<s>[INST] Integration involving $\sin x/x$ Let $f$ be a differentiable function satisfying:
$$\int_0^{f(x)}f^{-1}(t)dt-\int_0^x(\cos t-f(t))dt=0$$
and $f(\pi)=0$, Considering $g(x)=f(x)\forall x\in\mathbb R_0=\mathbb R+{0}$.
If $$\int_0^{\infty}(g(x))^3dx=A$$ and $$\int_0^{\infty}\frac{1-g(x)}{x^2}dx=\frac{kA}{k+1}$$ then k is?

First I did :
$$\int_0^{f(x)}f^{-1}(t)dt-\int_0^x(\cos t-f(t))dt=0\\\int_0^{f(x)}f^{-1}(t)dt+\int_0^xf(t)dt=\int_0^x\cos tdt\\xf(x)=\sin x$$
So $f(x)=\dfrac{\sin x}x$
But how can someone calculate?
$$\int_0^{\infty}\frac{\sin^3x}{x^3}dx$$
NB, limit to highschool level
 [/INST] 
But how can someone calculate $\quad\quad\quad\displaystyle\int_0^{\infty}\frac{\sin^3x}{x^3}dx$

Assuming that you know $$\int_0^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$$

Use$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\displaystyle\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$

And then Integrate by parts
For example
$$\int\frac{\sin^3xdx}{x^3}=\int\frac{3\sin xdx}{4x^3}-\int\frac{\sin xdx}{4x^3}$$
$$\int\frac{\sin xdx}{x^3}=-\frac{\sin x}{2x^2}-\int\frac{\cos x}{2x^2}dx$$
$$\int\frac{\cos x}{x^2}dx=-\frac{\cos x}{x}+\int\frac{\sin x}{x}dx$$
This method seems reasonable as $\text{Mathematica } 10.0.1.0 $ yields follwing Indefinite Integral
$$\int\frac{\sin^3x}{x^3}dx=\frac{-3 x^2 \text{Si}(x)+9 x^2 \text{Si}(3 x)-4 \sin ^2(x) (\sin (x)+3 x \cos (x))}{8 x^2}+c$$
Pat your back if you get

$$\large\int_0^{\infty}\frac{\sin^3x}{x^3}dx=\frac{3\pi}{8}$$

</s>