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

For fixed , the chain rule and the derivative of the hyperbolic secant give
Thus 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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