If the equality 

${\int }_0^x \frac{bt\cos 4t-a\sin 4t}{t^2}\mathrm{d}t=\frac{a\sin 4x}{x}-1$

holds for all x such that $0<x<\pi /4$ then a and b are given by

${\int }_0^x \frac{bt\cos 4t-a\sin 4t}{t^2}\mathrm{d}t$

$\begin{array}{l}=b{\int }_0^x\frac{\cos \left(4t\right)}{t}dt-a{\int }_0^x\frac{\sin \left(4t\right)}{t^2} dt=b{\int }_0^x \frac{\cos 4t}{t}\mathrm{d}t-a\left[-{\left.\frac{\sin 4t}{t}\right|}_0^x+4{\int }_0^x \frac{\cos 4t}{t}\mathrm{d}t\right]\\ =(b-4a){\int }_0^x \frac{\cos 4t}{t}\mathrm{d}t+\frac{a\sin 4x}{x}-4a\end{array}$

Thus $(b-4a){\int }_0^x\frac{x\cos 4t}{t}dt+ \frac{a\sin 4x}{x}-4a$

$=\frac{a\sin 4x}{x}-1$

i.e. $(b-4a){\int }_0^x\frac{x\cos 4t}{t}dt=4a-1$.

Since R.H.S. is independent of x, so we must have $b-4a=0$ and $4a-1=0$ i.e., $a=1/4,b=1$.