Find the general integral of the Riccati equation y'(1-sin…

Find the general integral of the Riccati equation $y'(1-\sin x\cos x)+y^2\cos x-y+\sin x=0$, given the particular integral $y_1=\cos x$.

Step 1: Verify the particular solution

With $y_1=\cos x$: $y_1'=-\sin x$, and $(-\sin x)(1-\sin x\cos x)+\cos^3x-\cos x+\sin x=\cos x(\sin^2x+\cos^2x-1)=0$. ✓

Step 2: Substitute $y=\cos x+1/u$

Writing the equation as $y'=P+Qy+Ry^2$ with $Q=\dfrac{1}{1-\sin x\cos x}$, $R=\dfrac{-\cos x}{1-\sin x\cos x}$, the standard Riccati reduction gives

\[u'+(Q+2Ry_1)u+R=0,\qquad Q+2Ry_1=\frac{1-2\cos^2x}{1-\sin x\cos x}=\frac{-\cos2x}{1-\sin x\cos x}.\]

So $u'-\dfrac{\cos2x}{1-\sin x\cos x}u=\dfrac{\cos x}{1-\sin x\cos x}$.

Step 3: Simplify and solve the linear equation

Since $1-\sin x\cos x=\tfrac12(2-\sin2x)$, the equation becomes $u'-\dfrac{2\cos2x}{2-\sin2x}u=\dfrac{2\cos x}{2-\sin2x}$. The integrating factor is $\mu(x)=2-\sin2x$ (since $\int\!\frac{-2\cos2x}{2-\sin2x}dx=\ln|2-\sin2x|$), giving

\[\frac{d}{dx}\big[(2-\sin2x)u\big]=2\cos x\quad\Longrightarrow\quad (2-\sin2x)u=2\sin x+K.\]