Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{{\sin }^{n}x}{{\sin }^{n}x+{\cos }^{n}x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{n}x}{{\sin }^{n}x+{\cos }^{n}x}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^n\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{{\sin }^n\left(\frac{\mathrm{\pi }}{2}-x\right)+{\cos }^n\left(\frac{\mathrm{\pi }}{2}-x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^{n}x}{{\cos }^{n}x+{\sin }^{n}x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\cos }^{\mathrm{n}}x}{{\sin }^{\mathrm{n}}\mathrm{x}+{\cos }^{\mathrm{n}}\mathrm{x}}\mathrm{d}\mathrm{x}...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{n}x}{{\sin }^{n}x+{\cos }^{n}x}+\frac{{\cos }^{n}x}{{\sin }^{n}x+{\cos }^{n}x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{{\sin }^{n}x+{\cos }^{n}x}{{\sin }^{n}x+{\cos }^{n}x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}1dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence}I=\frac{\mathrm{\pi }}{4} \end{gathered}$$