ex5(sin2x−2cos2x)\frac{e^{x}}{5}(\sin 2 x-2 \cos 2 x)5ex(sin2x−2cos2x)
e5xlog∣x∣=ze^{5 x} \log |x|=ze5xlog∣x∣=z
ex(sin2x+cos2x)e^{x}(\sin 2 x+\cos 2 x)ex(sin2x+cos2x)
2sin2ex2 \sin 2 e^{x}2sin2ex