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f(x)=sin(2x)*10^(-3x)−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

∫sin⁡(2x)⋅10−3x\int\sin\left(2x\right)\cdot10^{-3x}

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Final answer to the exercise

−81⋅ln⁡∣10∣4−16−81⋅ln⁡∣10∣4(sin⁡(2x)−3ln⁡∣10∣103x+−8cos⁡(2x)−81⋅ln⁡∣10∣4103x+4sin⁡(2x)27⋅ln⁡∣10∣3103x+2cos⁡(2x)−9⋅ln⁡∣10∣2⋅103x)+C0\frac{-81\cdot \ln\left|10\right|^{4}}{-16-81\cdot \ln\left|10\right|^{4}}\left(\frac{\sin\left(2x\right)}{-3\ln\left|10\right|10^{3x}}+\frac{-8\cos\left(2x\right)}{-81\cdot \ln\left|10\right|^{4}10^{3x}}+\frac{4\sin\left(2x\right)}{27\cdot \ln\left|10\right|^{3}10^{3x}}+\frac{2\cos\left(2x\right)}{-9\cdot \ln\left|10\right|^2\cdot 10^{3x}}\right)+C_0

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∫sin(2x)·10−3x
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