Evaluate : \[\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{100} \mathrm{x}}{\sin ^{100} \mathrm{x}+\cos ^{100} \mathrm{x}} d \mathrm{x} \]

$$\mathrm{I}=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{100} \mathrm{x}}{\sin ^{100} \mathrm{x}+\cos ^{100} \mathrm{x}} \mathrm{dx}$\displaystyle \mathrm{I}=2 \int_{0}^{\pi / 2} \frac{\sin ^{100} \mathrm{x}}{\sin ^{100} \mathrm{x}+\cos ^{100} \mathrm{x}} \mathrm{dx} \quad$ as $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\sin ^{100} \mathrm{x}}{\sin ^{100} \mathrm{x}+\cos ^{100} \mathrm{x}}$ is even\mathrm{I}=2 \int_{0}^{\pi / 2} \frac{\cos ^{100} \mathrm{x}}{\cos ^{100} \mathrm{x}+\sin ^{100} \mathrm{x}} \mathrm{dx} \quad \operatorname{using} \int_{0}^{a} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\int_{0}^{a} \mathrm{f}(a-\mathrm{x}) \mathrm{dx}$\displaystyle \mathrm{I}=\left.\mathrm{x}\right|_{0} ^{\pi / 2} \quad \Rightarrow \mathrm{I}=\frac{\pi}{2}$