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Today's calculation of Integral 282

Source: 1976 Chiba University entrance exam

January 24, 2008
calculusintegrationfunctiontrigonometrycalculus computations

Problem Statement

g(x) g(x) is a differentiable function for 0≤x≤π 0\leq x\leq \pi and g′(x) g'(x) is a continuous function for 0≤x≤π 0\leq x\leq \pi. Let f(x) \equal{} g(x)\sin x. Find g(x) g(x) such that \int_0^{\pi} \{f(x)\}^2dx \equal{} \int_0^{\pi}\{f'(x)\}^2dx.
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