$\underset{0}{\overset{\pi /4}{\int }}\frac{x^2}{{(x\sin x+\cos x)}^2}dx$

$$\begin{gathered} \mathrm{I}=\underset{0}{\overset{\mathrm{\pi }/4}{\int }}\frac{{\mathrm{x}}^2}{{(\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x})}^2} \\ =\underset{0}{\overset{\mathrm{\pi }/4}{\int }} \frac{\mathrm{x}}{\cos \mathrm{x}}.\frac{\mathrm{x} \cos \mathrm{x}}{{(\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x})}^2} \mathrm{dx} \\ [\underline{\mathrm{Note} :} \mathrm{J}=\int \frac{\cos \mathrm{x}}{{(\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x})}^2}\mathrm{dx} \\ \mathrm{Put} x\sin \mathrm{x}+\cos \mathrm{x}=\mathrm{t} \\ (\mathrm{x} \cos \mathrm{x}+\sin \mathrm{x}-\sin \mathrm{x})\mathrm{dx}=\mathrm{dt} \\ \mathrm{J}=\int \frac{\mathrm{dt}}{{\mathrm{t}}^2} =\frac{-1}{\mathrm{t}} \\ =\frac{-1}{\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x}}] \end{gathered}$$

$$\begin{gathered} I=\frac{\mathrm{x}}{\cos \mathrm{x}}.{\left(\frac{-1}{\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x}}\right)}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \\ -{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \left[\frac{\cancel{\cos \mathrm{x}+\mathrm{x} \mathrm{xin} \mathrm{x}}}{{\cos }^2\mathrm{x}}\times \frac{-1}{\cancel{\mathrm{x} \sin \mathrm{x}+\cos \mathrm{x}}}\right] \\ =\frac{{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}}\left[\frac{-1}{{\displaystyle \frac{{\displaystyle \frac{\mathrm{\pi }}{4}}+1}{\sqrt{2}}}}\right]+{\left(\tan \right)}_0^{\mathrm{\pi }/4} \\ =\frac{\mathrm{\pi }}{2}\left[\frac{-4}{\mathrm{\pi }+4}\right]+1 \\ =\frac{-2\mathrm{\pi }+\mathrm{\pi }+4}{\mathrm{\pi }+4}=\frac{4-\mathrm{\pi }}{4+\mathrm{\pi }} \end{gathered}$$