The slope of tangent at \((x, y)\) to a curve passing through \(\left(1, \frac{\pi}{4}\right)\) is \(\frac{y}{x}-\cos ^2 \frac{y}{x}\), then the equation of curve is

\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}-\cos ^2 \frac{y}{x}...(i)\)
Put \(y=v x\)...(ii)
\(\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{~d} x}...(iii)\)
Substituting (ii) and (iii) in (i), we get.
\(\mathrm{v}+x \frac{\mathrm{~d} v}{\mathrm{~d} x}=\mathrm{v}-\cos ^2 \mathrm{v} \quad \Rightarrow x \frac{\mathrm{dv}}{\mathrm{~d} x}=-\cos ^2 \mathrm{v}\)
Integrating on both sides, we get
\(\begin{aligned}
& \int \sec ^2 v \cdot d v=-\int \frac{d x}{x}+c \\
& \Rightarrow \tan v=-\log x+c \\
& \Rightarrow \tan \frac{y}{x}=-\log x+c...(iv)
\end{aligned}\)
Since the required curve passes through \(\left(1, \frac{\pi}{4}\right)\),
\(\begin{aligned}
& \tan \frac{\pi}{4}=-\log 1+\mathrm{c} \Rightarrow \mathrm{c}=1 \\
\therefore \quad & \tan \frac{y}{x}=-\log x+1 \\
& \Rightarrow \tan \frac{y}{x}=-\log x+\log \mathrm{e} \\
& \Rightarrow y=x \tan ^{-1}\left[\log \left(\frac{\mathrm{e}}{x}\right)\right]
\end{aligned}\)
...[From (iv)]