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

Exercise

∫sen(x)3x cos⁡2(x)dx\int sen\left(x\right)^3x\:\cos^2\left(x\right)dx

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Final answer to the exercise

(12x+14sin⁡(2x))sin⁡(x)3−32(x(12x−14sin⁡(2x))−14x2−18cos⁡(2x))+15cos⁡(x)3+3sin⁡(x)2cos⁡(x)310+C0\left(\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)\right)\sin\left(x\right)^3-\frac{3}{2}\left(x\left(\frac{1}{2}x-\frac{1}{4}\sin\left(2x\right)\right)-\frac{1}{4}x^2-\frac{1}{8}\cos\left(2x\right)\right)+\frac{1}{5}\cos\left(x\right)^{3}+\frac{3\sin\left(x\right)^{2}\cos\left(x\right)^{3}}{10}+C_0

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∫sen(x)3x cos2(x)dx
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