1. Given differential equation

$\frac{dy}{dx}+y\tan x=x\sec x, \qquad y(0)=1$

This is a linear differential equation of the form

$\frac{dy}{dx}+P(x)y=Q(x)$

with

$P(x)=\tan x, \qquad Q(x)=x\sec x.$

2. Find the integrating factor

The integrating factor is

$\mathrm{IF}=e^{\int \tan x\,dx}.$

Now,

$\int \tan x\,dx=\int \frac{\sin x}{\cos x}\,dx=-\ln(\cos x)=\ln(\sec x).$

Hence,

$\mathrm{IF}=e^{\ln(\sec x)}=\sec x.$

3. Multiply the equation by the integrating factor

Multiplying throughout by $\sec x$,

$\sec x\frac{dy}{dx}+y\sec x\tan x=x\sec^2 x.$

The left side becomes:

$\frac{d}{dx}(y\sec x)=x\sec^2 x.$

So,

$\frac{d}{dx}(y\sec x)=x\sec^2 x.$

4. Integrate both sides

$y\sec x=\int x\sec^2 x\,dx+C.$

Use integration by parts on

$\int x\sec^2 x\,dx.$

Let

$u=x, \quad dv=\sec^2 x\,dx.$

Then

$du=dx, \quad v=\tan x.$

Therefore,

$\int x\sec^2 x\,dx=x\tan x-\int \tan x\,dx.$

And since

$\int \tan x\,dx=\ln(\sec x),$

we get

$\int x\sec^2 x\,dx=x\tan x-\ln(\sec x).$

Thus,

$y\sec x=x\tan x-\ln(\sec x)+C.$

5. Use the initial condition

Given $y(0)=1$.

At $x=0$,

$\sec 0=1, \quad \tan 0=0, \quad \ln(\sec 0)=\ln 1=0.$

So,

$1\cdot 1=0-0+C \implies C=1.$

Hence the solution is

$y\sec x=x\tan x-\ln(\sec x)+1.$

Therefore,

$y=\cos x\,[x\tan x-\ln(\sec x)+1].$

6. Evaluate at $x=\frac{\pi}{6}$

We know:

$\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \qquad \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}, \qquad \sec \frac{\pi}{6}=\frac{2}{\sqrt{3}}.$

So,

$y\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\left[\frac{\pi}{6}\cdot \frac{1}{\sqrt{3}}-\ln\left(\frac{2}{\sqrt{3}}\right)+1\right].$

Now simplify term by term:

$\frac{\sqrt{3}}{2}\cdot \frac{\pi}{6\sqrt{3}}=\frac{\pi}{12},$

so

$y\left(\frac{\pi}{6}\right)=\frac{\pi}{12}-\frac{\sqrt{3}}{2}\ln\left(\frac{2}{\sqrt{3}}\right)+\frac{\sqrt{3}}{2}.$

Write the last two terms together: