I am asking to calculate the integral $$\int_{\ell} y \cos x d \ell$$ while $\ell$ is the graph of the function $\phi(x)=sin(x)$ in the domain $x \in [0,\frac{\pi}{2}]$ .
So what I understand is that $\frac{d\ell}{dx}=cos(x)$ so I can substitude it in my integral and getting
$$\int_{0}^{\frac{\pi}{2}} y \cos x\cdot \cos x dx$$
but the answer should be : $$\int_{0}^{\frac{\pi}{2}} \sin x \cos x \sqrt{1+\cos ^{2} x} d x$$
it is not clear for my how the variable $y$ just changed to $\sin x$ and how $d\ell$ calculated
Not quite true. What you call $d\ell/dx$ is, in fact, $d\phi/dx$. The length element $d\ell$ is the length of the graph curve if we step $dx$ in $x$ and $d\phi$ in $y$ directions. It makes (approximately) the right angle triangular, so the hypotenuse $d\ell$ can be calculated from the Pythagoras theorem $$ d\ell=\sqrt{d\phi^2+dx^2}=dx\sqrt{\phi'(x)^2+1}=dx\sqrt{\cos^2(x)+1}. $$ Finally, $y(x)$ along the graph is exactly $\phi(x)=\sin(x)$.