Answer
$y=\frac{4\left(\cos ^2\left(x\right)-2\ln \left|\cos \left(x\right)\right|\right)+C}{\sec \left(x\right)}$
Work Step by Step
Given that:
$y'+y\tan x=8\sin ^3x$
Use an integration factor of:
$$\mu (x)=e^{\int P(x)dx}$$
where $P(x)=\tan x$
$$\mu (x)=e^{\int \tan x dx}=e^{\log (\sec (x))}=\sec x$$
Multiply the entire equation by this factor.
$$\sec (x)(y'+y\tan x=8\sin ^3x)$$$$y'\sec x+y \tan x \sec x=8\sin^3x \sec x$$
Integrate each side.
$$\int y'\sec x+y \tan x \sec x=\int 8\sin^3x \sec x dx$$$$\sec (x)y =\int 8\sin^3x \sec x dx$$ $$\sec (x)y =8(-\ln|\cos x|+\frac{\cos^2x}{2})+C$$
Solve for y
$$y=\frac{4\left(\cos ^2\left(x\right)-2\ln \left|\cos \left(x\right)\right|\right)+C}{\sec \left(x\right)}$$
