Question #103215
Integrate sin³x dx
1
Expert's answer
2020-02-18T05:33:02-0500

Solution.

sinn(x)dx=cos(x)sinn1(x)n+n1nsin2+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

sin3(x)dx=13sin2(x)cos(x)+23sin(x)dx=sin2(x)cos(x)32cos(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

Answer:

112(cos(3x)9cos(x))+C\frac{1}{12}(cos(3x)-9cos(x)) + C


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