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

Using property of definite integration ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx,$ we get

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

Using $\sin \left(\frac{\mathrm{\pi }}{2}-x\right)=\cos x \& \cos \left(\frac{\mathrm{\pi }}{2}-x\right)=\sin x,$ 

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

On adding the equation $\left(i\right) \& \left(ii\right)$

$\Rightarrow 2I={\int }_0^{\frac{\pi }{2}}\frac{{\mathrm{s}\mathrm{i}\mathrm{n}}^{3}x+ {\mathrm{c}\mathrm{o}\mathrm{s}}^{3}x}{\sin x+\cos x}dx$

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

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

Now, using ${\sin }^{2}x+{\cos }^{2}x=1 \& 2\sin x\cos x=\sin 2x,$ we get

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

$\Rightarrow 2I= {\left[x+\frac{\cos 2x}{4}\right]}_0^{\frac{\pi }{2}}$

$\Rightarrow 2I=\left(\frac{\pi }{2}+\frac{\cos 2\left({\displaystyle \frac{\mathrm{\pi }}{2}}\right)}{4}\right)-\left(0+\frac{\cos 2\left({\displaystyle 0}\right)}{4}\right)$

$\Rightarrow 2I=\left(\frac{\pi }{2}-\frac{1}{4}\right)-\left(\frac{1}{4}\right)$

$\Rightarrow 2I=\frac{\pi }{2}-\frac{1}{2}$

$\Rightarrow I=\frac{\pi }{4}-\frac{1}{4}=\frac{\pi -1}{4}.$