 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/1a/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

I^{'} (x) = \int_{0}^{π} cos θ e^{x c o s θ} d θ .

On the other hand,

\frac{d}{d θ} (sin θ e^{x c o s θ}) = cos θ e^{x c o s θ} - x sin^{2} θ e^{x c o s θ} .

Its integral is zero because

sin θ

vanishes at both endpoints. Hence

I^{'} (x) = \int_{0}^{π} x sin^{2} θ e^{x c o s θ} d θ .

Differentiating once more,

I^{''} (x) = \int_{0}^{π} cos^{2} θ e^{x c o s θ} d θ .

For

x \neq = 0

, the preceding identity and

sin^{2} θ + cos^{2} θ = 1

give

I^{''} + \frac{1}{x} I^{'} - I = \int_{0}^{π} (cos^{2} θ + sin^{2} θ - 1) e^{x c o s θ} d θ = 0.

The equation extends through

x = 0

in its regular limiting form.

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