$I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^{3}x}{\sin x+\cos x}dx \dots \left(1\right)$

$\therefore I={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^3\left({\displaystyle \frac{\pi }{2}}-x\right)}{\sin \left(\frac{\pi }{2}-x\right)+\cos \left(\frac{\pi }{2}-x\right)}dx$

$\therefore I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{3}x}{\cos x+\sin x}dx \dots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we have

$2I={\int }_0^{\frac{\pi }{2}}\frac{{\sin }^{3}x+{\cos }^{3}x}{\sin x+\cos x}dx$

$\therefore 2I={\int }_0^{\frac{\pi }{2}}\frac{\left(\sin x+\cos x\right)\left({\sin }^{2}x+{\cos }^{2}x-\sin x·\cos x\right)}{\sin x+\cos x}dx$

$\therefore 2I={\int }_0^{\frac{\pi }{2}}\left(1-\sin x·\cos x\right)dx$

$\therefore 2I={\int }_0^{\frac{\pi }{2}}1dx-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}2\sin x·\cos xdx$

$\therefore 2I=\frac{\pi }{2}-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\sin 2xdx$

$\therefore 2I=\frac{\pi }{2}-\frac{1}{2}{\left[-\frac{\cos 2x}{2}\right]}_0^{\frac{\pi }{2}}$

$\therefore I=\frac{\pi }{4}+\frac{1}{8}\left(\cos \pi -\cos 0\right)$

$\therefore I=\frac{\pi }{4}+\frac{\left(-2\right)}{8}$

$\therefore I=\frac{\pi -1}{4}$