Official solution
\[I = \int\limits_0^{{\pi \over 2}} {{{(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}} + {{{\pi ^{\sin x}}(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}}} dx\]
\[I = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx} \]
\[I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}\]