Let $$g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx$$, where $f(x) = {\log _e}\left( {x + \sqrt {{x^2} + 1} } \right),x \in R$. Then which one of the following is correct?

$\because$ $f(x) = \ln \left( {x + \sqrt {{x^2} + 1} } \right)$<br><br>$\therefore$ $$f(x) + f( - x) = \ln \left( {\sqrt {{x^2} + 1} + x} \right) + \ln \left( {\sqrt {{x^2} + 1} - x} \right)$$<br><br>$\therefore$ $f(x) + f( - x) = 0$ .... (i)<br><br>$\because$ $$g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi \over 4}t + f(x)} \right)} dx$$<br><br>$$ = \int_0^{\pi /2} {\left\{ {\cos \left( {{\pi \over 4}t + f(x)} \right) + \cos \left( {{\pi \over 4}t + f( - x)} \right)} \right\}} dx$$<br><br>$$ = \int_0^{\pi /2} {\left\{ {\cos \left( {{{\pi t} \over 4} + f(x)} \right) + \cos \left( {{{\pi t} \over 4} - f(x)} \right)} \right\}dx} $$<br><br>$g(t) = 2\int_0^{\pi /2} {\cos {{\pi t} \over 4}.\cos (f(x))dx}$<br><br>$\therefore$ $g(1) = \sqrt 2 \int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx$ and <br><br>$g(0) = 2\int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx$<br><br>$\therefore$ $\sqrt 2 g(1) = g(0)$