I'd like to fine the value of $\int_{0}^{t} \frac{\cos2x}{\sqrt{t-x}}\, dx$, from my point of view Laplace is the best chose as follows
$$\int_{0}^{t} \frac{\cos2x}{\sqrt{t-x}}\, dx= L(\cos2x\star\frac{1}{\sqrt{x}})$$
Any suggestion? Thanks in advance
Just if you accept to consider other solutions than Laplace.
Let $$\sqrt{t-x}=u \implies x=t-u^2\implies dx=-2 du$$ which make $$\frac{\cos(2x)}{\sqrt{t-x}}\, dx=-2 \cos \left(2 \left(t-u^2\right)\right)\,du$$ Now, expanding $$\cos \left(2 \left(t-u^2\right)\right)=\cos(2t)\cos(2u^2)+\sin(2t)\sin(2u^2)$$ which lead to Fresnel integrals.
Provided $t>0$, the final result should be $$\int_{0}^{t} \frac{\cos(2x)}{\sqrt{t-x}}\, dx=\sqrt{\pi } \left(C\left(\frac{2 \sqrt{t}}{\sqrt{\pi }}\right) \cos (2 t)+S\left(\frac{2 \sqrt{t}}{\sqrt{\pi }}\right) \sin (2 t)\right)$$