Помогите решитьdy-cos(y-1)dx=0

Решите задачу:

$dy-cos(y-1)dx=0\\\\\frac{dy}{cos(y-1)}=dx\\\\\int \frac{dy}{cos(y-1)}=\int \frac{cos(y-1)}{cos^2(y-1)}dy=\int \frac{coe(y-1)}{cos^2(y-1)}dy=\int \frac{cos(y-1)}{1-sin^2(y-1)}dy=\\\\=[t=sin(y-1),dt=cos(y-1)dy]=\int \frac{dt}{1-t^2}=\frac{1}{2}ln|\frac{1+t}{1-t}|+C=\\\\=\frac{1}{2}ln|\frac{1+sin(y-1)}{1+sin(y-1)}|+C\\\\\\\frac{1}{2}ln|\frac{1+sin(y-1)}{1+sin(y-1)}|=x+C$

$\frac{1+sin \alpha }{1-sin \alpha }=\frac{1+cos(90^0- \alpha )}{1+cos(90^0- \alpha )}=\frac{2cos^2(45^0-\frac{ \alpha }{2})}{2sin^2(45^0-\frac{ \alpha }{2})}=ctg^2(45^0-\frac{\alpha}{2})$

$\frac{1}{2}ln|ctg^2(\frac{\pi}{4}-\frac{y-1}{2})|=x+C\\\\ln|ctg(\frac{\pi}{4}-\frac{y-1}{2})|=x+C$

$ctg(\frac{\pi}{4}- \alpha )=tg(\frac{\pi}{2}-(\frac{\pi}{4}- \alpha ))=tg(\frac{\pi}{4}+ \alpha )\\\\Otvet:ln|tg(\frac{\pi}{4}+\frac{y-1}{2})|=x+C$