Help with some double integral $$\int_{0}^{\pi/2}\int_{x}^{\pi/2}\frac{\cos y}{y} \, \operatorname{d}\!y\, \operatorname{d}\!x$$
$\iint_R2xy^2 \, \operatorname{d}\!A$ where R is the right half of the unit circle

That first one doesn't converge. Note that if it did, $$\begin{align}\int_0^{\pi/2}\frac{\cos y}y\,dy &= \int_0^{\pi/3}\frac{\cos y}y\,dy+\int_{\pi/3}^{\pi/2}\frac{\cos y}y\,dy\\ &\geq \int_0^{\pi/3}\frac{\cos y}y\,dy\\ &\geq \frac12\int_0^{\pi/3}\frac1y\,dy\\ &=\frac12\lim_{a\to 0^+}\int_a^{\pi/3}\frac1y\,dy.\end{align}$$ Check out that limit.
For the second one, note that in polar coordinates, the right half of the unit circle is $0\leq r\leq 1,-\frac\pi2\leq\theta\leq\frac\pi2.$ Use $x=r\cos\theta,y=r\sin\theta,dA=r\,dr\,d\theta$ to evaluate the second integral.

Since you've fixed your post, note that $$\int_0^{\pi/2}\int_x^{\pi/2}\frac{\cos y}y\,dy\,dx$$ is integrating the function $\frac{\cos y}y$ over the region bounded below by the line $y=x$ and above by the line $y=\frac\pi2,$ with $x$-values ranging from $0$ to $\pi/2$. Equivalently, though, it's bounded on the left by $x=0$ and on the right by the line $x=y$, with $y$-values ranging from $0$ to $\pi/2$. Hence, changing the order of integration gives us $$\begin{align}\int_0^{\pi/2}\int_x^{\pi/2}\frac{\cos y}y\,dy\,dx &= \int_0^{\pi/2}\int_0^y\frac{\cos y}y\,dx\,dy\\ &= \int_0^{\pi/2}\frac{\cos y}y\left(\int_0^y\,dx\right)\,dy.\end{align}$$ Since $\int_0^y\,dx=y,$ then the rest is easy.
