If $$y = y(x)$$ is the solution curve of the differential equation $$\frac{dy}{dx} + y\tan x = x\sec x$$, $$0 \leq x \leq \frac{\pi}{3}$$, $$y(0) = 1$$, then $$y\left(\frac{\pi}{6}\right)$$ is equal to

The given differential equation is
$$\frac{dy}{dx}+y\tan x = x\sec x,\qquad 0\le x\le\frac{\pi}{3},\qquad y(0)=1.$$

It is a linear first-order equation of the form $$\frac{dy}{dx}+P(x)\,y = Q(x)$$ with
$$P(x)=\tan x,\qquad Q(x)=x\sec x.$$

Step 1 : Integrating factor
The integrating factor (I.F.) is $$e^{\int P(x)\,dx}.$$
Since $$\int\tan x\,dx = -\ln\!\left(\cos x\right),$$ we get
$$\text{I.F.}=e^{-\ln(\cos x)}=\frac{1}{\cos x}=\sec x\qquad\bigl(0\le x\le \tfrac{\pi}{3}\bigr).$$

Step 2 : Multiply by the integrating factor
$$\sec x\,\frac{dy}{dx}+y\sec x\tan x = x\sec^2x.$$

Step 3 : Recognise the left side as a derivative
The left side is $$\frac{d}{dx}\bigl(y\sec x\bigr)$$ because
$$\frac{d}{dx}\bigl(y\sec x\bigr)=\sec x\,\frac{dy}{dx}+y\sec x\tan x.$$

Step 4 : Integrate
$$\frac{d}{dx}\bigl(y\sec x\bigr)=x\sec^2x\quad\Longrightarrow\quad y\sec x=\int x\sec^2x\,dx +C.$$

Compute the integral by parts:
Let $$u=x,\; dv=\sec^2x\,dx\;\Rightarrow\; du=dx,\; v=\tan x.$$
$$\int x\sec^2x\,dx = x\tan x-\int\tan x\,dx = x\tan x+\ln(\cos x)+C_1.$$

Hence
$$y\sec x = x\tan x +\ln(\cos x)+C.\tag{-1}$$

Step 5 : Apply the initial condition
At $$x=0, \; y(0)=1,\; \cos0=1,\; \tan0=0,$$ so $$1\cdot 1 = 0 +\ln1 +C\;\Longrightarrow\; C=1.$$

Therefore, the solution is
$$y\sec x = 1 + x\tan x + \ln(\cos x).$$

Step 6 : Express $$y(x)$$ explicitly
$$y(x)=\cos x\bigl[\,1 + x\tan x +\ln(\cos x)\bigr].$$

Step 7 : Evaluate at $$x=\dfrac{\pi}{6}$$
For $$x=\dfrac{\pi}{6}:$$
$$\cos\frac{\pi}{6}=\frac{\sqrt3}{2},\quad \tan\frac{\pi}{6}=\frac{1}{\sqrt3},\quad \ln\!\left(\cos\frac{\pi}{6}\right)=\ln\!\left(\frac{\sqrt3}{2}\right).$$

Substitute:
$$y\!\left(\frac{\pi}{6}\right)= \frac{\sqrt3}{2}\Bigl[\,1+\frac{\pi}{6}\cdot\frac{1}{\sqrt3}+\ln\!\Bigl(\frac{\sqrt3}{2}\Bigr)\Bigr].$$

Simplify term by term:
$$\frac{\sqrt3}{2}\cdot 1=\frac{\sqrt3}{2},$$ $$\frac{\sqrt3}{2}\cdot\frac{\pi}{6\sqrt3}=\frac{\pi}{12},$$ $$\frac{\sqrt3}{2}\ln\!\Bigl(\frac{\sqrt3}{2}\Bigr)=\frac{\sqrt3}{2}\ln\!\Bigl(\frac{\sqrt3}{2}\Bigr).$$

Group the constant 1 with the logarithm using $$1=\ln e:$$
$$\frac{\sqrt3}{2}\Bigl[\,1+\ln\!\Bigl(\frac{\sqrt3}{2}\Bigr)\Bigr] =\frac{\sqrt3}{2}\ln\!\Bigl(e\cdot\frac{\sqrt3}{2}\Bigr) =-\frac{\sqrt3}{2}\ln\!\Bigl(\frac{2}{e\sqrt3}\Bigr).$$

Hence
$$y\!\left(\frac{\pi}{6}\right)=\frac{\pi}{12}-\frac{\sqrt3}{2} \ln\!\Bigl(\frac{2}{e\sqrt3}\Bigr).$$

Step 8 : Match with the options
This value corresponds to Option A.
Therefore, $$y\!\left(\dfrac{\pi}{6}\right)=\frac{\pi}{12}-\frac{\sqrt3}{2}\log_e\frac{2}{e\sqrt3}.$$