Let g t = ∫ - π / 2 π / 2 cos π 4 t + f x d x , where f x = log e x + x 2 + 1 , x…

We have,

$f\left(x\right)={\log }_e\left(x+\sqrt{x^2+1}\right),x\in R$

$\Rightarrow f\left(x\right)={\log }_e\left(\frac{\left(x+\sqrt{x^2+1}\right)\left(x-\sqrt{x^2+1}\right)}{x-\sqrt{x^2+1}}\right)$

$\Rightarrow f\left(x\right)={\log }_e\left(\frac{-1}{x-\sqrt{x^2+1}}\right)$

$\Rightarrow f\left(x\right)={\log }_e\left(\frac{1}{\sqrt{x^2+1}-x}\right)$

$\Rightarrow f\left(-x\right)={\log }_e\left(\frac{1}{\sqrt{x^2+1}+x}\right)$

$\Rightarrow f\left(-x\right)={\log }_e{\left(\sqrt{x^2+1}+x\right)}^{-1}$

$\Rightarrow f\left(-x\right)=-{\log }_e\left(\sqrt{x^2+1}+x\right)$

$\Rightarrow f\left(-x\right)=-f\left(x\right)$

Hence, $f\left(x\right)$ is an odd function.

Now,

$g\left(t\right)={\int }_{-\pi /2}^{\pi /2}\left(\cos \frac{\pi }{4}t+f\left(x\right)\right)dx$

$\Rightarrow g\left(t\right)=\cos \frac{\pi }{4}t{\int }_{-\pi /2}^{\pi /2}1 dx+{\int }_{-\pi /2}^{\pi /2}f\left(x\right)dx$

$\Rightarrow g\left(t\right)=\pi \cos \frac{\pi }{4}t+{\int }_{-\pi /2}^{\pi /2}f\left(x\right)dx$

$\Rightarrow g\left(t\right)=\pi \cos \frac{\pi }{4}t+0$

$\because f\left(x\right)$ is an odd function.

$\therefore g\left(1\right)=\frac{\pi }{\sqrt{2}}\Rightarrow \sqrt{2}g\left(1\right)=\mathrm{\pi }$

$g\left(0\right)=\pi$