Solution: Given, $\frac{d y}{d x}+x \sin 2 y=x^{3} \cos ^{2} y$

Dividing both sides by cos $^{2} \mathrm{x}$

$\sec ^{2} x \frac{d y}{d x}+x \sec ^{2} x \sin 2 y=x^{3}$

$\sec ^{2} x \frac{d y}{d x}+2 x \tan y=x^{3} \ldots \ldots \ldots \ldots(1)$

Put tan $y=v$ and differentiate w.r.t. x

$\sec ^{2} x \frac{d y}{d x}=\frac{d v}{d x}$

Hence, from ( 1 ), we get $\frac{d v}{d x}+2 v \cdot x=x^{3}$

$\therefore P=2 x$ And $Q=x^{3}$

$\therefore \int P d x=\int 2 x d x=x^{2}$

$\therefore$ I.F. $=e^{\int P d x}=e^{\int 2 x d x}=e^{x^{2}}$

$\therefore$ The solution is $v \ e^{x^{2}}=\int e^{x^{2}} x^{3} d x+c$

To find the integral put $x^{2}=t, x d x=\frac{a t}{2}$,

$\therefore I=\int e^{t} \cdot t \cdot \frac{d t}{2}=\frac{1}{2}\left[t e^{t}-\int e^{t} . d t\right] \ldots \ldots \ldots[ By \ parts ]$

$\therefore I=\frac{1}{2}\left[t e^{t}-e^{t}\right]=\frac{1}{2} e^{t}(t-1)=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)$

$\therefore$ The solution is $v \ e^{x^{2}}=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c$

$\therefore \tan y \ e^{x^{2}}=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c$