Find the general integral of the linear equation y'+ycos…

Find the general integral of the linear equation $y'+y\cos x=\sin x\cos x$.

Step 1: Apply the integrating factor

With $\mu(x)=e^{\sin x}$, $\dfrac{d}{dx}(e^{\sin x}y)=e^{\sin x}\sin x\cos x$.

Step 2: Integrate the right-hand side

Substituting $u=\sin x$, $du=\cos x\,dx$: $\displaystyle\int e^{u}u\,du=e^{u}(u-1)+C=e^{\sin x}(\sin x-1)+C$.

Step 3: Solve for $y$

So $e^{\sin x}y=e^{\sin x}(\sin x-1)+C$, giving

\[\boxed{\,y=\sin x-1+Ce^{-\sin x}.}\]