Step 1: Set up exactness.
Write the equation as $M\,dx+N\,dy=0$ with $M=2y\cos x - xy\sin x$ and $N=2x\cos x$. After multiplying by a factor $\mu$, we need $\frac{\partial(\mu M)}{\partial y}=\frac{\partial(\mu N)}{\partial x}$.
Step 2: Try $\mu = xy$.
Then $\mu M = 2xy^2\cos x - x^2y^2\sin x$ and $\mu N = 2x^2y\cos x$. Differentiating, $\frac{\partial(\mu M)}{\partial y}=4xy\cos x - 2x^2y\sin x$ and $\frac{\partial(\mu N)}{\partial x}=4xy\cos x - 2x^2y\sin x$. They match, so $xy$ works.
Step 3: Try $\mu=\sqrt{\sec x}$.
With $\mu=(\cos x)^{-1/2}$, a short check of the two partial derivatives again gives the same expression on both sides. So $\sqrt{\sec x}$ also makes the equation exact.
Step 4: Rule out the others.
For $\mu=\tfrac{1}{xy}$ the difference of the two partials comes out as $\tfrac{2\sin x}{y}$, not zero. For $\mu=\sec x$ the difference is $-x\tan x$, not zero. So these two fail.
Step 5: Conclusion.
The integrating factors are $xy$ and $\sqrt{\sec x}$.
\[ \boxed{B \text{ and } D} \]