 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/13/a/ii/solution

sin x cos x

e^{- s i n x}

\frac{d}{d x} (y e^{- s i n x}) = sin x cos x e^{- s i n x} .

u = sin x

inside the antiderivative. Then

\int u e^{- u} d u = - (u + 1) e^{- u} + C, y = - (sin x + 1) + C e^{s i n x} .

The initial value forces

C = 1

, hence

y (x) = e^{s i n x} - sin x - 1.

The resulting expression is explicit and valid for all real

x

; the substitution does not impose a global monotonicity condition on

sin x

when verifying an antiderivative.

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