1. Given differential equation

   $\sec y\,\frac{dy}{dx}+2x\sin y=x^3\cos y, \qquad y(1)=0$

2. Rewrite in a more useful form

   Multiply through by $\cos y$:

   $\frac{dy}{dx}+2x\sin y\cos y=x^3\cos^2 y$

   Now let

   $u=\sin y$

   Then

   $\frac{du}{dx}=\cos y\,\frac{dy}{dx}$

   But from the original equation,

   $\sec y\,\frac{dy}{dx}=\frac{1}{\cos y}\frac{dy}{dx}$

   so multiplying the original equation by $\cos y$ gives directly:

   $\frac{du}{dx}+2xu=x^3(1-u^2)$

   However, this is not linear in $u$. So instead, observe a better substitution.

3. Use the substitution $z=\tan y$?

   From the original equation:

   $\sec y\,\frac{dy}{dx}+2x\sin y=x^3\cos y$

   Note that

   $\frac{d}{dx}(\ln(\sec y+\tan y))=\sec y\frac{dy}{dx}$

   but that does not simplify the remaining terms nicely.

   A better observation is to multiply the original equation by $\sec y$:

   $\sec^2 y\,\frac{dy}{dx}+2x\tan y=x^3$

   Now let

   $t=\tan y$

   Then

   $\frac{dt}{dx}=\sec^2 y\,\frac{dy}{dx}$

   Hence the differential equation becomes

   $\frac{dt}{dx}+2xt=x^3$

   This is a linear differential equation.

4. Solve the linear ODE

   $\frac{dt}{dx}+2xt=x^3$

   Integrating factor:

   $\text{I.F.}=e^{\int 2x\,dx}=e^{x^2}$

   Therefore,

   $\frac{d}{dx}\left(t e^{x^2}\right)=x^3 e^{x^2}$

   Integrate both sides:

   $t e^{x^2}=\int x^3 e^{x^2}\,dx + C$

   Let $u=x^2$, so $du=2x\,dx$. Then

   $\int x^3 e^{x^2}\,dx=\int x^2\cdot x e^{x^2}\,dx$

   With $u=x^2$, this becomes

   $\frac12\int u e^u\,du$

   and

   $\int u e^u\,du=e^u(u-1)$

   So,

   $\int x^3 e^{x^2}\,dx=\frac12 e^{x^2}(x^2-1)$

   Hence

   $t e^{x^2}=\frac12 e^{x^2}(x^2-1)+C$

   Thus,

   $t=\frac{x^2-1}{2}+Ce^{-x^2}$

5. Use the initial condition

   Since $t=\tan y$ and $y(1)=0$,

   $t(1)=\tan 0=0$

   Therefore,

   $0=\frac{1-1}{2}+Ce^{-1}$

   which gives

   $C=0$

   So,

   $\tan y=\frac{x^2-1}{2}$

6. Find $y(\sqrt{3})$

   Substitute $x=\sqrt{3}$:

   $\tan y(\sqrt{3})=\frac{3-1}{2}=1$

   Therefore,

   $y(\sqrt{3})=\tan^{-1}(1)=\frac{\pi}{4}$

   using the branch consistent with $y(1)=0$.

7. Check options

   • A: $\frac{\pi}{6}$ — incorrect
   • B: $\frac{\pi}{12}$ — incorrect
   • C: $\frac{\pi}{3}$ — incorrect
   • D: $\frac{\pi}{4}$ — correct

Therefore, the required value is $\boxed{\frac{\pi}{4}}$.