You probably haven't received any help yet because the way you have posted your DE is difficult to read.
Is this your DE?
$\displaystyle \begin{align*} \cos^2{(x)} \sin{(x)} \,\frac{\mathrm{d}y}{\mathrm{d}x} = y\cos^3{(x)} + 1 \end{align*}$
You probably haven't received any help yet because the way you have posted your DE is difficult to read.
Is this your DE?
$\displaystyle \begin{align*} \cos^2{(x)} \sin{(x)} \,\frac{\mathrm{d}y}{\mathrm{d}x} = y\cos^3{(x)} + 1 \end{align*}$
Rewrite it as
$\displaystyle \begin{align*} \cos^2{(x)}\sin{(x)}\,\frac{\mathrm{d}y}{\mathrm{d }x} &= y\,\cos^3{(x)} + 1 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y\,\cos^3{(x)} + 1}{\cos^2{(x)}\sin{(x)}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= y\,\frac{\cos{(x)}}{\sin{(x)}} + \frac{1}{\cos^2{(x)}\sin{(x)}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} - \frac{\cos{(x)}}{\sin{(x)}}\,y &= \frac{1}{\cos^2{(x)}\sin{(x)}} \end{align*}$
This is now first order linear, so you can solve the DE using an integrating factor.