\(\int \frac{\mathrm{d} x}{3-2 \cos 2 x}=\frac{\tan ^{-1}(\mathrm{f}(x))}{\sqrt{5}}+\mathrm{c}\), (where c is constant of integration), then \(f(\pi / 4)\) has the value

\( \text { Let } \mathrm{I}=\int \frac{\mathrm{d} x}{3-2 \cos 2 x} \)
\( \text { Put } \tan x=\mathrm{t} \)
\( \therefore x=\tan ^{-1} \mathrm{t}\)
\( \therefore \mathrm{d} x=\frac{\mathrm{dt}}{1+\mathrm{t}^2} \)
\( \cos 2 x=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2} \)
\( \therefore \mathrm{I} =\int \frac{\frac{\mathrm{dt}}{1+\mathrm{t}^2}}{3-2\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)} \)
\( =\int \frac{\mathrm{dt}}{3\left(1+\mathrm{t}^2\right)-2\left(1-\mathrm{t}^2\right)} \)
\( =\int \frac{\mathrm{dt}}{3+3 \mathrm{t}^2-2+2 \mathrm{t}^2} \)
\( =\int \frac{\mathrm{dt}}{(\sqrt{5} \mathrm{t})^2+(1)^2} \)
\( =\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \mathrm{t})+\mathrm{c}\)
\(\therefore I=\frac{1}{\sqrt{5}} \tan ^{-1} \sqrt{5} \tan x+c\)
Comparing with \(\frac{\tan ^{-1} \mathrm{f}(x)}{\sqrt{5}}\), we get
\( \mathrm{f}(x)=\sqrt{5} \tan x \)
\( \therefore \mathrm{f}\left(\frac{\pi}{4}\right)=\sqrt{5} \tan \frac{\pi}{4}=\sqrt{5}\)