Answer: \(\frac{\mathrm dy}{\mathrm dx}=\tan^2x\); \(\int_0^{\pi/4}\tan^2x\,\mathrm dx=1-\frac{\pi}{4}\).

Set up the integral using the correct limits or antiderivative first. When definite integration is used, evaluate the upper limit and lower limit separately before subtracting.

Since \(y=\tan x-x\),

\(\frac{\mathrm dy}{\mathrm dx}=\sec^2x-1.\)

Using \(\sec^2x=1+\tan^2x\), this becomes

\(\frac{\mathrm dy}{\mathrm dx}=\tan^2x.\)

Hence \(\int\tan^2x\,\mathrm dx=\tan x-x+c\).

Therefore

\(\int_0^{\pi/4}\tan^2x\,\mathrm dx=[\tan x-x]_0^{\pi/4}=1-\frac{\pi}{4}.\)

This gives the required answer: \(\frac{\mathrm dy}{\mathrm dx}=\tan^2x\); \(\int_0^{\pi/4}\tan^2x\,\mathrm dx=1-\frac{\pi}{4}\).