Find a function $f(x)$ such that$\int_0^txf(x)f(\sqrt{t^2-x^2})dx=\sin(t^2)$ I am looking for a function that satisfies the integral equation $\int_0^txf(x)f(\sqrt{t^2-x^2})dx=\sin(t^2)$.  It's some sort of convolution.
If we denote the above equation by $ f(t)*f(t)=\sin(t^2)$ I was also able to show that $f(t)*(2\cos(t^2))=t^2f(t).$
The definition of $ f(t)*g(t)=\int_0^txf(x)g(\sqrt{t^2-x^2})dx$.  
I also got $f(t)*\dfrac{f'(t)}{t}+f(0)f(t)=2\cos(t^2)$ through differentiation ($t>0$). That's all.

Using the hint
$\int_0^t g(X)g(t^2-X)dX=2\sin(t^2)$ we can write $\mathcal{L} \left \{ g(t^2) *g(t^2) \right\}= \mathcal{L} \left \{  2\sin(t^2 ) \right \}$. 
Let $T=t^2$ and $ G(s)=\mathcal{L}\{g(T)\}$. Then  $ G(s)G(s)=\mathcal{L} \left \{  2\sin(T ) \right \}=\dfrac{2}{s^2+1}$.
Therefore $ G(s) =\dfrac{ \sqrt{2}}{\sqrt{s^2+1}} $ and $g(T)=\sqrt{2}\mathcal{L}^{-1} \left \{ \dfrac{1}{\sqrt{s^2+1}} \right \}=\sqrt{2}J_0(T) $.
Thus the answer is $f(t)=g(t^2)=\sqrt{2}J_0(t^2).$  
