∫e^ysiny dy
=-∫e^y d(cosy)
=-[e^y*cosy-∫cosy d(e^y)]
=∫cosy*e^y dy-e^ycosy
=∫e^y d(siny)-e^ycosy
=e^ysiny-∫siny d(e^y)-e^ycosy
=e^y(siny-cosy)-∫e^ysiny dy
所以 2∫e^ysiny dy = e^y(siny-cosy)
∫e^ysiny dy = e^y(siny-cosy)/2 + C
∫(e^y)siny dy=?∫e^ysiny dy=?书上的答案是 (1/2)(e^y)(siny-cosy)+C实在
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