Official solution
I = \[\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx} \] .....(1)
\[ \Rightarrow \] I = \[\int\limits_0^{{\pi \over 2}} {{{{{\cos }^3}x} \over {\sin x + \cos x}}} dx\] ......(2)
Adding those two
2I = \[\int\limits_0^{{\pi \over 2}} {{{si{n^3}x + {{\cos }^3}x} \over {\sin x + \cos x}}} dx\]
= \[\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x + \cos x} \right)\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} \over {\sin x + \cos x}}} dx\]
= \[\int\limits_0^{{\pi \over 2}} {\left( {si{n^2}x + {{\cos }^2}x - \sin x\cos x} \right)} dx\]
= \[\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\sin 2x} \over 2}} \right)} dx\]
= \[\left[ {x + {{\cos 2x} \over 4}} \right]_0^{{\pi \over 2}}\]
= \[\left( {{\pi \over 2} - {1 \over 4}} \right) - \left( {{1 \over 4}} \right)\]
\[ \therefore \] 2I = \[\left( {{\pi \over 2} - {1 \over 2}} \right)\]
\[ \therefore \] I = \[{{\pi - 1} \over 4}\]