frac{{\sin }^{3}t}{t^3}dt$
$I=\frac{1}{3}I+\frac{4}{9}A$
$\frac{2}{3}I=\frac{4A}{9}\Rightarrow I=\frac{2A}{3}\Rightarrow a+b=5$