If $y=y(x)$ is the solution curve of the differential equation

$\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$, then $y\left(\frac{\pi}{6}\right)$ is equal to

$\frac{d y}{d x}+y \tan x=x \sec x$ <br/><br/>$\therefore $ I.F $=e^{\int \tan x d x}=\sec x$ <br/><br/>$\Rightarrow y \sec x=\int x \sec ^{2} x d x$ <br/><br/>$\Rightarrow y \sec x=x \tan x-\ln |\sec x|+c$ <br/><br/>Given, $y(0)=1$ <br/><br/>$\Rightarrow 1=c$ <br/><br/>$\therefore$ $ y \sec x=x \tan x-\ln |\sec x|+1$ <br/><br/>$\Rightarrow y=x \sin x-\cos x \ln |\sec x|+cosx$ <br/><br/>Now, at $x=\pi / 6$, we have <br/><br/>$$ \begin{aligned} & y=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e \frac{2}{\sqrt{3}}+\frac{\sqrt{3}}{2} \\\\ & =\frac{\pi}{12}-\frac{\sqrt{3}}{2}\left[\log _e \frac{2}{\sqrt{3}}-\log _e e\right]\\\\ &=\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right) \end{aligned} $$