Linear First Order

This is a first-order linear equation \frac{dy}{dx} + P y = Q with P = \cot x and Q = 2\cos x, solved by the integrating factor \mu = e^{\int \cot x,dx} = e^{\ln \sin x} = \sin x. Multiplying the equation by \sin x makes the left side an exact derivative: \frac{d}{dx}(y\sin x) = 2\cos x\sin x = \sin 2x. Integrating both sides gives y\sin x = \int \sin 2x,dx = -\frac{\cos 2x}{2} + C. Apply the condition y = 1 at x = \pi/2. Since \cos\pi = -1, the term -\frac{\cos 2x}{2} evaluates to -\frac{\cos\pi}{2} = \frac{1}{2}, and \sin(\pi/2) = 1, so 1 = \frac{1}{2} + C and therefore C = \frac{1}{2}. Substituting back gives y\sin x = -\frac{\cos 2x}{2} + \frac{1}{2} = \frac{1 - \cos 2x}{2}, hence y = \frac{1 - \cos 2x}{2\sin x}, equivalently \frac{-\cos 2x}{2\sin x} + \frac{1}{2\sin x}. Option y = \sin x\cos x drops the constant. Option y = \cos 2x\sin x mishandles the integrating factor. Option \frac{\cos 2x}{2\sin x} omits the constant term. This is the standard linear-ODE pattern. As a final check, the double-angle identity 1 - \cos 2x = 2\sin^2 x gives y = \frac{2\sin^2 x}{2\sin x} = \sin x, so y(\pi/2) = \sin(\pi/2) = 1, confirming the initial condition is satisfied.