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.

$$\int_0^t f(x)f(\sqrt{t^2-x^2})xdx=\sin(t^2)$$
HINT :
Let $\quad f(x)=g(x^2)$
$$\int_0^t g(x^2)g(t^2-x^2)\frac{d(x^2)}{2}=\sin(t^2)$$
Let $\quad\begin{cases} x^2=X \\t^2=T\end{cases}$
$$\int_0^{\sqrt{T}} g(X)g(T-X)dX=2\sin(T)$$
