若函数f(x)的二阶导数连续,且满足f"(x)-f(x)=x,求f(x)cos xdx.

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问题 若函数f(x)的二阶导数连续,且满足f"(x)-f(x)=x,求f(x)cos xdx.

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答案由f(x)=f"(x)一x,可得 [*]f(x)cos xdx=[*]f"(x)cos xdx—[*]xcos xdx =[*]cos xd[f'(x)]一0=cos xf'(x)[*](一sin x)f'(x)dx =一f'(π)+f'(一π)十sin xf(x)[*]cos xf(x)dx =一[f'(π)一f'(一π)]一[*]f(x)cos xdx, [*]f(x)cos xdx=一[*]

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