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Solve: g(s)=f(s)+\lambda \int _{0}^{{2\pi }}\sin(s)\cos(t)g(t)dt\,

g(s)=f(s)+\lambda \sin(s)\int _{0}^{{2\pi }}\cos(t)g(t)dt\,

Let c=\int _{0}^{{2\pi }}\cos(t)g(t)dt\,.

Write a new equation:

c=\int _{0}^{{2\pi }}\cos(t)\left[f(t)+\lambda \sin(t)c\right]dt\,

c=\int _{0}^{{2\pi }}\cos(t)f(t)dt+\lambda c\int _{0}^{{2\pi }}\cos(t)\sin(t)dt\,

The last integrand is odd and the limits of integration are a multiple of its period, so the integral equals 0.

c=\int _{0}^{{2\pi }}\cos(t)f(t)dt\,


g(s)=f(s)+\lambda \sin(s)c\,

the answer is

g(s)=f(s)+\lambda \sin(s)\int _{0}^{{2\pi }}\cos(t)f(t)dt\,

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