<s>[INST] If $\int_ {0}^ \infty (\frac{ \sin x}{x})^3 = A$, then $\int_ {0}^ \infty \frac{ x- \sin x}{x^3} = ??$ 
If $\int_ {0}^ \infty (\frac{ \sin x}{x})^3 = A$, then $\int_ {0}^ \infty \frac{ x- \sin x}{x^3} = k A $. Find the value of k

The main problem is how to get started as in the given integral , there is a term of  $(\sin x)^3$, where as in the integral to be computed , there is no cube. Please help :)
Edit :I'm still in highschool so don't know how to compute this integral and that is why the first one was given
 [/INST] As $\sin^3 x=(3\sin x-\sin 3x)/4$, we have
\begin{align}
A&=\int_{0}^{\infty}\frac{(3x-\sin 3x)-3(x-\sin x)}{4x^3}\,dx
\\ &=\frac{1}{4}\int_{0}^{\infty}\frac{3x-\sin 3x}{x^3}\,dx-\frac{3}{4}\int_{0}^{\infty}\frac{x-\sin x}{x^3}\,dx
\\ &=\frac{9}{4}\int_{0}^{\infty}\frac{y-\sin y}{y^3}\,dy-\frac{3}{4}\int_{0}^{\infty}\frac{y-\sin y}{y^3}\,dy
\\ &=\frac{3}{2}\int_{0}^{\infty}\frac{x-\sin x}{x^3}\,dx.
\end{align}
Thus $k=2/3$.
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