设f(x)=x-∫(0,π)f(x)cosxdx,求f(X)

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设f(x)=x-∫(0,π)f(x)cosxdx,求f(X)设f(x)=x-∫(0,π)f(x)cosxdx,求f(X)设f(x)=x-∫(0,π)f(x)cosxdx,求f(X)f(x)=x-∫(0

设f(x)=x-∫(0,π)f(x)cosxdx,求f(X)
设f(x)=x-∫(0,π)f(x)cosxdx,求f(X)

设f(x)=x-∫(0,π)f(x)cosxdx,求f(X)
f(x) = x - ∫(0~π) f(x) * cosx dx
f'(x) = 1
∫(0~π) f(x) * cosx dx
= (0~π) f(x) dsinx
= f(x) * sinx |(0~π) - ∫(0~π) f'(x) * sinx dx
= - ∫(0~π) f'(x) * sinx dx
= - (0~π) sinx dx
= cosx |(0~π)
= - 1 - 1 = - 2
∴f(x) = x - (- 2)
=> f(x) = x + 2