The answer can be written as:
#-\sin^3(x)/9-\frac{2}{3}\sin(x)+C#
Using trigonometric identities and substitution:
#\int\sin^{3}(x) dx=\int(1-\cos^{2}(x))\sin(x)dx=\int (u^2-1) du=u^3/3-u+C=\cos^3(x)/3-\cos(x)+C#, where #u=\cos(x)# and #du=-sin(x)dx#.
Using the same techniques,
#\int(\cos^3(x)/3-\cos(x))dx=\frac{1}{3}\int(1-sin^2(x))\cos(x)dx-\int\cos(x)dx=\frac{1}{3}\int(1-u^2)du-sin(x)+C=(u/3-u^3/9)-sin(x)+C=sin(x)/3-sin^3(x)/9-\sin(x)+C=-\sin^3(x)/9-\frac{2}{3}\sin(x)+C#, where this time #u=\sin(x)# and #du=\cos(x) dx#.
