If $\int\limits_0^x \frac{b t \cos 4 t-a \sin 4 t}{t^2} d t=\frac{a \sin 4 x}{x}$ for all $x \neq 0$ then $a$ and $b$ are given by

$a=\frac{1}{4}, b=1$

We have,

$\int\limits_0^x \frac{b t \cos 4 t-a \sin 4 t}{t^2} d t=\frac{a \sin 4 x}{x}$

Differentiating both sides w.r. to $x$, we get

$\frac{b x \cos 4 x-a \sin 4 x}{x^2}=\frac{4 a x \cos 4 x-a \sin 4 x}{x^2}$ for all $x \neq 0 $

$\Rightarrow (b-4 a) x \cos 4 x=0$ for all $x \neq 0$

$\Rightarrow b-4 a=0 \Rightarrow b=4 a$

Clearly, $a=\frac{1}{4}$ and $b=1$ satisfy it.