Past exam of the mathematical methods I course of the University of Oxford / 2015 / 3 / Solution

An integrating factor for the first-order linear differential equation is $\mu (x)=e^{\sin x}$. Multiplication givesSubstitute $u=\sin x$ in the antiderivative: $\int 2u{e}^{u}du=2(u-1)e^u$. ThereforeDifferentiating verifies the equation everywhere on the real line; no division by $\cos x$ restricts the solution.