Exact equation with trig function

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The differential equation $$(\tan{x}-\sin{x}\sin{y}) dx+\cos{x}\cos{y} dy=0$$ is determined to be exact. The solution involves integrating the function $f_x(x,y)=\tan(x)-\sin(x)\sin(y)$ with respect to $x$, yielding the result $-\ln|\cos(x)|+\cos(x)\sin(y)=c$. The logarithmic term arises from the integration process, specifically from the integral of $\tan(x)$, which is $-\ln|\cos(x)|$.

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Determine whether the given differential equation is exact and if so solve it.

$$(\tan{x}-\sin{x}\sin{y}) dx+\cos{x}\cos{y} dy=0$$

I got [math]\cos{x}\sin{y}+\sec^2{x}=c[/math] but the answer key has [math]-ln|\cos{x}|+\cos{x}\sin{y}=c[/math] where did $$ln$$ come from?
 
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find_the_fun said:
Determine whether the given differential equation is exact and if so solve it.

$$(\tan(x)-\sin(x)\sin(y)) \, dx+\cos(x)\cos(y) \, dy=0$$

I got [math]\cos(x)\sin(y)+\sec^2(x)=c[/math] but the answer key has [math]-\ln|\cos(x)|+\cos(x)\sin(y)=c.[/math] Where did the $$\ln$$ come from?

So we start out with
$$f_x(x,y)=\tan(x)-\sin(x)\sin(y).$$
Integrating w.r.t. $x$ yields $-\ln|\cos(x)|+\cos(x)\sin(y)$. Then you do the usual differentiation w.r.t. $y$, compare with $\cos(x) \cos(y)$, etc. That's where the logarithm came from.
 

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