$\begin{align}  & \int{{{\sin }^{3}}xdx}=\int{{{\sin }^{2}}x\sin xdx} \\ & \int{{{\sin }^{3}}x}=\int{(1-{{\cos }^{2}}x)\sin xdx} \\ & \int{{{\sin }^{3}}x}=\int{(\sin x}-\sin x{{\cos }^{2}}x)dx \\ & \text{Let }u=\cos x \\ & \frac{du}{dx}=-\sin x,\text{ }dx=-\frac{du}{\sin x} \\ & \int{{{\sin }^{3}}xdx}=-\cos x-\int{\sin x\cdot {{u}^{2}}(-\tfrac{du}{\sin x})} \\ & \int{{{\sin }^{3}}xdx}=-\cos x+\int{{{u}^{2}}du}=-\cos x+\frac{{{u}^{3}}}{3}du \\ & \int{{{\sin }^{3}}xdx}=-\cos x+\frac{{{\cos }^{3}}x}{3}+C \\\end{align}$