The answer can be written as:
-\sin^3(x)/9-\frac{2}{3}\sin(x)+C−sin3(x)9−23sin(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∫sin3(x)dx=∫(1−cos2(x))sin(x)dx=∫(u2−1)du=u33−u+C=cos3(x)3−cos(x)+C, where u=\cos(x)u=cos(x) and du=-sin(x)dxdu=−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.
