Answer to Question #103215 in Calculus for Patrick Kariuki

Question

Integrate sin³x dx

Answer 1

Solution.

$\int sin^n(x)dx = -\frac{cos(x)sin^{n-1}(x)}{n} + \frac{n-1}{n}\int sin^{-2+n}(x)dx \ , n=3$

$\int sin^3(x)dx = -\frac{1}{3}sin^2(x)cos(x) + \frac{2}{3}\int sin(x)dx = -\frac{sin^2(x)cos(x)}{3} -\frac{2cos(x)}{3} + C$

Answer:

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