$\begin{align}  & \int{{{\sin }^{5}}x{{\cos }^{3}}xdx}=\int{{{\sin }^{4}}x{{\cos }^{3}}x\sin xdx} \\ & \text{Let }u=\cos x,\text{ }du=-\sin xdx \\ & \int{{{\sin }^{5}}x{{\cos }^{3}}xdx}=\int{{{\sin }^{4}}x\cdot {{u}^{3}}(-du)} \\ & \int{{{\sin }^{5}}x{{\cos }^{3}}xdu}=-\int{{{({{\sin }^{2}}x)}^{2}}{{u}^{3}}du} \\ & \int{{{\sin }^{5}}x{{\cos }^{3}}xdx}=-\int{{{(1-{{\cos }^{2}}x)}^{2}}{{u}^{3}}du} \\ & \int{{{\sin }^{5}}x\cos xdx}=-\int{{{(1-{{u}^{2}})}^{2}}{{u}^{3}}du} \\ & \int{{{\sin }^{5}}x\cos xdx}=-\int{(1-2{{u}^{2}}+{{u}^{4}}){{u}^{3}}du} \\ & \int{{{\sin }^{5}}x\cos xdx}=-\int{({{u}^{3}}-2{{u}^{5}}+{{u}^{7}})du} \\ & \int{{{\sin }^{5}}x\cos xdx}=-\left[ \frac{{{u}^{4}}}{4}-\frac{2{{u}^{6}}}{6}+\frac{{{u}^{8}}}{8} \right]+C \\ & \int{{{\sin }^{5}}x{{\cos }^{4}}xdx}=-\frac{{{\cos }^{4}}x}{4}+\frac{{{\cos }^{6}}x}{2}-\frac{{{\cos }^{8}}x}{8}+C \\\end{align}$