1. **State the problem:** We are given the differential equation: $$ (\sin y - 2y e^x \sin x) dx + (\cos y + 2 e^x \cos x) dy = 0 $$ We want to find an integrating factor using Case 1 of the method, where the integrating factor depends only on $x$. 2. **Identify $M(x,y)$ and $N(x,y)$:** $$ M = \sin y - 2 y e^x \sin x $$ $$ N = \cos y + 2 e^x \cos x $$ 3. **Compute partial derivatives:** $$ \frac{\partial M}{\partial y} = \cos y - 2 e^x \sin x $$ $$ \frac{\partial N}{\partial x} = 0 + 2 e^x \cos x - 2 e^x \sin x = 2 e^x (\cos x - \sin x) $$ 4. **Calculate the expression:** $$ \frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) = \frac{\cos y - 2 e^x \sin x - 2 e^x (\cos x - \sin x)}{\cos y + 2 e^x \cos x} $$ Simplify numerator: $$ \cos y - 2 e^x \sin x - 2 e^x \cos x + 2 e^x \sin x = \cos y - 2 e^x \cos x $$ So the expression becomes: $$ \frac{\cos y - 2 e^x \cos x}{\cos y + 2 e^x \cos x} $$ 5. **Check if this expression depends only on $x$:** It depends on $y$ through $\cos y$, so it is not a function of $x$ alone. 6. **Conclusion:** Since the expression is not a function of $x$ only, Case 1 integrating factor depending only on $x$ does not apply here. **Therefore, no integrating factor of the form $\mu(x)$ exists by Case 1.**