$ \int_{0}^{\pi /2}{\frac{{{\sin }^{100}}\,x}{{{\sin }^{100}}\,x+\,{{\cos }^{100}}x}}\,\,dx $
$ =\int_{0}^{\pi /2}{\frac{{{\sin }^{100}}}{{{\sin }^{100}}x+{{\sin }^{100}}\left( \frac{\pi }{2}-x \right)}}\,\,dx $
$ =\frac{\frac{\pi }{2}-0}{2}=\frac{\pi }{4} $
$ \left[ \because \,\,\int_{a}^{b}{\frac{f(x)\,dx}{f(x)+f(a+b-x)}=\frac{b-a}{2}} \right] $
Alternate $ I=\int_{0}^{\pi /2}{\frac{{{\sin }^{100}}x}{{{\sin }^{100}}x+{{\cos }^{100}}x}\,\,\,dx} $ .. (i)
$ I=\int_{0}^{\pi /2}{\frac{{{\sin }^{100}}\,\left( \frac{\pi }{2}-x \right)}{{{\sin }^{100}}\left( \frac{\pi }{2}-x \right)-{{\cos }^{100}}\left( \frac{\pi }{2}-x \right)}}\,\,\,dx $
$ \left[ \begin{align} & \because \,\,by\,\,define\,\,\text{integral property}\text{.} \\ & \int_{a}^{0}{f(x)\,dx=\int_{0}^{a}{f(a-x)\,dx}} \\ \end{align} \right] $
$ I=\int_{0}^{\pi /2}{\frac{{{\cos }^{100}}x}{{{\cos }^{100}}x+{{\sin }^{100}}x}}\,dx $ .. (ii)
On adding Eqs. (i) and (ii), we get
$ 2I=\int_{0}^{\pi /2}{1\,\,dx\,=\frac{\pi }{2}-0\,\,\Rightarrow \,I=\frac{\pi }{4}} $