Solution.
∫sinn(x)dx=−cos(x)sinn−1(x)n+n−1n∫sin−2+n(x)dx ,n=3\int sin^n(x)dx = -\frac{cos(x)sin^{n-1}(x)}{n} + \frac{n-1}{n}\int sin^{-2+n}(x)dx \ , n=3∫sinn(x)dx=−ncos(x)sinn−1(x)+nn−1∫sin−2+n(x)dx ,n=3
∫sin3(x)dx=−13sin2(x)cos(x)+23∫sin(x)dx=−sin2(x)cos(x)3−2cos(x)3+C\int sin^3(x)dx = -\frac{1}{3}sin^2(x)cos(x) + \frac{2}{3}\int sin(x)dx = -\frac{sin^2(x)cos(x)}{3} -\frac{2cos(x)}{3} + C∫sin3(x)dx=−31sin2(x)cos(x)+32∫sin(x)dx=−3sin2(x)cos(x)−32cos(x)+C
Answer:
112(cos(3x)−9cos(x))+C\frac{1}{12}(cos(3x)-9cos(x)) + C121(cos(3x)−9cos(x))+C
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