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

Exercise

∫(x3+2x2)sin⁡(2x)dx\int\left(x^3+2x^2\right)\sin\left(2x\right)dx

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Final answer to the exercise

(−12)x3cos⁡(2x)−x2cos⁡(2x)+34x2sin⁡(2x)+xsin⁡(2x)+34xcos⁡(2x)+12cos⁡(2x)−38sin⁡(2x)+C0\left(-\frac{1}{2}\right)x^3\cos\left(2x\right)-x^2\cos\left(2x\right)+\frac{3}{4}x^{2}\sin\left(2x\right)+x\sin\left(2x\right)+\frac{3}{4}x\cos\left(2x\right)+\frac{1}{2}\cos\left(2x\right)-\frac{3}{8}\sin\left(2x\right)+C_0

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∫(x3+2x2)sin(2x)dx
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