Consider the given differential equation. sin‌y‌
dy
dx
=cos‌y(1−x‌cos‌y) sin‌y‌
dy
dx
=cos‌y−xcos2y
sin‌y
cos2y
dy
dx
=
1
cos‌y
−x sec‌y‌tan‌y‌
dy
dx
=sec‌y−x Put, sec‌y=t sec‌y‌tan‌y‌
dy
dx
=
dt
dx
So,
dt
dx
=t−x
dt
dx
+(−t)=(−x) The integrating factor is given by, ‌IF=e∫−dx =e−x So, the required solution is, t(IF)=∫(−x)(IF)dx+c sec‌y(e−x)=∫(−x)(e−x)dx+c sec‌y=x+1+cex