Explanation

(d)
Given, $f^{''}(x) = \tan^{2}{x = (1 - \sec^{2}{x)}}$
On integrating both sides, we get
$f^{'}(x) = x - \tan{x + c_{1}}$
$f^{'}(0) = 0 - 0 + c_{1}\ \Rightarrow c_{1} = 0$
$\Rightarrow f^{'}(x) = x - \tan x$
Again integrating both sides, we get
$\Rightarrow f(x) = \frac{x^{2}}{2} - \log{\sec{x + c_{2}}}$
$\Rightarrow f(0) = 0 - \log{1 + c_{2}\ \Rightarrow c_{2} = 0}$
$\therefore f(x) = \frac{x^{2}}{2} - \log{\sec x}$