How do you work out $\int e^{itx} \sin^2(t) / t^2 dt$?
Given the integral is with respect to $t$, I'm thinking this is about the inversion formula, but the exponent would have to be $-itx$. (In that case, the result would be the density for the sum of two independent uniformly distributed random variables on $[-1, 1]$.)
The other thing is that, if you swap $x$ and $t$, the integral might be a characteristic function, but I don't think $\sin^2(x) / x^2$ can be a density (it integrates to greater than 1).
The Fourier transform of the law of the symmetric uniform distribution on $[-1,1]$ is given by $$\frac12\int^1_{-1}e^{-itx}\,dx=\frac{\sin t}{t}$$
The Fourier transform of the distribution of the sum of two i.i.d symmetric uniform random variables $X_1,X_2$ on $[-1,1]$ is (I leave details to the OP) $$\phi_{X_1+X_2}(t)=E[e^{it(X_1+X_2)}]=\frac{\sin^2}{t^2}$$
Since $\phi_{X_1+X_2}\in L_1(\mathbb{R})$ (I leave details to the OP), the inversion formula states that $$\frac{1}{2\pi}\int^\infty_{-\infty}e^{-itx}\frac{\sin^2}{t^2}\,dt=f(x)$$ where $f$ is the density of the law of $X_1+X_2$, that is (I leave details to the OP) \begin{align} f(x)&=\frac14\big(\mathbb{1}_{(-1,1)}*\mathbb{1}_{(-1,1)}\big)(x)=\frac{1}{4}\int^1_{-1}\mathbb{1}_{(-1,1)}(x-y)\,dy\\ &=\frac14(2-|x|)_+ \end{align}