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f(x)=(sinx)^2+∫_{0~π/2}f(t)dt a=∫_{0~π/2}f(t)dt とすると f(x)=(sinx)^2+a f(t)=(sint)^2+a a=∫_{0~π/2}f(t)dt =∫_{0~π/2}{(sint)^2+a}dt =∫_{0~π/2}{[{1-cos(2t)}/2]+a}dt ={([t-{sin(2t)/2]/2)+at}_{0~π/2} =(π/4)+(aπ/2) 2a(2-π)=π a=π/{4-(2π)} f(x)={(sinx)^2}+[π/{4-(2π)}]