The value of ∫ - π 2 π 2 1 + sin 2 x 1 + π sin x d x is :

$I={\int }_a^{b}f\left(x\right)dx$

$f(a+b-x)=f\left(x\right)$

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{1+{\sin }^{2}x}{1+{\pi }^{\sin x}}dx\dots \left(\mathrm{i}\right)$

$I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{1+{\sin }^{2}x}{1+{\pi }^{-\sin x}}dx...\left(\mathrm{ii}\right)$

Add equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$2I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\left(1+{\sin }^{2}x\right)dx$

$2I=\pi +\frac{1}{2}{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\left(1-\cos 2x\right)dx$

$2I=\pi +\frac{1}{2}{\left[x-\frac{\sin 2x}{2}\right]}_{\frac{-\pi }{2}}^{\frac{\pi }{2}}$

$2I=\pi +\frac{1}{2}\left[\left(\frac{\pi }{2}-0\right)-\left(\frac{-\pi }{2}-0\right)\right]$

$2I=\pi +\frac{1}{2}\pi$

$2I=\frac{3\pi }{2}$

$\Rightarrow I=\frac{3\pi }{4}$