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(sinx)^4的原函數 =1/4∫{[1-cos2x]^2}d(x) =1/4∫(1-2cos2x+cos^2 2x)dx =1/4∫dx - 1/4∫cos2x d(2x)+ 1/4∫(cosx)^2 2xdx =3/8x-1/4sin2x+1/32sin4x+C
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