Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/8/c/i/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2005 Paper 1 8 c i Solution by
Codex 0 Created 2026-10-18 Updated 2026-10-20
For fixed , the chain rule and the derivative of the hyperbolic secant giveThus the stated antiderivative holds, with its integration constant independent of but allowed to depend on . At the integrand is zero and the antiderivative is simply constant; the displayed formula with is meant for nonzero .
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