Solve $\sin(x)\cos(2y)\;dx+\cos(x)\sin(2y)\;dy=0$ with $y(0)=\frac{\pi}2$ with separation of variable method. Problem: Solve $\sin(x)\cos(2y)\;dx+\cos(x)\sin(2y)\;dy=0$ with $y(0)=\frac{\pi}2$
My Steps

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*$\sin(x)\cos(2y)\;dx+\cos(x)\sin(2y)\;dy=0$

*$\bigg(\frac{\sin(x)}{\cos(x)}\bigg)dx+\bigg(\frac{sin(2y)}{cos(2y)}\bigg)dy=0$

*$\tan(x)dx+\tan(2y)dy=0$

*$\int\frac{\sin(x)}{\cos(x)}dx+\int\frac{\sin(2y)}{\cos(2y)}dy=0$

*Let $u=\cos(x)$
$du=-\sin(x)$
$-du=\sin(x)$
let $u=\cos(2y)$
$du=-2\sin(2y)$
$-\frac{du}{2}=\sin(2y)$

*$-\int\frac1udu-\frac12\int\frac1udu=0$
However, Claimed Answer Was
$$\boxed{-\ln(\cos(x))-\frac12\ln(\cos(2y)=c}$$
No matter what I try here I don't get the above answer. I keep getting an expression with secant functions. What did I do wrong? How should I continue?
Edit: The answer I got was $\cos(2y)=c\cdot\sec^2(x)$.
I got this my adding $\ln(\cos(x))$ to both sides giving me
$$-\frac12\ln(\cos(2y)=c+\ln(\cos(x))$$
I then multiply both sides by $-2$, which I get;
$\ln(\cos(2y)=c+(-2\ln(\cos(x)))$, where $c$ is a constant so it stays the same even though multiplied by $-2$.
$e^{\ln(\cos(2y)}=e^{(c+\ln(\cos^-2(x))}$
I get $\cos(2y)=c\cdot\sec^2(x)$, That is my answer which is completely wrong, so I stopped here.
 A: You're saying the answer $\cos(2y)=c \cdot \sec^2x$ is wrong. This is the correct answer. Remember that $\sec x = \frac{1}{\cos x}.$ 
Another way to arrive at their answer would be take your original result:
$$
\begin{align}
-\ln(\cos x)-\frac{1}{2} \cdot \ln(\cos2y) &= c_0 \\
2\ln(\cos x)+\ln (\cos2y) &= c_1 \\
\ln(\cos^2 x \cdot \cos2y) &= c_1 \\
\exp(\ln(\cos^2 x \cdot \cos2y)) &= \exp(c_1) \\
\cos^2 x \cdot \cos2y &= c_2.
\end{align}$$
A: $$\sin(x) \cos(2y)dx+\cos(x) \sin(2y)dy=0, \ \ \ y(0)=\frac{\pi}{2}$$
$$\sin(x) \cos(2y)dx+\cos(x) \sin(2y)dy=0 \\ \Rightarrow \sin(x) \cos(2y)dx=-\cos(x) \sin(2y)dy \\ \Rightarrow \frac{\sin(x)}{\cos(x)}dx=-\frac{\sin(2y)}{\cos(2y)}dy \\ \Rightarrow \int \frac{\sin(x)}{\cos(x)}dx=-\int \frac{\sin(2y)}{\cos(2y)}dy \\ \Rightarrow -\ln{|\cos(x)|}=\frac{1}{2}\ln{|\cos(2y)|} +c\\ \Rightarrow \ln{|\cos(x)|^{-1}}=\ln{|\cos(2y)|^{\frac{1}{2}}} +c\\ \Rightarrow e^{\ln{|\cos(x)|^{-1}}}=e^{\ln{|\cos(2y)|^{\frac{1}{2}}} +c }\\ \Rightarrow |\cos(x)|^{-1}=C|\cos(2y)|^{\frac{1}{2}}, \text{ where } C=e^c \\ \Rightarrow |\frac{1}{\cos(x)}|=C|\cos(2y)|^{\frac{1}{2}} \\ \Rightarrow \frac{1}{\cos^2(x)}=C\cos(2y)$$
Applying the initial condition we have the following:
$$\frac{1}{\cos^2 \left (\frac{\pi}{2} \right )}=C\cos(2 \cdot 0) \Rightarrow C=1$$
Therefore, $$\frac{1}{\cos^2(x)}=\cos(2y) \Rightarrow arc \left ( \frac{1}{\cos^2(x)} \right )=2y \\ \Rightarrow y=\frac{1}{2}arc \left ( \frac{1}{\cos^2(x)} \right )$$
