| 似たような問題を見つけてわかってきました 下の答えであってますか?
問:∫{2(e^(2x))*sin(x)}dx
I = ∫{2(e^(2x))*sin(x)}dx = ∫{(e^(2x))'*sin(x)}dx = e^(2x) * sin(x) - ∫{e^(2x) * cos(x)}dx = e^(2x) * sin(x) - ∫[{(1/2) * e^(2x)}' * cos(x)]dx = e^(2x) * sin(x) - (1/2) * e^(2x) * cos(x) + (1/2)∫{e^(2x)*sin(x)}dx = e^(2x) * (sin(x) - (1/2) * cos(x)) + (1/4) * I
(3/4) * I = 2e^(2x) * (sin(x) - (1/2) * cos(x)) I = (8/3)e^(2x) * (sin(x) - (1/2) * cos(x))
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