Just Integrate

$\begin{aligned} f(x) & = \int \frac{\color{#3D99F6}{\sec x}}{2 \color{#D61F06}{\cos(2x)}-1}dx \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\sec x = \frac{1}{\cos x}} \quad \color{#D61F06}{\cos (2x) = 2\cos^2 x -1 } \\ & = \int \frac{1}{\color{#3D99F6}{\cos x} (2[\color{#D61F06}{2\cos^2 x - 1}]-1)}dx \\ & = \int \frac{1}{\color{#3D99F6}{4\cos^3x-3\cos x}}dx \quad \quad \quad \quad \small \color{#3D99F6}{\cos (3x) = 4\cos^3x-3\cos x} \\ & = \int \frac{1}{\color{#3D99F6}{\cos (3x)}} dx \quad \quad \quad \quad \quad \quad \quad \quad \space \small \color{#3D99F6}{\text{Let } t = \tan \left(\frac{3x}{2} \right) \quad \Rightarrow \cos(3x) = \frac{1-t^2}{1+t^2} \quad \Rightarrow dt = \frac{3}{2}(1+t^2)dx} \\ & = \frac{2}{3} \int \frac{1}{1-t^2} dt \\ & =\frac{2}{3} \int \frac{1}{(1-t)(1+t)} dt \\ & = \frac{1}{3}\left(\int \frac{1}{1-t} dt + \int \frac{1}{1+t} dt \right) \\ & = \frac{1}{3}\left( - \ln (1-t) + \ln (1+t) \right) + c \quad \small \color{#3D99F6}{c \text{ is a constant}} \\ & = \frac{1}{3}\left( \ln \left[ \frac{1+t}{1-t} \right] \right) + c \\ & = \frac{1}{3}\left( \ln \left[ \frac{(1+t)(1+t)}{(1-t)(1+t)} \right] \right) + c \\ & = \frac{1}{3}\left( \ln \left[ \frac{1 + 2t + t^2}{1-t^2} \right] \right) + c \\ & = \frac{1}{3}\left( \ln \left[ \frac{2t}{1-t^2} + \frac{1+t^2}{1-t^2} \right] \right) + c \\ \\ & = \frac{1}{3}\left( \ln \left[ \tan (3x) + \sec (3x) \right] \right) + c \end{aligned}$

It is given that $f(0) = 1$ , then:

$\begin{aligned} f(0) & = \frac{1}{3}\left( \ln \left[ \tan 0 + \sec 0 \right] \right) + c \\ & = \frac{1}{3} \ln 1 + c \& = 0 + c \\ \Rightarrow c & = 1 \end{aligned}$

And we have:

$\begin{aligned} f\left( \frac{\pi}{12} \right) & = \frac{1}{3}\left( \ln \left[ \tan \left( \frac{\pi}{4} \right) + \sec \left( \frac{\pi}{4} \right) \right] \right) + 1 \\ & = \frac{1}{3} \ln (1 + \sqrt{2}) + 1 \\ & = \boxed{1.294} \end{aligned}$