Three Unknowns in One Line?

Begin with the substitution $u=\frac{t}{x}$ , yielding :

$\displaystyle \lim_{x\to 0} \int_{0}^1 \frac{x^3u^2 du}{(x-\sin x)\sqrt{a+xu}} =1$

The $xu$ in the square-root is neglected , as it tends to zero. Now work on the limit of $x$ alone:

$\displaystyle \lim_{x\to 0} \frac{x^3}{x-\sin x}$

It is quite easy; Apply L'Hospital three times to get the value of $6$ . So our initial equation became:

$\displaystyle \int_{0}^1 \frac{6u^2 du}{\sqrt{a}} = 1$

Now nothing but easy integration and solving an equation gives $\boxed{a=4}$ .